Comprehensive Lecture Notes on Linear Algebra and Trigonometry

Course Structure and Assessment Policy

  • Course Title: Linear Algebra and Trigonometry (GL/ES/GM – 153)

  • Academic Institution: University of Mines and Technology, Tarkwa, School of Railway and Infrastructural Development, Department of Mathematical Sciences

  • Assessment Framework:

    • Continuous Assessment (40%40\%

    • Attendance: Random attendance checks constitute 10%10\% of the continuous assessment.

    • Assignments and Presentations: Group assignments, individual assignments, and group presentations constitute 10%10\% of the continuous assessment.

    • Quizzes and Class Tests: Announced and unannounced quizzes constitute 20%20\% of the continuous assessment.

    • Note: The course instructor determines the exact frequency and schedule of all assignments and quizzes.

    • End of Semester Examination (60%60\%

    • A comprehensive written examination evaluating all covered topics.

    • Student Evaluation of Instructor: Administered at the end of the semester via a structured questionnaire to assess instructional quality and course delivery.

Complex Numbers: Foundations and Algebraic Structure

  • Historical Evolution of Number Systems:

    • Natural numbers (N\mathbb{N}) were augmented with zero to form whole numbers, and negative numbers were introduced to form integers (Z\mathbb{Z}).

    • Linear equations of the form qx=pqx = p necessitated rational numbers (Q\mathbb{Q}).

    • Incommensurable geometric lengths necessitated irrational numbers; rationals and irrationals together form real numbers (R\mathbb{R}).

    • Real numbers cannot solve equations such as x2+1=0x^2 + 1 = 0 or x2=−1x^2 = -1. This necessitated extending the real system to complex numbers (C\mathbb{C}).

  • Fundamental Definition of the Imaginary Unit:

    • The imaginary unit ii satisfies the identity:     i2=−1i^2 = -1     i=−1i = \sqrt{-1}

    • Solutions to x2+1=0x^2 + 1 = 0 are given by x=±−1=±ix = \pm \sqrt{-1} = \pm i.

    • Square roots of negative quantities factor as −k=−1×k=−1k=ik\sqrt{-k} = \sqrt{-1 \times k} = \sqrt{-1}\sqrt{k} = i\sqrt{k}.

  • Standard Algebraic Form of a Complex Number:

    • Any number expressible as z=a+biz = a + bi, where a,b∈Ra, b \in \mathbb{R} and i=−1i = \sqrt{-1}, is a complex number.

    • Real Part: Expressed as Re(z)=a\text{Re}(z) = a.

    • Imaginary Part: Expressed as Im(z)=b\text{Im}(z) = b.

    • Purely Imaginary Numbers: If Re(z)=a=0\text{Re}(z) = a = 0, then z=biz = bi is called purely imaginary (e.g., 3i3i, −10i-10i, πi\pi i).

    • Real Numbers as a Subset: When Im(z)=b=0\text{Im}(z) = b = 0, z=a∈Rz = a \in \mathbb{R}, confirming R⊂C\mathbb{R} \subset \mathbb{C}.

  • Equality of Complex Numbers:

    • Two complex numbers z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di are equal if and only if their corresponding real parts and imaginary parts are identical:     a+bi=c+di  ⟺  a=candb=da + bi = c + di \iff a = c \quad \text{and} \quad b = d

  • Quadratic Equations with Complex Roots:

    • For ax2+bx+c=0ax^2 + bx + c = 0, the quadratic formula is:     x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    • Worked Example: Solve x2−6x+13=0x^2 - 6x + 13 = 0:

    • Identify coefficients: a=1a = 1, b=−6b = -6, c=13c = 13.

    • Substitute into formula:       x=−(−6)±(−6)2−4(1)(13)2(1)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)}       x=6±36−522=6±−162x = \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2}       x=6±4i2=3±2ix = \frac{6 \pm 4i}{2} = 3 \pm 2i

Complex Conjugates and Geometric Representation

  • Complex Conjugate Definition:

    • For any complex number z=a+biz = a + bi, its complex conjugate is denoted by zˉ\bar{z} and defined as:     zˉ=a−bi\bar{z} = a - bi

    • Conjugate pairs arise naturally as roots of real polynomials (e.g., 3+2i3 + 2i and 3−2i3 - 2i).

  • Properties of the Complex Conjugate:

    • A complex number is real if and only if it equals its conjugate: z=zˉ  ⟺  z∈Rz = \bar{z} \iff z \in \mathbb{R}.

    • Verification: If z=5=5+0iz = 5 = 5 + 0i, then zˉ=5−0i=5\bar{z} = 5 - 0i = 5.

    • A complex number is purely imaginary if and only if its conjugate is its negative: zˉ=−z\bar{z} = -z.

    • Verification: If z=7i=0+7iz = 7i = 0 + 7i, then zˉ=0−7i=−7i=−z\bar{z} = 0 - 7i = -7i = -z.

    • The involution property holds: the conjugate of a conjugate is the original number:     (zˉ)‾=z\overline{(\bar{z})} = z

    • Distribution over addition:     z1+z2‾=zˉ1+zˉ2\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2

    • Distribution over multiplication:     z1⋅z2‾=zˉ1⋅zˉ2\overline{z_1 \cdot z_2} = \bar{z}_1 \cdot \bar{z}_2

    • Distribution over division:     (z1z2)‾=zˉ1zˉ2\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z}_1}{\bar{z}_2}

  • Geometrical Representation: The Complex Plane (Argand Diagram):

    • A complex number z=a+biz = a + bi corresponds to an ordered pair of real numbers (a,b)(a, b) plotted in a Cartesian coordinate system.

    • Horizontal Axis (xx-axis): The Real Axis (Re\text{Re}-axis), representing aa.

    • Vertical Axis (yy-axis): The Imaginary Axis (Im\text{Im}-axis), representing bb.

    • Points z=a+biz = a + bi and zˉ=a−bi\bar{z} = a - bi represent reflections across the real axis.

Fig. 1.1 Argand Diagram
  • Modulus (Absolute Value) of a Complex Number:

    • The modulus represents the Euclidean distance from the origin O(0,0)O(0, 0) to the point P(a,b)P(a, b).

    • By the Pythagorean theorem on right triangle OMPOMP with base OM=aOM = a and perpendicular MP=bMP = b:     ∣z∣=OM2+MP2=a2+b2|z| = \sqrt{OM^2 + MP^2} = \sqrt{a^2 + b^2}

    • Algebraic relation with conjugate:     zzˉ=(a+bi)(a−bi)=a2−(bi)2=a2+b2=∣z∣2z \bar{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2

  • Essential Modulus Inequalities and Properties:

    • ∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1| |z_2|

    • ∣z1z2∣=∣z1∣∣z2∣\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}, provided z2≠0z_2 \neq 0

  • Triangle Inequality: ∣z1+z2∣≤∣z1∣+∣z2∣|z_1 + z_2| \le |z_1| + |z_2|

  • Reverse Triangle Inequality: ∣z1−z2∣≥∣z1∣−∣z2∣|z_1 - z_2| \ge |z_1| - |z_2|

Complex Arithmetic

  • Addition of Complex Numbers:

    • Add corresponding real and imaginary parts:     (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i

    • Example 1: If z1=6+9iz_1 = 6 + 9i and z2=12+7iz_2 = 12 + 7i:     z1+z2=(6+12)+(9+7)i=18+16iz_1 + z_2 = (6 + 12) + (9 + 7)i = 18 + 16i

    • Example 2: If z1=4+7iz_1 = 4 + 7i and z2=1−6iz_2 = 1 - 6i:     z1+z2=(4+1)+(7−6)i=5+iz_1 + z_2 = (4 + 1) + (7 - 6)i = 5 + i

  • Subtraction of Complex Numbers:

    • Subtract corresponding real and imaginary parts:     (a+bi)−(c+di)=(a−c)+(b−d)i(a + bi) - (c + di) = (a - c) + (b - d)i

    • Example 1: If z1=6+9iz_1 = 6 + 9i and z2=12+7iz_2 = 12 + 7i:     z1−z2=(6−12)+(9−7)i=−6+2iz_1 - z_2 = (6 - 12) + (9 - 7)i = -6 + 2i

    • Example 2: If z1=4+7iz_1 = 4 + 7i and z2=1−6iz_2 = 1 - 6i:     z1−z2=(4−1)+(7−(−6))i=3+13iz_1 - z_2 = (4 - 1) + (7 - (-6))i = 3 + 13i

  • Multiplication of Complex Numbers:

    • Perform polynomial binomial expansion applying i2=−1i^2 = -1:     (a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

    • Example: Evaluate (3+2i)(4+5i)(3 + 2i)(4 + 5i):     (3+2i)(4+5i)=3(4+5i)+2i(4+5i)=12+15i+8i+10i2=12+23i−10=2+23i(3 + 2i)(4 + 5i) = 3(4 + 5i) + 2i(4 + 5i) = 12 + 15i + 8i + 10i^2 = 12 + 23i - 10 = 2 + 23i

  • Division of Complex Numbers:

    • Multiply numerator and denominator by the complex conjugate of the denominator:     a+bic+di=(a+bi)(c−di)(c+di)(c−di)=(ac+bd)+(bc−ad)ic2+d2=ac+bdc2+d2+(bc−adc2+d2)i\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \left(\frac{bc - ad}{c^2 + d^2}\right)i

    • Example: Divide 3+i3 + i by 4−2i4 - 2i:     3+i4−2i=(3+i)(4+2i)(4−2i)(4+2i)=12+6i+4i+2i216−4i2=12+10i−216+4=10+10i20=12+12i\frac{3 + i}{4 - 2i} = \frac{(3 + i)(4 + 2i)}{(4 - 2i)(4 + 2i)} = \frac{12 + 6i + 4i + 2i^2}{16 - 4i^2} = \frac{12 + 10i - 2}{16 + 4} = \frac{10 + 10i}{20} = \frac{1}{2} + \frac{1}{2}i

  • Extraction of Square Roots of Complex Numbers:

    • To find x+yi\sqrt{x + yi}, set a+bi=x+yia + bi = \sqrt{x + yi} and square both sides:     (a+bi)2=a2−b2+2abi=x+yi(a + bi)^2 = a^2 - b^2 + 2abi = x + yi

    • Equate real and imaginary components:     a2−b2=xa^2 - b^2 = x     2ab=y2ab = y

    • Worked Example: Compute 35−12i\sqrt{35 - 12i}:

    • Let 35−12i=a+bi\sqrt{35 - 12i} = a + bi.

    • Expanding: a2−b2+2abi=35−12ia^2 - b^2 + 2abi = 35 - 12i.

    • Equate components: a2−b2=35a^2 - b^2 = 35 and 2ab=−12  ⟹  ab=−62ab = -12 \implies ab = -6.

    • Solving the system yields a=6,b=−1a = 6, b = -1 or a=−6,b=1a = -6, b = 1.

    • Therefore, the square roots are ±(6−i)\pm(6 - i).

Polar Form, Euler's Formula, and Complex Exponentials

  • Geometric Coordinate Conversion:

    • Let point P(a,b)P(a, b) represent z=a+biz = a + bi, at distance r=∣z∣r = |z| from the origin making angle θ\theta with the positive real axis:     a=rcos⁡(θ)a = r\cos(\theta)     b=rsin⁡(θ)b = r\sin(\theta)     r=a2+b2r = \sqrt{a^2 + b^2}     tan⁡(θ)=ba  ⟹  θ=tan⁡−1(ba)\tan(\theta) = \frac{b}{a} \implies \theta = \tan^{-1}\left(\frac{b}{a}\right)

  • Definition of Argument:

    • The argument θ=arg⁡(z)\theta = \arg(z) is the angle between the positive real axis and the position vector of the complex number.

  • Polar Form Definition:

    • The polar representation of a complex number is:     z=r(cos⁡(θ)+isin⁡(θ))z = r(\cos(\theta) + i\sin(\theta))

Fig 1.3 Argand Diagram
  • Multiplication and Division in Polar Form:

    • Multiplication Rule: Multiply moduli, add arguments:     z1z2=r1r2[(cos⁡(θ)cos⁡(ϕ)−sin⁡(θ)sin⁡(ϕ))+i(sin⁡(θ)cos⁡(ϕ)+cos⁡(θ)sin⁡(ϕ))]z_1 z_2 = r_1 r_2 [(\cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi)) + i(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))]     z1z2=r1r2[cos⁡(θ+ϕ)+isin⁡(θ+ϕ)]z_1 z_2 = r_1 r_2 [\cos(\theta + \phi) + i\sin(\theta + \phi)]

    • Example: If z1=3(cos⁡(30∘)+isin⁡(30∘))z_1 = 3(\cos(30^\circ) + i\sin(30^\circ)) and z2=4(cos⁡(60∘)+isin⁡(60∘))z_2 = 4(\cos(60^\circ) + i\sin(60^\circ)):     z1z2=3×4[cos⁡(30∘+60∘)+isin⁡(30∘+60∘)]=12[cos⁡(90∘)+isin⁡(90∘)]=12iz_1 z_2 = 3 \times 4 [\cos(30^\circ + 60^\circ) + i\sin(30^\circ + 60^\circ)] = 12[\cos(90^\circ) + i\sin(90^\circ)] = 12i

    • Division Rule: Divide moduli, subtract argument of denominator from numerator:     z1z2=r1r2[cos⁡(θ−ϕ)+isin⁡(θ−ϕ)]\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta - \phi) + i\sin(\theta - \phi)]

  • Euler's Formula:

    • For any real number θ\theta:     eiθ=cos⁡(θ)+isin⁡(θ)e^{i\theta} = \cos(\theta) + i\sin(\theta)

  • Exponential Form of a Complex Number:

    • Expressed as:     z=reiθz = re^{i\theta}

    • Modulus verification:     ∣z∣=∣reiθ∣=∣r∣⋅∣cos⁡(θ)+isin⁡(θ)∣=rcos⁡2(θ)+sin⁡2(θ)=r|z| = |re^{i\theta}| = |r| \cdot |\cos(\theta) + i\sin(\theta)| = r\sqrt{\cos^2(\theta) + \sin^2(\theta)} = r

    • Since z=r(cos⁡(θ)+isin⁡(θ))z = r(\cos(\theta) + i\sin(\theta)) traces a circle of radius rr, z=reiθz = re^{i\theta} serves as a parametric representation of a circle of radius rr.

De Moivre’s Theorem and Complex Roots

  • De Moivre's Theorem Statement:

    • If z=r(cos⁡(θ)+isin⁡(θ))z = r(\cos(\theta) + i\sin(\theta)) and nn is a positive integer, then:     zn=[r(cos⁡(θ)+isin⁡(θ))]n=rn(cos⁡(nθ)+isin⁡(nθ))z^n = [r(\cos(\theta) + i\sin(\theta))]^n = r^n(\cos(n\theta) + i\sin(n\theta))

    • Powers of a complex number are obtained by raising the modulus to the nn-th power and multiplying the argument by nn.

  • Simplification Examples:

    • Simplify (cos⁡(θ)−isin⁡(θ))4(\cos(\theta) - i\sin(\theta))^4:     (cos⁡(θ)−isin⁡(θ))4=(cos⁡(−θ)+isin⁡(−θ))4=cos⁡(−4θ)+isin⁡(−4θ)=cos⁡(4θ)−isin⁡(4θ)(\cos(\theta) - i\sin(\theta))^4 = (\cos(-\theta) + i\sin(-\theta))^4 = \cos(-4\theta) + i\sin(-4\theta) = \cos(4\theta) - i\sin(4\theta)

    • Simplify (sin⁡(θ)−icos⁡(θ))7(\sin(\theta) - i\cos(\theta))^7:     (sin⁡(θ)−icos⁡(θ))7=[−i(cos⁡(θ)+isin⁡(θ))]7=(−i)7(cos⁡(θ)+isin⁡(θ))7=i(cos⁡(7θ)+isin⁡(7θ))=icos⁡(7θ)−sin⁡(7θ)(\sin(\theta) - i\cos(\theta))^7 = [-i(\cos(\theta) + i\sin(\theta))]^7 = (-i)^7(\cos(\theta) + i\sin(\theta))^7 = i(\cos(7\theta) + i\sin(7\theta)) = i\cos(7\theta) - \sin(7\theta)

    • Simplify (cos⁡(2θ)+isin⁡(2θ))3(cos⁡(2θ)−isin⁡(2θ))4\frac{(\cos(2\theta) + i\sin(2\theta))^3}{(\cos(2\theta) - i\sin(2\theta))^4}:     cos⁡(6θ)+isin⁡(6θ)cos⁡(−8θ)+isin⁡(−8θ)=cos⁡(6θ−(−8θ))+isin⁡(6θ−(−8θ))=cos⁡(14θ)+isin⁡(14θ)\frac{\cos(6\theta) + i\sin(6\theta)}{\cos(-8\theta) + i\sin(-8\theta)} = \cos(6\theta - (-8\theta)) + i\sin(6\theta - (-8\theta)) = \cos(14\theta) + i\sin(14\theta)

  • Expressing Powers in x+iyx + iy Form:

    • Example 1: Evaluate [2(cos⁡(π2)+isin⁡(π2))]6\left[\sqrt{2}\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right)\right]^6:     (2)6(cos⁡(6π2)+isin⁡(6π2))=8(cos⁡(3π)+isin⁡(3π))=8(−1+0i)=−8(\sqrt{2})^6 \left(\cos\left(\frac{6\pi}{2}\right) + i\sin\left(\frac{6\pi}{2}\right)\right) = 8(\cos(3\pi) + i\sin(3\pi)) = 8(-1 + 0i) = -8

    • Example 2: Evaluate (1+3i)10(1 + \sqrt{3}i)^{10}:

    • Modulus: r=12+(3)2=4=2r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.

    • Argument: θ=tan⁡−1(3)=π3\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.

    • In polar form: 1+3i=2(cos⁡(π3)+isin⁡(π3))1 + \sqrt{3}i = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right).

    • Applying theorem:       210(cos⁡(10π3)+isin⁡(10π3))=1024(−12−i32)=−512−5123i2^{10}\left(\cos\left(\frac{10\pi}{3}\right) + i\sin\left(\frac{10\pi}{3}\right)\right) = 1024\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -512 - 512\sqrt{3}i

    • Example 3: Evaluate (12+12i)10\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right)^{10}:

    • Polar form: r=1r = 1, θ=π4\theta = \frac{\pi}{4}.

    • Result: cos⁡(10π4)+isin⁡(10π4)=cos⁡(5π2)+isin⁡(5π2)=0+i=i\cos\left(\frac{10\pi}{4}\right) + i\sin\left(\frac{10\pi}{4}\right) = \cos\left(\frac{5\pi}{2}\right) + i\sin\left(\frac{5\pi}{2}\right) = 0 + i = i.

  • Roots of a Complex Number:

    • For z=r(cos⁡(θ)+isin⁡(θ))z = r(\cos(\theta) + i\sin(\theta)) and integer n≥1n \ge 1, there are nn distinct nn-th roots given by:     wk=r1/n[cos⁡(θ+2kπn)+isin⁡(θ+2kπn)],k=0,1,2,…,n−1w_k = r^{1/n}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right], \quad k = 0, 1, 2, \dots, n - 1

    • Geometric distribution: All nn roots lie equally spaced on a circle of radius r1/nr^{1/n} centered at the origin, with angular separation 2πn\frac{2\pi}{n}.

    • Worked Example: Find the sixth roots of z=−8z = -8:

    • Polar form of −8-8: r=8r = 8, θ=π\theta = \pi, so z=8(cos⁡(π)+isin⁡(π))z = 8(\cos(\pi) + i\sin(\pi)).

    • Sixth roots have radius r1/6=81/6=(23)1/6=21/2=2r^{1/6} = 8^{1/6} = (2^3)^{1/6} = 2^{1/2} = \sqrt{2}.

    • General formula: wk=2[cos⁡(π+2kπ6)+isin⁡(π+2kπ6)]w_k = \sqrt{2}\left[\cos\left(\frac{\pi + 2k\pi}{6}\right) + i\sin\left(\frac{\pi + 2k\pi}{6}\right)\right] for k=0,1,2,3,4,5k = 0, 1, 2, 3, 4, 5.

    • Evaluating roots:

      • k=0k = 0: w0=2(cos⁡(π6)+isin⁡(π6))=2(32+12i)=62+22iw_0 = \sqrt{2}\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) = \sqrt{2}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i

      • k=1k = 1: w1=2(cos⁡(3π6)+isin⁡(3π6))=2(cos⁡(π2)+isin⁡(π2))=2iw_1 = \sqrt{2}\left(\cos\left(\frac{3\pi}{6}\right) + i\sin\left(\frac{3\pi}{6}\right)\right) = \sqrt{2}\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = \sqrt{2}i

      • k=2k = 2: w2=2(cos⁡(5π6)+isin⁡(5π6))=2(−32+12i)=−62+22iw_2 = \sqrt{2}\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right) = \sqrt{2}\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = -\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i

      • k=3k = 3: w3=2(cos⁡(7π6)+isin⁡(7π6))=2(−32−12i)=−62−22iw_3 = \sqrt{2}\left(\cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right)\right) = \sqrt{2}\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i

      • k=4k = 4: w4=2(cos⁡(9π6)+isin⁡(9π6))=2(cos⁡(3π2)+isin⁡(3π2))=−2iw_4 = \sqrt{2}\left(\cos\left(\frac{9\pi}{6}\right) + i\sin\left(\frac{9\pi}{6}\right)\right) = \sqrt{2}\left(\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right) = -\sqrt{2}i

      • k=5k = 5: w5=2(cos⁡(11π6)+isin⁡(11π6))=2(32−12i)=62−22iw_5 = \sqrt{2}\left(\cos\left(\frac{11\pi}{6}\right) + i\sin\left(\frac{11\pi}{6}\right)\right) = \sqrt{2}\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i

Applications of De Moivre’s Theorem

  • Summation of Trigonometric Series:

    • Consider the series:     C=∑r=1ncos⁡(rθ)=cos⁡(θ)+cos⁡(2θ)+⋯+cos⁡(nθ)C = \sum_{r=1}^n \cos(r\theta) = \cos(\theta) + \cos(2\theta) + \dots + \cos(n\theta)     S=∑r=1nsin⁡(rθ)=sin⁡(θ)+sin⁡(2θ)+⋯+sin⁡(nθ)S = \sum_{r=1}^n \sin(r\theta) = \sin(\theta) + \sin(2\theta) + \dots + \sin(n\theta)

    • Let z=cos⁡(θ)+isin⁡(θ)z = \cos(\theta) + i\sin(\theta). By De Moivre's theorem, zr=cos⁡(rθ)+isin⁡(rθ)z^r = \cos(r\theta) + i\sin(r\theta).

    • Express combined sum as a geometric progression:     C+iS=∑r=1nzr=z+z2+⋯+zn=z(1−zn)1−zC + iS = \sum_{r=1}^n z^r = z + z^2 + \dots + z^n = \frac{z(1 - z^n)}{1 - z}

    • Algebraic manipulation of 1−zn1 - z^n and 1−z1 - z:     1−z=1−cos⁡(θ)−isin⁡(θ)=2sin⁡2(θ2)−2isin⁡(θ2)cos⁡(θ2)=−2isin⁡(θ2)[cos⁡(θ2)+isin⁡(θ2)]1 - z = 1 - \cos(\theta) - i\sin(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) - 2i\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) = -2i\sin\left(\frac{\theta}{2}\right)\left[\cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right)\right]     1−zn=−2isin⁡(nθ2)[cos⁡(nθ2)+isin⁡(nθ2)]1 - z^n = -2i\sin\left(\frac{n\theta}{2}\right)\left[\cos\left(\frac{n\theta}{2}\right) + i\sin\left(\frac{n\theta}{2}\right)\right]

    • Dividing the expressions and equating real and imaginary parts yields:     ∑r=1ncos⁡(rθ)=cos⁡(n+12θ)sin⁡(nθ2)sin⁡(θ2)\sum_{r=1}^n \cos(r\theta) = \frac{\cos\left(\frac{n+1}{2}\theta\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}     ∑r=1nsin⁡(rθ)=sin⁡(n+12θ)sin⁡(nθ2)sin⁡(θ2)\sum_{r=1}^n \sin(r\theta) = \frac{\sin\left(\frac{n+1}{2}\theta\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}

  • Multiple Angle Polynomial Expansions:

    • Using cos⁡(nθ)+isin⁡(nθ)=(cos⁡(θ)+isin⁡(θ))n\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n, expand the right side via the Binomial Theorem and equate real and imaginary parts.

    • Worked Example: Expand cos⁡(3θ)\cos(3\theta) and sin⁡(3θ)\sin(3\theta):

    • Let c=cos⁡(θ)c = \cos(\theta) and s=sin⁡(θ)s = \sin(\theta).

    • Expansion:       (cos⁡(θ)+isin⁡(θ))3=c3+3c2(is)+3c(is)2+(is)3=c3+3ic2s−3cs2−is3=(c3−3cs2)+i(3c2s−s3)(\cos(\theta) + i\sin(\theta))^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2s - 3cs^2 - is^3 = (c^3 - 3cs^2) + i(3c^2s - s^3)

    • Real part (cos⁡(3θ)\cos(3\theta)):       cos⁡(3θ)=c3−3cs2=cos⁡3(θ)−3cos⁡(θ)sin⁡2(θ)=cos⁡3(θ)−3cos⁡(θ)(1−cos⁡2(θ))=4cos⁡3(θ)−3cos⁡(θ)\cos(3\theta) = c^3 - 3cs^2 = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) = \cos^3(\theta) - 3\cos(\theta)(1 - \cos^2(\theta)) = 4\cos^3(\theta) - 3\cos(\theta)

    • Imaginary part (sin⁡(3θ)\sin(3\theta)):       sin⁡(3θ)=3c2s−s3=3(1−sin⁡2(θ))sin⁡(θ)−sin⁡3(θ)=3sin⁡(θ)−4sin⁡3(θ)\sin(3\theta) = 3c^2s - s^3 = 3(1 - \sin^2(\theta))\sin(\theta) - \sin^3(\theta) = 3\sin(\theta) - 4\sin^3(\theta)

Matrix Fundamentals and Special Matrices

  • Matrix Definition and Dimensions:

    • A matrix is a rectangular array of elements arranged in rows and columns.

    • An m×nm \times n matrix contains mm horizontal rows and nn vertical columns.

    • Matrix element notation: aija_{ij} specifies the entry in row ii and column jj.

  • Classification of Special Matrices:

    • Square Matrix: A matrix with an equal number of rows and columns (m=nm = n).

    • Principal Diagonal: The entries a11,a22,…,anna_{11}, a_{22}, \dots, a_{nn} of a square matrix.

    • Trace of a Matrix: The scalar sum of all elements along the principal diagonal:     tr(A)=∑i=1naii\text{tr}(A) = \sum_{i=1}^n a_{ii}

    • Example: For A=(369121518212427)A = \begin{pmatrix} 3 & 6 & 9 \\ 12 & 15 & 18 \\ 21 & 24 & 27 \end{pmatrix}, tr(A)=3+15+27=45\text{tr}(A) = 3 + 15 + 27 = 45.

    • Sub-Matrix: Any matrix obtained by deleting a subset of rows and/or columns from a given matrix.

    • Zero Matrix (00): A matrix in which every entry is zero.

    • Identity Matrix (II): A square matrix with ones along the principal diagonal and zeros elsewhere:     I3=(100010001)I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

    • Transpose of a Matrix (ATA^T): Obtained by exchanging rows with columns ((AT)ij=Aji(A^T)_{ij} = A_{ji}).

    • Conjugate Transpose (A∗A^* or AHA^H): For a complex matrix, obtained by transposing and taking the complex conjugate of each entry ((A∗)ij=aji‾(A^*)_{ij} = \overline{a_{ji}}).

    • Example: If A=(45+3i2−i2)A = \begin{pmatrix} 4 & 5 + 3i \\ 2 - i & 2 \end{pmatrix}, then A∗=(42+i5−3i2)A^* = \begin{pmatrix} 4 & 2 + i \\ 5 - 3i & 2 \end{pmatrix}.

    • Symmetric Matrix: A square matrix identical to its transpose: A=ATA = A^T (aij=ajia_{ij} = a_{ji} for all i,ji, j).

    • Diagonal Matrix: A matrix whose off-diagonal entries are all zero (aij=0a_{ij} = 0 for i≠ji \neq j).

  • Triangular Matrices:

    • Upper Triangular Matrix: All entries below the principal diagonal are zero (aij=0a_{ij} = 0 for i>ji > j).

    • Lower Triangular Matrix: All entries above the principal diagonal are zero (aij=0a_{ij} = 0 for i<ji < j).

Matrix Algebra and Operational Properties

  • Matrix Addition and Subtraction:

    • Matrices must share identical dimensions (m×nm \times n).

    • Operations occur entrywise:     (A±B)ij=aij±bij(A \pm B)_{ij} = a_{ij} \pm b_{ij}

    • Properties of Matrix Addition:

    • Commutative: A+B=B+AA + B = B + A

    • Associative: (A+B)+C=A+(B+C)(A + B) + C = A + (B + C)

    • Additive Identity: A+0=AA + 0 = A

    • Additive Inverse: A+(−A)=0A + (-A) = 0

    • Scalar Distribution: k(tA)=(kt)Ak(tA) = (kt)A, (k+t)A=kA+tA(k + t)A = kA + tA, k(A+B)=kA+kBk(A + B) = kA + kB

  • Scalar Multiplication:

    • Multiplying a matrix by a scalar kk scales each entry:     (kA)ij=k⋅aij(kA)_{ij} = k \cdot a_{ij}

  • Matrix Multiplication:

    • Inner Dimension Compatibility: The product ABAB is defined if and only if the number of columns in AA equals the number of rows in BB.

    • If AA is of size m×nm \times n and BB is of size n×kn \times k, the product C=ABC = AB is of size m×km \times k with entries:     cij=∑r=1nairbrjc_{ij} = \sum_{r=1}^n a_{ir} b_{rj}

    • Worked Example: Compute ABAB for A=(012345)A = \begin{pmatrix} 0 & 1 & 2 \\ 3 & 4 & 5 \end{pmatrix} (2×32 \times 3) and B=(67891011)B = \begin{pmatrix} 6 & 7 \\ 8 & 9 \\ 10 & 11 \end{pmatrix} (3×23 \times 2):     c11=0(6)+1(8)+2(10)=28c_{11} = 0(6) + 1(8) + 2(10) = 28     c12=0(7)+1(9)+2(11)=31c_{12} = 0(7) + 1(9) + 2(11) = 31     c21=3(6)+4(8)+5(10)=18+32+50=100c_{21} = 3(6) + 4(8) + 5(10) = 18 + 32 + 50 = 100     c22=3(7)+4(9)+5(11)=21+36+55=112c_{22} = 3(7) + 4(9) + 5(11) = 21 + 36 + 55 = 112     AB=(2831100112)AB = \begin{pmatrix} 28 & 31 \\ 100 & 112 \end{pmatrix}

  • Properties of Matrix Multiplication:

    • Non-Commutative: In general, AB≠BAAB \neq BA.

  • Multiplication Order Terminology:

    • In ABAB, AA is post-multiplied by BB, or BB is pre-multiplied by AA.

  • Associative Property: A(BC)=(AB)CA(BC) = (AB)C

  • Distributive Properties: A(B+C)=AB+ACA(B + C) = AB + AC and (A+B)C=AC+BC(A + B)C = AC + BC

  • Scalar Compatibility: k(AB)=(kA)B=A(kB)k(AB) = (kA)B = A(kB)

  • Multiplicative Identity: For any square matrix AA, IA=AI=AIA = AI = A.

Determinants, Adjoints, and Inverses

  • Determinant of a Matrix:

    • The determinant is a scalar value associated with square matrices, representing the oriented volume scaling factor of the linear transformation.

    • For a 2×22 \times 2 matrix A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}:     det⁡(A)=∣A∣=ad−bc\det(A) = |A| = ad - bc

  • Fundamental Properties of Determinants:

    • If two rows or columns are equal or linearly dependent, det⁡(A)=0\det(A) = 0.

    • Invariance under transpose: det⁡(AT)=det⁡(A)\det(A^T) = \det(A).

    • Multiplicative property: det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B).

    • Triangular matrices: The determinant of an upper or lower triangular matrix is the product of its principal diagonal elements:     det⁡(A)=∏i=1naii\det(A) = \prod_{i=1}^n a_{ii}

  • Minors and Cofactors:

    • Minor (MijM_{ij}

    • The determinant of the (n−1)×(n−1)(n - 1) \times (n - 1) sub-matrix formed by deleting the ii-th row and jj-th column of AA.

    • Cofactor (CijC_{ij}

    • The signed minor given by:       Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}

    • Sign pattern forms an alternating checkerboard: (+−+−+−+−+)\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}.

    • Laplace Expansion: The determinant of an n×nn \times n matrix along the ii-th row is:     det⁡(A)=∑j=1naijCij\det(A) = \sum_{j=1}^n a_{ij} C_{ij}

  • Adjoint (Adjugate) of a Matrix (Adj(A)\text{Adj}(A)

    • The transpose of the cofactor matrix CC:     Adj(A)=CT\text{Adj}(A) = C^T

  • Matrix Inversion:

    • The inverse of an n×nn \times n matrix AA is the unique matrix A−1A^{-1} such that AA−1=A−1A=IAA^{-1} = A^{-1}A = I.

    • Existence: A−1A^{-1} exists if and only if det⁡(A)≠0\det(A) \neq 0 (non-singular matrix).

    • Formula via Adjoint:     A−1=1det⁡(A)Adj(A)A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)

  • Comprehensive Worked Example of Inversion:

    • Let A=(235416140)A = \begin{pmatrix} 2 & 3 & 5 \\ 4 & 1 & 6 \\ 1 & 4 & 0 \end{pmatrix}.

    • Compute Minors:     M11=det⁡(1640)=−24,M12=det⁡(4610)=−6,M13=det⁡(4114)=15M_{11} = \det\begin{pmatrix} 1 & 6 \\ 4 & 0 \end{pmatrix} = -24, \quad M_{12} = \det\begin{pmatrix} 4 & 6 \\ 1 & 0 \end{pmatrix} = -6, \quad M_{13} = \det\begin{pmatrix} 4 & 1 \\ 1 & 4 \end{pmatrix} = 15     M21=det⁡(3540)=−20,M22=det⁡(2510)=−5,M23=det⁡(2314)=5M_{21} = \det\begin{pmatrix} 3 & 5 \\ 4 & 0 \end{pmatrix} = -20, \quad M_{22} = \det\begin{pmatrix} 2 & 5 \\ 1 & 0 \end{pmatrix} = -5, \quad M_{23} = \det\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = 5     M31=det⁡(3516)=13,M32=det⁡(2546)=−8,M33=det⁡(2341)=−10M_{31} = \det\begin{pmatrix} 3 & 5 \\ 1 & 6 \end{pmatrix} = 13, \quad M_{32} = \det\begin{pmatrix} 2 & 5 \\ 4 & 6 \end{pmatrix} = -8, \quad M_{33} = \det\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} = -10

    • Form Cofactor Matrix CC:     C=(−2461520−5−5138−10)C = \begin{pmatrix} -24 & 6 & 15 \\ 20 & -5 & -5 \\ 13 & 8 & -10 \end{pmatrix}

    • Transpose to find Adjoint Adj(A)=CT\text{Adj}(A) = C^T:     Adj(A)=(−2420136−5815−5−10)\text{Adj}(A) = \begin{pmatrix} -24 & 20 & 13 \\ 6 & -5 & 8 \\ 15 & -5 & -10 \end{pmatrix}

    • Compute Determinant along first row:     det⁡(A)=a11C11+a12C12+a13C13=2(−24)+3(6)+5(15)=−48+18+75=45\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 2(-24) + 3(6) + 5(15) = -48 + 18 + 75 = 45

    • Compute Inverse:     A−1=145(−2420136−5815−5−10)=(−815491345215−1984513−19−29)A^{-1} = \frac{1}{45}\begin{pmatrix} -24 & 20 & 13 \\ 6 & -5 & 8 \\ 15 & -5 & -10 \end{pmatrix} = \begin{pmatrix} -\frac{8}{15} & \frac{4}{9} & \frac{13}{45} \\ \frac{2}{15} & -\frac{1}{9} & \frac{8}{45} \\ \frac{1}{3} & -\frac{1}{9} & -\frac{2}{9} \end{pmatrix}

Row Echelon Forms, Vector Independence, and Rank

  • Row Echelon Form (REF):

    • Conditions for REF:

    • The first non-zero element in each non-zero row (leading entry or pivot) is 1.

    • Each leading entry is located in a column strictly to the right of the leading entry in the preceding row.

    • Any rows consisting entirely of zeros are grouped at the bottom.

  • Reduced Row Echelon Form (RREF):

    • Conditions for RREF:

    • Satisfies all conditions of Row Echelon Form.

    • Each column containing a leading 1 has zeros everywhere else in that column.

    • Uniqueness: While multiple REF matrices can represent a given matrix, the RREF is strictly unique.

  • Elementary Row Operations:

    • Type I: Interchanging two rows (Ri↔RjR_i \leftrightarrow R_j).

    • Type II: Multiplying a row by a non-zero scalar (Ri→kRiR_i \to kR_i).

    • Type III: Adding a scalar multiple of one row to another (Ri→Ri+kRjR_i \to R_i + kR_j).

  • Linear Combinations and Vector Independence:

    • A vector v\mathbf{v} is a linear combination of vectors u1,u2,…,uk\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_k if scalars c1,c2,…,ckc_1, c_2, \dots, c_k exist such that:     v=c1u1+c2u2+⋯+ckuk\mathbf{v} = c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + \dots + c_k\mathbf{u}_k

    • Linear Independence: A set of vectors is linearly independent if:     c1u1+c2u2+⋯+ckuk=0  ⟹  c1=c2=⋯=ck=0c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + \dots + c_k\mathbf{u}_k = \mathbf{0} \implies c_1 = c_2 = \dots = c_k = 0

    • Linear Dependence: Holds if at least one vector can be written as a linear combination or scalar multiple of the others.

  • Rank of a Matrix:

    • Definition 1 (Determinantal Rank): The size rr of the largest square sub-matrix with a non-zero determinant.

    • Definition 2 (Row/Column Rank): The maximum number of linearly independent row vectors (or column vectors) in the matrix, equivalent to the number of non-zero rows in its Row Echelon Form.

    • Upper Bound: For an r×cr \times c matrix, rank(A)≤min⁡(r,c)\text{rank}(A) \le \min(r, c).

    • Full Rank: An m×nm \times n matrix is full rank if rank(A)=min⁡(m,n)\text{rank}(A) = \min(m, n).

Systems of Linear Equations

  • Linear System Representation:

    • A system of nn equations in nn unknowns:     a11x1+a12x2+⋯+a1nxn=y1a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = y_1     a21x1+a22x2+⋯+a2nxn=y2a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = y_2     ⋮\vdots     an1x1+an2x2+⋯+annxn=yna_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = y_n

    • Matrix equation form: Ax=yAx = y, where AA is the n×nn \times n coefficient matrix, xx is the column vector of unknowns, and yy is the column vector of constants.

    • Homogeneous Systems: If y=0y = \mathbf{0}, the system is homogeneous; otherwise, it is non-homogeneous.

  • Method 1: Matrix Inversion Method

    • Multiply both sides by A−1A^{-1}:     A−1(Ax)=A−1y  ⟹  Ix=A−1y  ⟹  x=A−1yA^{-1}(Ax) = A^{-1}y \implies Ix = A^{-1}y \implies x = A^{-1}y

    • Worked Example: Solve the system:     x1+2x2+2x3=1x_1 + 2x_2 + 2x_3 = 1     2x1+2x2+2x3=22x_1 + 2x_2 + 2x_3 = 2     2x1+2x2+x3=32x_1 + 2x_2 + x_3 = 3

    • Write in matrix form Ax=yAx = y:     (122222221)(x1x2x3)=(123)\begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

    • Compute A−1A^{-1}:     A−1=(−1101−1.5101−1)A^{-1} = \begin{pmatrix} -1 & 1 & 0 \\ 1 & -1.5 & 1 \\ 0 & 1 & -1 \end{pmatrix}

    • Compute solution vector:     (x1x2x3)=(−1101−1.5101−1)(123)=(−1(1)+1(2)+0(3)1(1)−1.5(2)+1(3)0(1)+1(2)−1(3))=(11−1)\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -1 & 1 & 0 \\ 1 & -1.5 & 1 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -1(1) + 1(2) + 0(3) \\ 1(1) - 1.5(2) + 1(3) \\ 0(1) + 1(2) - 1(3) \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}     x1=1,x2=1,x3=−1x_1 = 1, \quad x_2 = 1, \quad x_3 = -1

  • Method 2: Cramer’s Rule

    • Determinant evaluation of individual variables: Let D=det⁡(A)D = \det(A). Let DxjD_{x_j} be the determinant formed by replacing the jj-th column of AA with vector yy.

    • Variable formula:     xj=DxjDx_j = \frac{D_{x_j}}{D}

    • Evaluation for the above system:

    • Coefficient Determinant: D=det⁡(122222221)=2D = \det\begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 1 \end{pmatrix} = 2

    • Dx1=det⁡(122222321)=2  ⟹  x1=22=1D_{x_1} = \det\begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 2 \\ 3 & 2 & 1 \end{pmatrix} = 2 \implies x_1 = \frac{2}{2} = 1

    • Dx2=det⁡(112222231)=2  ⟹  x2=22=1D_{x_2} = \det\begin{pmatrix} 1 & 1 & 2 \\ 2 & 2 & 2 \\ 2 & 3 & 1 \end{pmatrix} = 2 \implies x_2 = \frac{2}{2} = 1

    • Dx3=det⁡(121222223)=−2  ⟹  x3=−22=−1D_{x_3} = \det\begin{pmatrix} 1 & 2 & 1 \\ 2 & 2 & 2 \\ 2 & 2 & 3 \end{pmatrix} = -2 \implies x_3 = \frac{-2}{2} = -1

  • Method 3: Gaussian Elimination

    • Form augmented matrix: [A∣y][A | y].

    • Perform elementary row operations to reduce AA to an upper triangular form.

    • Augmented Matrix Reduction:     (122122222213)\begin{pmatrix} 1 & 2 & 2 & 1 \\ 2 & 2 & 2 & 2 \\ 2 & 2 & 1 & 3 \end{pmatrix}

    • Row operations: R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−2R1R_3 \to R_3 - 2R_1:       (12210−2−200−2−31)\begin{pmatrix} 1 & 2 & 2 & 1 \\ 0 & -2 & -2 & 0 \\ 0 & -2 & -3 & 1 \end{pmatrix}

    • Scale row 2 (R2→−12R2R_2 \to -\frac{1}{2}R_2):       (122101100−2−31)\begin{pmatrix} 1 & 2 & 2 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{pmatrix}

    • Eliminate entry in row 3 (R3→R3+2R2R_3 \to R_3 + 2R_2):       (1221011000−11)\begin{pmatrix} 1 & 2 & 2 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{pmatrix}

    • Scale row 3 (R3→−R3R_3 \to -R_3):       (12210110001−1)\begin{pmatrix} 1 & 2 & 2 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1 \end{pmatrix}

    • Back-Substitution:

    • From row 3: x3=−1x_3 = -1

    • From row 2: x2+x3=0  ⟹  x2−1=0  ⟹  x2=1x_2 + x_3 = 0 \implies x_2 - 1 = 0 \implies x_2 = 1

    • From row 1: x1+2(1)+2(−1)=1  ⟹  x1=1x_1 + 2(1) + 2(-1) = 1 \implies x_1 = 1

Eigenvalues and Eigenvectors

  • Fundamental Definition:

    • Let AA be an n×nn \times n matrix. A scalar λ\lambda is an eigenvalue of AA if there exists a non-zero vector v\mathbf{v} such that:     Av=λvA\mathbf{v} = \lambda \mathbf{v}

    • Vector v\mathbf{v} is called an eigenvector corresponding to eigenvalue λ\lambda.

  • Characteristic Equation and Polynomial:

    • Rewrite eigenvalue equation as:     (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}

    • For non-trivial solutions (v≠0\mathbf{v} \neq \mathbf{0}), the matrix (A−λI)(A - \lambda I) must be singular:     p(λ)=det⁡(A−λI)=0p(\lambda) = \det(A - \lambda I) = 0

    • p(λ)p(\lambda) is the characteristic polynomial of AA, and its roots are the eigenvalues.

  • Worked Example:

    • Find eigenvalues and eigenvectors of A=(2−4−1−1)A = \begin{pmatrix} 2 & -4 \\ -1 & -1 \end{pmatrix}:

    • Characteristic Determinant:     det⁡(A−λI)=det⁡(2−λ−4−1−1−λ)=(2−λ)(−1−λ)−(−4)(−1)\det(A - \lambda I) = \det\begin{pmatrix} 2 - \lambda & -4 \\ -1 & -1 - \lambda \end{pmatrix} = (2 - \lambda)(-1 - \lambda) - (-4)(-1)     p(λ)=(λ2−λ−2)−4=λ2−λ−6=(λ−3)(λ+2)=0p(\lambda) = (\lambda^2 - \lambda - 2) - 4 = \lambda^2 - \lambda - 6 = (\lambda - 3)(\lambda + 2) = 0

    • Eigenvalues: λ1=3\lambda_1 = 3 and λ2=−2\lambda_2 = -2.

    • Finding Eigenvector for λ1=3\lambda_1 = 3:     (2−3−4−1−1−3)(v1v2)=(−1−4−1−4)(v1v2)=(00)\begin{pmatrix} 2 - 3 & -4 \\ -1 & -1 - 3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -1 & -4 \\ -1 & -4 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}     −v1−4v2=0  ⟹  v1=−4v2-v_1 - 4v_2 = 0 \implies v_1 = -4v_2

    • Setting v2=1v_2 = 1 yields eigenvector v1=(−41)\mathbf{v}_1 = \begin{pmatrix} -4 \\ 1 \end{pmatrix}.

    • Finding Eigenvector for λ2=−2\lambda_2 = -2:     (2−(−2)−4−1−1−(−2))(v1v2)=(4−4−11)(v1v2)=(00)\begin{pmatrix} 2 - (-2) & -4 \\ -1 & -1 - (-2) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 & -4 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}     −v1+v2=0  ⟹  v1=v2-v_1 + v_2 = 0 \implies v_1 = v_2

    • Setting v2=1v_2 = 1 yields eigenvector v2=(11)\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

Vector Foundations and Vector Algebra

  • Vector Definition and Representation:

    • A vector is a quantity characterized by both magnitude and direction.

    • Directed line segment representation: PQ⃗\vec{PQ} has initial point PP and terminal point QQ.

    • Scalar quantities: Characterized solely by magnitude (e.g., mass, distance, speed, temperature, energy).

    • Vector quantities: Require magnitude and direction (e.g., velocity, acceleration, force, momentum).

  • Classifications of Vectors:

    • Zero Vector (Null Vector, 0\mathbf{0}): Magnitude of zero with undefined or arbitrary direction.

    • Unit Vector (a^\hat{\mathbf{a}}

    • Magnitude of 1 unit:       a^=a∣a∣\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}

    • Equal Vectors: Possess identical magnitude and direction regardless of spatial origin.

    • Collinear Vectors: Vectors lying along the same straight line or parallel lines.

    • Concurrent Vectors: Vectors whose lines of action intersect at a common point.

    • Coplanar Vectors: Vectors situated within the same geometric plane.

  • Laws of Vector Algebra:

    • Commutative Law of Addition: a+b=b+a\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}

    • Associative Law of Addition: (a+b)+c=a+(b+c)(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})

    • Associative Law of Scalar Multiplication: m(na)=(mn)am(n\mathbf{a}) = (mn)\mathbf{a}

    • Distributive Laws: (m+n)a=ma+na(m + n)\mathbf{a} = m\mathbf{a} + n\mathbf{a} and m(a+b)=ma+mbm(\mathbf{a} + \mathbf{b}) = m\mathbf{a} + m\mathbf{b}

  • Geometric Addition Principles:

    • Parallelogram Law: If two vectors a\mathbf{a} and b\mathbf{b} originate from a common point, forming adjacent sides of a parallelogram, their resultant a+b\mathbf{a} + \mathbf{b} is the directed diagonal originating from that common point.

    • Triangle Law: If two vectors a\mathbf{a} and b\mathbf{b} are arranged head-to-tail, their resultant is the directed segment completing the triangle from the initial point of the first to the terminal point of the second.

    • Polygon Law: For n−1n-1 vectors represented head-to-tail along an nn-sided polygon, their resultant is the closing side drawn in reverse order.

  • Position Vectors and Midpoints:

    • The position vector of point AA relative to origin OO is OA⃗=a\vec{OA} = \mathbf{a}.

    • Displacement vector between points AA and BB:     AB⃗=OB⃗−OA⃗=b−a\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}

    • Midpoint Position Vector: If MM bisects ABAB:     OM⃗=OA⃗+12AB⃗=a+12(b−a)=12(a+b)\vec{OM} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})

Coordinate Vectors and Direction Cosines

  • Component Representation in Three Dimensions:

    • Using a right-handed orthogonal system with unit basis vectors i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} along the x,y,zx, y, z axes:     a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}

    • Magnitude of Vector a\mathbf{a}:     ∣a∣=a12+a22+a32|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

    • Example: Given a=3i−j+4k\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}, b=5j+2k\mathbf{b} = 5\mathbf{j} + 2\mathbf{k}, c=i−2j−k\mathbf{c} = \mathbf{i} - 2\mathbf{j} - \mathbf{k}:

    • Resultant: a+b+c=(3+0+1)i+(−1+5−2)j+(4+2−1)k=4i+2j+5k\mathbf{a} + \mathbf{b} + \mathbf{c} = (3 + 0 + 1)\mathbf{i} + (-1 + 5 - 2)\mathbf{j} + (4 + 2 - 1)\mathbf{k} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}

    • Magnitude of a−b\mathbf{a} - \mathbf{b}: a−b=3i−6j+2k\mathbf{a} - \mathbf{b} = 3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}, so ∣a−b∣=32+(−6)2+22=9+36+4=49=7|\mathbf{a} - \mathbf{b}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7.

  • Direction Angles and Direction Cosines:

    • The direction angles α,β,γ∈[0,π]\alpha, \beta, \gamma \in [0, \pi] are the angles between non-zero vector a\mathbf{a} and the positive x,y,zx, y, z coordinate axes.

    • Direction Cosines: Denoted by ℓ,μ,ρ\ell, \mu, \rho:     ℓ=cos⁡(α)=a1∣a∣\ell = \cos(\alpha) = \frac{a_1}{|\mathbf{a}|}     μ=cos⁡(β)=a2∣a∣\mu = \cos(\beta) = \frac{a_2}{|\mathbf{a}|}     ρ=cos⁡(γ)=a3∣a∣\rho = \cos(\gamma) = \frac{a_3}{|\mathbf{a}|}

    • Fundamental Theorem: The sum of the squares of the direction cosines equals 1:     ℓ2+μ2+ρ2=cos⁡2(α)+cos⁡2(β)+cos⁡2(γ)=1\ell^2 + \mu^2 + \rho^2 = \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1

    • Proof:     ℓ2+μ2+ρ2=(a1∣a∣)2+(a2∣a∣)2+(a3∣a∣)2=a12+a22+a32∣a∣2=∣a∣2∣a∣2=1\ell^2 + \mu^2 + \rho^2 = \left(\frac{a_1}{|\mathbf{a}|}\right)^2 + \left(\frac{a_2}{|\mathbf{a}|}\right)^2 + \left(\frac{a_3}{|\mathbf{a}|}\right)^2 = \frac{a_1^2 + a_2^2 + a_3^2}{|\mathbf{a}|^2} = \frac{|\mathbf{a}|^2}{|\mathbf{a}|^2} = 1

    • Unit Vector Relation: The components of a unit vector are its direction cosines:     a^=ℓi+μj+ρk\hat{\mathbf{a}} = \ell\mathbf{i} + \mu\mathbf{j} + \rho\mathbf{k}

    • Worked Example: Find direction cosines for a=i−2j+3k\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}:

    • Magnitude: ∣a∣=12+(−2)2+32=1+4+9=14|\mathbf{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}.

    • Direction cosines:       ℓ=114,μ=−214,ρ=314\ell = \frac{1}{\sqrt{14}}, \quad \mu = -\frac{2}{\sqrt{14}}, \quad \rho = \frac{3}{\sqrt{14}}

    • Verification: (114)2+(−214)2+(314)2=1+4+914=1\left(\frac{1}{\sqrt{14}}\right)^2 + \left(-\frac{2}{\sqrt{14}}\right)^2 + \left(\frac{3}{\sqrt{14}}\right)^2 = \frac{1 + 4 + 9}{14} = 1.

Vector Multiplication: Dot and Cross Products

  • Scalar (Dot) Product:

    • Definition: For vectors a\mathbf{a} and b\mathbf{b} with angle θ∈[0,π]\theta \in [0, \pi] between them:     a⋅b=∣a∣∣b∣cos⁡(θ)\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)

    • Algebraic Component Form:     a⋅b=a1b1+a2b2+a3b3\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

    • Properties of Dot Product:

    • Commutative: a⋅b=b⋅a\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}

    • Distributive: a⋅(b+c)=a⋅b+a⋅c\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}

    • Basis Orthogonality: i⋅i=j⋅j=k⋅k=1\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1 and i⋅j=j⋅k=k⋅i=0\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0

    • Perpendicularity Condition: Two non-zero vectors are perpendicular if and only if a⋅b=0\mathbf{a} \cdot \mathbf{b} = 0.

    • Parallel Condition: If a∥b\mathbf{a} \parallel \mathbf{b}, then a⋅b=∣a∣∣b∣\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|

  • Physical and Geometric Applications of Dot Product:

    • Angle Between Vectors:     cos⁡(θ)=a⋅b∣a∣∣b∣=a1b1+a2b2+a3b3a12+a22+a32b12+b22+b32\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2}\sqrt{b_1^2 + b_2^2 + b_3^2}}

    • Distance of Point from Origin: d=r⋅r=∣r∣d = \sqrt{\mathbf{r} \cdot \mathbf{r}} = |\mathbf{r}|

    • Work Done by a Constant Force:     W=F⋅r=∣F∣∣r∣cos⁡(θ)W = \mathbf{F} \cdot \mathbf{r} = |\mathbf{F}||\mathbf{r}|\cos(\theta)

    • Worked Example: A constant force of 5 N5\,\text{N} in direction i+j−2k\mathbf{i} + \mathbf{j} - 2\mathbf{k} displaces an object from (1,0,1)(1, 0, 1) to (3,4,−1)(3, 4, -1).

    • Unit direction vector: n^=i+j−2k12+12+(−2)2=i+j−2k6\hat{\mathbf{n}} = \frac{\mathbf{i} + \mathbf{j} - 2\mathbf{k}}{\sqrt{1^2 + 1^2 + (-2)^2}} = \frac{\mathbf{i} + \mathbf{j} - 2\mathbf{k}}{\sqrt{6}}

    • Force vector: F=5n^=56(i+j−2k)\mathbf{F} = 5\hat{\mathbf{n}} = \frac{5}{\sqrt{6}}(\mathbf{i} + \mathbf{j} - 2\mathbf{k})

    • Displacement vector: r=(3−1)i+(4−0)j+(−1−1)k=2i+4j−2k\mathbf{r} = (3-1)\mathbf{i} + (4-0)\mathbf{j} + (-1-1)\mathbf{k} = 2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}

    • Work Done:       W=F⋅r=56[1(2)+1(4)+(−2)(−2)]=56[2+4+4]=506 JW = \mathbf{F} \cdot \mathbf{r} = \frac{5}{\sqrt{6}}[1(2) + 1(4) + (-2)(-2)] = \frac{5}{\sqrt{6}}[2 + 4 + 4] = \frac{50}{\sqrt{6}}\,\text{J}

  • Vector (Cross) Product:

    • Definition: Generates a vector perpendicular to both a\mathbf{a} and b\mathbf{b}:     a×b=∣a∣∣b∣sin⁡(θ) n^\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin(\theta)\,\hat{\mathbf{n}}

    • Where n^\hat{\mathbf{n}} is the unit normal vector specified by the right-hand rule.

    • Properties of Cross Product:

    • Anti-Commutative: a×b=−(b×a)\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})

    • Distributive: a×(b+c)=a×b+a×c\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}

    • Parallel Condition: a×b=0\mathbf{a} \times \mathbf{b} = \mathbf{0} if and only if a∥b\mathbf{a} \parallel \mathbf{b}

    • Cyclic Basis Properties:       i×i=j×j=k×k=0\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}       i×j=k,j×k=i,k×i=j\mathbf{i} \times \mathbf{j} = \mathbf{k}, \quad \mathbf{j} \times \mathbf{k} = \mathbf{i}, \quad \mathbf{k} \times \mathbf{i} = \mathbf{j}       j×i=−k,k×j=−i,i×k=−j\mathbf{j} \times \mathbf{i} = -\mathbf{k}, \quad \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \quad \mathbf{i} \times \mathbf{k} = -\mathbf{j}

    • Determinant Representation:     a×b=∣ijka1a2a3b1b2b3∣=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}

  • Geometric Applications of Cross Product:

    • Area of Parallelogram with adjacent side vectors a\mathbf{a} and b\mathbf{b}:     Area=∣a×b∣\text{Area} = |\mathbf{a} \times \mathbf{b}|

    • Area of Triangle with adjacent side vectors a\mathbf{a} and b\mathbf{b}:     Area=12∣a×b∣\text{Area} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|

    • Angle between vectors using cross product:     sin⁡(θ)=∣a×b∣∣a∣∣b∣\sin(\theta) = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}

Fundamental Trigonometric Ratios and Evaluations

  • Right Triangle Trigonometric Ratios:

    • For an acute angle θ\theta (0∘<θ<90∘0^\circ < \theta < 90^\circ) in a right triangle with adjacent side aa, opposite side bb, and hypotenuse cc:     sin⁡(θ)=OppositeHypotenuse=bc\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{b}{c}     cos⁡(θ)=AdjacentHypotenuse=ac\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{a}{c}     tan⁡(θ)=OppositeAdjacent=ba\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{b}{a}

  • Reciprocal Ratios:

    • Defined as:     csc⁡(θ)=1sin⁡(θ)=cb\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{c}{b}     sec⁡(θ)=1cos⁡(θ)=ca\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{c}{a}     cot⁡(θ)=1tan⁡(θ)=ab\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{a}{b}

  • Geometric Derivation for Special Angles:

    • Evaluation at 45∘45^\circ (π4\frac{\pi}{4}

    • Constructed using an isosceles right triangle with side lengths 1,11, 1, giving hypotenuse 12+12=2\sqrt{1^2 + 1^2} = \sqrt{2}.     sin⁡(45∘)=12,cos⁡(45∘)=12,tan⁡(45∘)=1\sin(45^\circ) = \frac{1}{\sqrt{2}}, \quad \cos(45^\circ) = \frac{1}{\sqrt{2}}, \quad \tan(45^\circ) = 1     csc⁡(45∘)=2,sec⁡(45∘)=2,cot⁡(45∘)=1\csc(45^\circ) = \sqrt{2}, \quad \sec(45^\circ) = \sqrt{2}, \quad \cot(45^\circ) = 1

    • Evaluation at 60∘60^\circ (π3\frac{\pi}{3}) and 30∘30^\circ (π6\frac{\pi}{6}

    • Constructed by bisecting an equilateral triangle of side length 2, creating a right triangle with base 1, hypotenuse 2, and height 22−12=3\sqrt{2^2 - 1^2} = \sqrt{3}.

    • For 60∘60^\circ:       sin⁡(60∘)=32,cos⁡(60∘)=12,tan⁡(60∘)=3\sin(60^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(60^\circ) = \frac{1}{2}, \quad \tan(60^\circ) = \sqrt{3}       csc⁡(60∘)=23,sec⁡(60∘)=2,cot⁡(60∘)=13\csc(60^\circ) = \frac{2}{\sqrt{3}}, \quad \sec(60^\circ) = 2, \quad \cot(60^\circ) = \frac{1}{\sqrt{3}}

    • For 30∘30^\circ:       sin⁡(30∘)=12,cos⁡(30∘)=32,tan⁡(30∘)=13\sin(30^\circ) = \frac{1}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \tan(30^\circ) = \frac{1}{\sqrt{3}}       csc⁡(30∘)=2,sec⁡(30∘)=23,cot⁡(30∘)=3\csc(30^\circ) = 2, \quad \sec(30^\circ) = \frac{2}{\sqrt{3}}, \quad \cot(30^\circ) = \sqrt{3}

    • Boundary Angle Evaluations (0∘0^\circ and 90∘90^\circ

    • At 0∘0^\circ: sin⁡(0∘)=0\sin(0^\circ) = 0, cos⁡(0∘)=1\cos(0^\circ) = 1, tan⁡(0∘)=0\tan(0^\circ) = 0, csc⁡(0∘)=∞\csc(0^\circ) = \infty, sec⁡(0∘)=1\sec(0^\circ) = 1, cot⁡(0∘)=∞\cot(0^\circ) = \infty.

    • At 90∘90^\circ: sin⁡(90∘)=1\sin(90^\circ) = 1, cos⁡(90∘)=0\cos(90^\circ) = 0, tan⁡(90∘)=∞\tan(90^\circ) = \infty, csc⁡(90∘)=1\csc(90^\circ) = 1, sec⁡(90∘)=∞\sec(90^\circ) = \infty, cot⁡(90∘)=0\cot(90^\circ) = 0.

Analytical Trigonometric Identities

  • Pythagorean Identities:   sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1   1+tan⁡2(θ)=sec⁡2(θ)1 + \tan^2(\theta) = \sec^2(\theta)   1+cot⁡2(θ)=csc⁡2(θ)1 + \cot^2(\theta) = \csc^2(\theta)

  • Even-Odd Identities:

    • Odd Functions (f(−x)=−f(x)f(-x) = -f(x)     sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta)     tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta)     csc⁡(−θ)=−csc⁡(θ)\csc(-\theta) = -\csc(\theta)     cot⁡(−θ)=−cot⁡(θ)\cot(-\theta) = -\cot(\theta)

    • Even Functions (f(−x)=f(x)f(-x) = f(x)     cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta)     sec⁡(−θ)=sec⁡(θ)\sec(-\theta) = \sec(\theta)

  • Quotient Identities:   tan⁡(θ)=sin⁡(θ)cos⁡(θ),cot⁡(θ)=cos⁡(θ)sin⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}

  • Sum and Difference Identities:

    • Cosine Identities:     cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)     cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)

    • Sine Identities:     sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)     sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)

    • Tangent Identities:     tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}     tan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)tan⁡(B)\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}

  • Double Angle Identities:

    • Double Angle for Sine:     sin⁡(2A)=2sin⁡(A)cos⁡(A)\sin(2A) = 2\sin(A)\cos(A)

    • Double Angle for Cosine (Three Equivalent Forms):     cos⁡(2A)=cos⁡2(A)−sin⁡2(A)\cos(2A) = \cos^2(A) - \sin^2(A)     cos⁡(2A)=1−2sin⁡2(A)\cos(2A) = 1 - 2\sin^2(A)     cos⁡(2A)=2cos⁡2(A)−1\cos(2A) = 2\cos^2(A) - 1

    • Double Angle for Tangent:     tan⁡(2A)=2tan⁡(A)1−tan⁡2(A)\tan(2A) = \frac{2\tan(A)}{1 - \tan^2(A)}

  • Half Angle Identities:

    • Derived by substituting A=D2A = \frac{D}{2} into the double angle cosine formulas:     cos⁡(D2)=±1+cos⁡(D)2\cos\left(\frac{D}{2}\right) = \pm\sqrt{\frac{1 + \cos(D)}{2}}     sin⁡(D2)=±1−cos⁡(D)2\sin\left(\frac{D}{2}\right) = \pm\sqrt{\frac{1 - \cos(D)}{2}}

  • Factor Formulas (Sum-to-Product Identities):

    • Sum for Sine:     sin⁡(X)+sin⁡(Y)=2sin⁡(X+Y2)cos⁡(X−Y2)\sin(X) + \sin(Y) = 2\sin\left(\frac{X + Y}{2}\right)\cos\left(\frac{X - Y}{2}\right)

    • Sum for Cosine:     cos⁡(X)+cos⁡(Y)=2cos⁡(X+Y2)cos⁡(X−Y2)\cos(X) + \cos(Y) = 2\cos\left(\frac{X + Y}{2}\right)\cos\left(\frac{X - Y}{2}\right)

Spherical Trigonometry

  • Core Definitions and Geometry of the Sphere:

    • A spherical triangle is formed by the intersection of three great circle arcs on the surface of a sphere.

    • Vertices are labeled A,B,CA, B, C with interior vertex angles α,β,γ\alpha, \beta, \gamma.

    • Sides are labeled a,b,ca, b, c, measured as angles subtended by the arcs at the sphere's center.

Fig 4.1 spherical triangle
  • Angular Sum and Spherical Excess:

    • Unlike planar triangles whose interior angle sum is exactly π\pi, the sum of angles of a spherical triangle satisfies:     π<α+β+γ<3π\pi < \alpha + \beta + \gamma < 3\pi

    • Spherical Excess (EE): The positive angular quantity defined by:     E=α+β+γ−πE = \alpha + \beta + \gamma - \pi

  • Surface Area of a Spherical Triangle:

    • For a sphere of radius RR, with spherical excess EE measured in radians:     Area=R2E\text{Area} = R^2 E

  • Right Spherical Triangles (γ=π2=90∘\gamma = \frac{\pi}{2} = 90^\circ

    • The following fundamental identities govern right spherical triangles:     sin⁡(a)=sin⁡(α)sin⁡(c)=tan⁡(b)cot⁡(β)\sin(a) = \sin(\alpha)\sin(c) = \tan(b)\cot(\beta)     sin⁡(b)=sin⁡(β)sin⁡(c)=tan⁡(a)cot⁡(α)\sin(b) = \sin(\beta)\sin(c) = \tan(a)\cot(\alpha)     cos⁡(α)=cos⁡(a)sin⁡(β)=tan⁡(b)cot⁡(c)\cos(\alpha) = \cos(a)\sin(\beta) = \tan(b)\cot(c)     cos⁡(β)=cos⁡(b)sin⁡(α)=tan⁡(a)cot⁡(c)\cos(\beta) = \cos(b)\sin(\alpha) = \tan(a)\cot(c)     cos⁡(c)=cot⁡(α)cot⁡(β)=cos⁡(a)cos⁡(b)\cos(c) = \cot(\alpha)\cot(\beta) = \cos(a)\cos(b)

  • General Spherical Triangles:

    • Spherical Law of Sines:     sin⁡(a)sin⁡(α)=sin⁡(b)sin⁡(β)=sin⁡(c)sin⁡(γ)\frac{\sin(a)}{\sin(\alpha)} = \frac{\sin(b)}{\sin(\beta)} = \frac{\sin(c)}{\sin(\gamma)}

    • Spherical Law of Cosines for Sides:     cos⁡(a)=cos⁡(b)cos⁡(c)+sin⁡(b)sin⁡(c)cos⁡(α)\cos(a) = \cos(b)\cos(c) + \sin(b)\sin(c)\cos(\alpha)     cos⁡(b)=cos⁡(a)cos⁡(c)+sin⁡(a)sin⁡(c)cos⁡(β)\cos(b) = \cos(a)\cos(c) + \sin(a)\sin(c)\cos(\beta)     cos⁡(c)=cos⁡(a)cos⁡(b)+sin⁡(a)sin⁡(b)cos⁡(γ)\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(\gamma)

    • Spherical Law of Cosines for Angles:     cos⁡(α)=−cos⁡(β)cos⁡(γ)+sin⁡(β)sin⁡(γ)cos⁡(a)\cos(\alpha) = -\cos(\beta)\cos(\gamma) + \sin(\beta)\sin(\gamma)\cos(a)     cos⁡(β)=−cos⁡(α)cos⁡(γ)+sin⁡(α)sin⁡(γ)cos⁡(b)\cos(\beta) = -\cos(\alpha)\cos(\gamma) + \sin(\alpha)\sin(\gamma)\cos(b)     cos⁡(γ)=−cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(c)\cos(\gamma) = -\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\cos(c)

  • Semi-Perimeter and Half-Angle Tangent Formulas:

    • Let semi-perimeter of sides be s=a+b+c2s = \frac{a + b + c}{2}. The vertex half-angle is given by:     tan⁡(α2)=sin⁡(s−b)sin⁡(s−c)sin⁡(s)sin⁡(s−a)\tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{\sin(s - b)\sin(s - c)}{\sin(s)\sin(s - a)}}

    • Let semi-sum of vertex angles be σ=α+β+γ2\sigma = \frac{\alpha + \beta + \gamma}{2}. The side half-arc is given by:     tan⁡(a2)=−cos⁡(σ)cos⁡(σ−α)cos⁡(σ−β)cos⁡(σ−γ)\tan\left(\frac{a}{2}\right) = \sqrt{\frac{-\cos(\sigma)\cos(\sigma - \alpha)}{\cos(\sigma - \beta)\cos(\sigma - \gamma)}}