Trigonometry Notes

Angles

When a straight line rotates around a point, it forms an angle measurable in degrees or radians. A full rotation returns to the start at 360 degrees (360360^{\circ}). Each degree has 60 minutes (6060'), and each minute has 60 seconds (6060''). A straight angle is half of a full rotation (180180^{\circ}), and a right angle is half of a straight angle (9090^{\circ}).

  • Acute angle: Less than 9090^{\circ}.
  • Obtuse angle: Greater than 9090^{\circ}.

Converting Degrees, Minutes, and Seconds to Decimal Degrees

To convert an angle from degrees, minutes, and seconds to decimal degrees, use the following formula:
Decimal Degrees=Degrees+Minutes60+Seconds3600\text{Decimal Degrees} = \text{Degrees} + \frac{\text{Minutes}}{60} + \frac{\text{Seconds}}{3600}
Example: Convert 45361845^{\circ}36'18'' to decimal form.
453618=45+3660+183600=45.60545^{\circ}36'18'' = 45^{\circ} + \frac{36}{60}^{\circ} + \frac{18}{3600}^{\circ} = 45.605^{\circ}

Converting Decimal Degrees to Degrees, Minutes, and Seconds

To convert an angle from decimal degrees to degrees, minutes, and seconds, use the following method:

  1. The whole number part of the decimal is the degrees.
  2. Multiply the decimal part by 60 to get minutes.
  3. The whole number part of the result is the minutes.
  4. Multiply the decimal part of the minutes by 60 to get seconds.

Example: Convert 18.47818.478^{\circ} to degrees, minutes, and seconds.
18.478=18+0.478×60=18+28.68=18+28+0.68×60=18+28+4118.478^{\circ} = 18^{\circ} + 0.478 \times 60' = 18^{\circ} + 28.68' = 18^{\circ} + 28' + 0.68 \times 60'' = 18^{\circ} + 28' + 41''

Radians

A radian is the angle subtended at the center of a circle by an arc equal in length to the radius of the circle. If a line of length rr rotates about one end so that the other end describes an arc of length rr, the line is said to have rotated through 1 radian.

Radians and Degrees Conversion

A full angle (360360^{\circ}) is equal to 2π2\pi radians. Therefore, to convert between degrees and radians, use the following relationships:

  • Degrees to Radians: Multiply by π180\frac{\pi}{180}.
  • Radians to Degrees: Multiply by 180π\frac{180}{\pi}.

Therefore:
360=2π radians360^{\circ} = 2\pi \text{ radians}
1 radian=180π57.29581 \text{ radian} = \frac{180}{\pi} \approx 57.2958^{\circ}

Radian Conversion Examples
  1. Convert 3030^{\circ} to radians: 30×π180=π630^{\circ} \times \frac{\pi}{180} = \frac{\pi}{6} radians
  2. Convert 120120^{\circ} to radians: 120×π180=2π3120^{\circ} \times \frac{\pi}{180} = \frac{2\pi}{3} radians
  3. Convert 270270^{\circ} to radians: 270×π180=3π2270^{\circ} \times \frac{\pi}{180} = \frac{3\pi}{2} radians

Triangles

Triangles are defined by their shape (angles) and size (side lengths). Similar triangles have the same shape (equal angles) but can be different sizes.

Properties of Similar Triangles

In similar triangles, the ratios of corresponding sides are equal.
If triangles ABCABC and ABCA'B'C' are similar, then:
ABAB=ACAC=BCBC\frac{AB}{A'B'} = \frac{AC}{A'C'} = \frac{BC}{B'C'}
Knowing the ratios of sides in one triangle allows deductions about any similar triangle.

Trigonometric Ratios

In a right-angled triangle ABCABC with angle θ\theta at vertex BB, where ACAC is the opposite side, BCBC is the adjacent side, and ABAB is the hypotenuse, the trigonometric ratios are defined as:

  • Sine: sinθ=oppositehypotenuse=ACAB\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB}
  • Cosine: cosθ=adjacenthypotenuse=BCAB\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AB}
  • Tangent: tanθ=oppositeadjacent=ACBC\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC}

Reciprocal Ratios

There are three reciprocal trigonometric ratios:

  • Secant: secθ=1cosθ\sec \theta = \frac{1}{\cos \theta}
  • Cosecant: cscθ=1sinθ\csc \theta = \frac{1}{\sin \theta}
  • Cotangent: cotθ=1tanθ\cot \theta = \frac{1}{\tan \theta}

Pythagoras' Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
a2+b2=c2a^2 + b^2 = c^2
where cc is the length of the hypotenuse, and aa and bb are the lengths of the other two sides. The side length corresponds to the lowercase of the opposite angle.

Special Triangles

45-45-90 Triangle

In a right-angled isosceles triangle with angles 4545^{\circ}, 4545^{\circ}, and 9090^{\circ}, the side lengths are in the ratio 1:1:2\sqrt{2}.
sin(45)=cos(45)=12\sin(45^{\circ}) = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}
sin(π4)=cos(π4)=12\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}

30-60-90 Triangle

In a half equilateral triangle with angles 3030^{\circ}, 6060^{\circ}, and 9090^{\circ}, the side lengths are in the ratio 1:3\sqrt{3}:2.
sin(π6)=cos(π3)=12\sin(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) = \frac{1}{2}
sin(π3)=cos(π6)=32\sin(\frac{\pi}{3}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}

The Fundamental Trigonometrical Identity

For any angle θ\theta:
cos2θ+sin2θ=1\cos^2 \theta + \sin^2 \theta = 1

Derivation

Starting with Pythagoras' theorem: a2+b2=c2a^2 + b^2 = c^2 Divide both sides by c2c^2:
(ac)2+(bc)2=1(\frac{a}{c})^2 + (\frac{b}{c})^2 = 1
Since ac=cosθ\frac{a}{c} = \cos \theta and bc=sinθ\frac{b}{c} = \sin \theta:
cos2θ+sin2θ=1\cos^2 \theta + \sin^2 \theta = 1

Two More Identities

Dividing the fundamental identity by cos2θ\cos^2 \theta:
cos2θcos2θ+sin2θcos2θ=1cos2θ\frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
1+tan2θ=sec2θ1 + \tan^2 \theta = \sec^2 \theta

Dividing the fundamental identity by sin2θ\sin^2 \theta:
cos2θsin2θ+sin2θsin2θ=1sin2θ\frac{\cos^2 \theta}{\sin^2 \theta} + \frac{\sin^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}
cot2θ+1=csc2θ\cot^2 \theta + 1 = \csc^2 \theta

Identities for Compound Angles

  • Cosine of a sum: cos(θ+φ)=cosθcosφsinθsinφ\cos(\theta + \varphi) = \cos \theta \cos \varphi - \sin \theta \sin \varphi
  • Cosine of a difference: cos(θφ)=cosθcosφ+sinθsinφ\cos(\theta - \varphi) = \cos \theta \cos \varphi + \sin \theta \sin \varphi

Sums and Differences of Angles

  • cos(θ±φ)=cosθcosφsinθsinφ\cos(\theta \pm \varphi) = \cos \theta \cos \varphi \mp \sin \theta \sin \varphi
  • sin(θ±φ)=sinθcosφ±cosθsinφ\sin(\theta \pm \varphi) = \sin \theta \cos \varphi \pm \cos \theta \sin \varphi
  • tan(θ±φ)=tanθ±tanφ1tanθtanφ\tan(\theta \pm \varphi) = \frac{\tan \theta \pm \tan \varphi}{1 \mp \tan \theta \tan \varphi}

Double Angles

  • sin2θ=2sinθcosθ\sin 2\theta = 2 \sin \theta \cos \theta
  • cos2θ=cos2θsin2θ=2cos2θ1=12sin2θ\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta
  • tan2θ=2tanθ1tan2θ\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}

Sums and Differences of Trigonometric Functions

  • sinθ+sinφ=2sin(θ+φ2)cos(θφ2)\sin \theta + \sin \varphi = 2 \sin(\frac{\theta + \varphi}{2}) \cos(\frac{\theta - \varphi}{2})
  • sinθsinφ=2cos(θ+φ2)sin(θφ2)\sin \theta - \sin \varphi = 2 \cos(\frac{\theta + \varphi}{2}) \sin(\frac{\theta - \varphi}{2})
  • cosθ+cosφ=2cos(θ+φ2)cos(θφ2)\cos \theta + \cos \varphi = 2 \cos(\frac{\theta + \varphi}{2}) \cos(\frac{\theta - \varphi}{2})
  • cosθcosφ=2sin(θ+φ2)sin(θφ2)\cos \theta - \cos \varphi = -2 \sin(\frac{\theta + \varphi}{2}) \sin(\frac{\theta - \varphi}{2})

Products of Ratios

  • 2sinθcosφ=sin(θ+φ)+sin(θφ)2 \sin \theta \cos \varphi = \sin(\theta + \varphi) + \sin(\theta - \varphi)
  • 2cosθcosφ=cos(θ+φ)+cos(θφ)2 \cos \theta \cos \varphi = \cos(\theta + \varphi) + \cos(\theta - \varphi)
  • 2sinθsinφ=cos(θφ)cos(θ+φ)2 \sin \theta \sin \varphi = \cos(\theta - \varphi) - \cos(\theta + \varphi)

Trigonometric Functions

The trigonometric ratios, originally defined for acute angles in right-angled triangles, can be extended to cover any angle while retaining the properties of the original functions.

Consider a unit circle (radius = 1) centered at the origin. A line segment OAOA of unit length rotates around the origin OO in an anticlockwise direction, with point AA tracing the circle.

Sine Function

For angles θ\theta where 0 < \theta < \frac{\pi}{2}, sinθ=ABOA=AB\sin \theta = \frac{AB}{OA} = AB. This means the sine of an angle is equal to the height of point AA above the x-axis.

The sine function, sinθ\sin \theta, gives the height of AA above BB for any angle 0θ0 \leq \theta \leq \infty.

When AA is below the x-axis, the height is negative.

Cosine Function

For angles θ\theta where 0 < \theta < \frac{\pi}{2}, cosθ=OBOA=OB\cos \theta = \frac{OB}{OA} = OB. This means the cosine of an angle is equal to the distance from the origin OO to point BB on the x-axis.

The cosine function, cosθ\cos \theta, gives the distance from OO to BB for any angle θ\theta.

Tangent Function

The tangent function is defined as the ratio of sine to cosine:
tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}
Because cosθ=0\cos \theta = 0 whenever θ\theta is an odd multiple of π2\frac{\pi}{2}, the tangent function is undefined at these points, resulting in vertical asymptotes in its graph.

Inverse Trigonometric Functions

The inverse trigonometric functions are obtained by reflecting the graphs of the trigonometric functions in the line y=xy = x.

However, these reflections are not functions because there is more than one value of yy corresponding to a given value of xx. To make them functions, the graphs are cut off to obtain single-valued functions.

Period

A periodic function is one whose output repeats itself over a regular interval of the input, called the period of the function.

  • Sine and cosine functions repeat every 2π2\pi radians so they are periodic with period 2π2\pi radians.
  • The tangent function repeats every π\pi radians so it is periodic with period π\pi radians.
Finding the Period

For a function like sin(3θ)\sin(3\theta), the period TT is found by solving 3(T)=2π3(T) = 2\pi, which gives
T=2π3T = \frac{2\pi}{3}
This means there are 3 cycles of oscillation within the 2π2\pi period.

Amplitude

The amplitude of a periodic function is the difference between the maximum and minimum values of the output, divided by two. For example, the cosine function ranges from -1 to +1, and its average value is zero. Thus its amplitude is 1.

Phase

The phase difference of a periodic function is the interval of the input by which the output leads or lags behind the reference function. For example, y=sin(x+π4)y = \sin(x + \frac{\pi}{4}) has the same shape as y=sin(x)y = \sin(x), but it leads it by π4\frac{\pi}{4} radians. So it has a phase difference of π4\frac{\pi}{4} compared to sin(x)\sin(x).

Oscillatory Sinusoidal Waves

An oscillatory wave can be described by:
f(t)=Asin(ωt+φ)f(t) = A \sin(\omega t + \varphi)
Where:

  • AA is the amplitude.
  • ω\omega is the angular frequency.
  • φ\varphi is the phase.
Wave Properties
  • Period: T=2πωT = \frac{2\pi}{\omega}
  • Frequency: f=ω2π=1Tf = \frac{\omega}{2\pi} = \frac{1}{T}
  • The wave crosses the t-axis at: t=φωt = -\frac{\varphi}{\omega}
Spatial Variation

In space, a wave can be described as:
f(x)=Asin(kx+φ)f(x) = A \sin(kx + \varphi)
Where:

  • xx is the distance.
  • kk is the angular wavenumber.
Wavelength
  • Wavelength: λ=2πk\lambda = \frac{2\pi}{k}

It can vary in both space and time:
f(x,t)=Asin(kxωt+φ)f(x, t) = A \sin(kx - \omega t + \varphi)

Addition of Waveforms

Case 1: Waves with the Same Angular Frequency

If y<em>1=asin(ωt)y<em>1 = a \sin(\omega t) and y</em>2=bcos(ωt)y</em>2 = b \cos(\omega t), then their sum is:
y=y<em>1+y</em>2=asin(ωt)+bcos(ωt)y = y<em>1 + y</em>2 = a \sin(\omega t) + b \cos(\omega t)
To express this as Csin(ωt+φ)C \sin(\omega t + \varphi), we can use the compound angle formula:
Csin(ωt+φ)=Csin(ωt)cos(φ)+Ccos(ωt)sin(φ)C \sin(\omega t + \varphi) = C \sin(\omega t) \cos(\varphi) + C \cos(\omega t) \sin(\varphi)
asin(ωt)+bcos(ωt)=Csin(ωt)cos(φ)+Ccos(ωt)sin(φ)a \sin(\omega t) + b \cos(\omega t) = C \sin(\omega t) \cos(\varphi) + C \cos(\omega t) \sin(\varphi)

Determining Amplitude and Phase

Ccos(φ)amp;=a Csin(φ)amp;=b\begin{aligned} C \cos(\varphi) &amp;= a \ C \sin(\varphi) &amp;= b \end{aligned}
Squaring and adding these equations:
C2(sin2(φ)+cos2(φ))=a2+b2C^2(\sin^2(\varphi) + \cos^2(\varphi)) = a^2 + b^2
C=a2+b2C = \sqrt{a^2 + b^2}
Dividing the equations:
Csin(φ)Ccos(φ)=tan(φ)=ba\frac{C \sin(\varphi)}{C \cos(\varphi)} = \tan(\varphi) = \frac{b}{a}
φ=tan1(ba)\varphi = \tan^{-1}(\frac{b}{a})

Case 2: Waves with Different Angular Frequency

If y<em>1=sin(ω</em>1t)y<em>1 = \sin(\omega</em>1 t) and y<em>2=sin(ω</em>2t)y<em>2 = \sin(\omega</em>2 t), then their sum is:
y=y<em>1+y</em>2=sin(ω<em>1t)+sin(ω</em>2t)y = y<em>1 + y</em>2 = \sin(\omega<em>1 t) + \sin(\omega</em>2 t)
To simplify this, we use the sum-to-product formula:
sin(α)+sin(β)=2sin(α+β2)cos(αβ2)\sin(\alpha) + \sin(\beta) = 2 \sin(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})
Letting α=ω<em>1t\alpha = \omega<em>1 t and β=ω</em>2t\beta = \omega</em>2 t:
sin(ω<em>1t)+sin(ω</em>2t)=2sin((ω<em>1+ω</em>2)t2)cos((ω<em>1ω</em>2)t2)\sin(\omega<em>1 t) + \sin(\omega</em>2 t) = 2 \sin(\frac{(\omega<em>1 + \omega</em>2)t}{2}) \cos(\frac{(\omega<em>1 - \omega</em>2)t}{2})

Beat Frequency

A special case occurs when ω<em>1ω</em>2\omega<em>1 \approx \omega</em>2. The difference ω<em>1ω</em>2\omega<em>1 - \omega</em>2 is close to 0, resulting in beats.

Addition of Waveforms Examples

Find a solution of y=y<em>1+y</em>2y = y<em>1 + y</em>2:

  1. y<em>1=5cos(ωt)y<em>1 = 5 \cos(\omega t) and y</em>2=2sin(ωt)y</em>2 = -2 \sin(\omega t)
  2. y<em>1=4cos(5t)y<em>1 = 4 \cos(5t) and y</em>2=3sin(5t)y</em>2 = 3 \sin(5t)
  3. y<em>1=sin(6t)y<em>1 = \sin(6t) and y</em>2=sin(8t)y</em>2 = \sin(8t)