VCE Mathematical Methods Units 1 & 2: Trigonometric Identities, Relationships, and Simplifying Equations

Warm-Up: Cyclical Revision

• Problem (Tech-Active): Simplify the logarithmic expression $2\log_{10}(4) - 2\log_{10}(5) + \log_{10}(50)$.

  • Options:

  • A. $4\log_{10}(2)$

  • B. $5\log_{10}(2)$

  • C. $15$

  • D. $32$

• Step-by-Step Solution:

  • Apply the power law of logarithms ($k\log_b(a) = \log_b(a^k)$):     $2\log_{10}(4) - 2\log_{10}(5) + \log_{10}(50) = \log_{10}(4^2) - \log_{10}(5^2) + \log_{10}(50)$     $= \log_{10}(16) - \log_{10}(25) + \log_{10}(50)$

  • Apply the product and quotient laws of logarithms ($\log_b(m) - \log_b(n) + \log_b(p) = \log_b\left(\frac{m \times p}{n}\right)$):     $= \log_{10}\left(\frac{16 \times 50}{25}\right)$

  • Simplify the arithmetic inside the logarithm:     $= \log_{10}\left(16 \times 2\right) = \log_{10}(32)$

  • Express $32$ as a power of $2$ ($32 = 2^5$):     $= \log_{10}(2^5) = 5\log_{10}(2)$

  • Correct Answer: B. $5\log_{10}(2)$

Key Vocabulary & Concepts

• Hypotenuse: The longest side of a right-angled triangle, situated directly opposite the right angle ($90^\circ$ or $\frac{\pi}{2}$ radians).

• Pythagorean Identity: The fundamental identity relating the sine and cosine of an angle derived from the Pythagorean theorem on a unit circle: $\sin^2(x) + \cos^2(x) = 1$.

• Complementary Angles: Any pair of angles whose sum equals $90^\circ$ (or $\frac{\pi}{2}$ radians).

• Supplementary Angles: Any pair of angles whose sum equals $180^\circ$ (or $\pi$ radians).

Calculating Pronumerals on the Unit Circle

Using the Pythagorean theorem ($a^2 + b^2 = c^2$) where the hypotenuse of a right-angled triangle on the unit circle is $1$ unit:

• Problem 1: Solve for $y$ given a horizontal side length of $\frac{\sqrt{3}}{2}$ and hypotenuse $1$:   $y^2 = 1 - \left(\frac{\sqrt{3}}{2}\right)^2$   $y^2 = 1 - \frac{3}{4}$   $y^2 = \frac{1}{4}$   $y = \frac{1}{2}$

• Problem 2: Solve for $x$ given a vertical side length of $\frac{1}{\sqrt{2}}$ and hypotenuse $1$:

  $x^2 = 1 - \left(\frac{1}{\sqrt{2}}\right)^2$   $x^2 = 1 - \frac{1}{2}$   $x^2 = \frac{1}{2}$   $x = \frac{1}{\sqrt{2}}$

The Pythagorean Identity

• Fundamental Formula:   $\sin^2(x) + \cos^2(x) = 1$

• Algebraic Rearrangements:

  • Expressing sine in terms of cosine:     $\sin^2(x) = 1 - \cos^2(x)$

  • Expressing cosine in terms of sine:     $\cos^2(x) = 1 - \sin^2(x)$

• Derivation on the Unit Circle:

  • Any point on the unit circle is given by coordinates $(x, y) = (\cos(x), \sin(x))$.

  • Since the radius $r = 1$, applying the equation of a circle $x^2 + y^2 = r^2$ directly yields $\cos^2(x) + \sin^2(x) = 1$.

Simplifying Trigonometric Expressions

• Worked Example 1a: Simplify $2\sin^2(x) + 2\cos^2(x)$.

  • Factor out common terms:     $2\sin^2(x) + 2\cos^2(x) = 2(\sin^2(x) + \cos^2(x))$

  • Substitute the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$:     $= 2 \times 1 = 2$

• Worked Example 1b: Show that $\frac{3 - 3\cos^2(x)}{\sin(x)\cos(x)} = 3\tan(x)$.

  • Factor the numerator:     $3 - 3\cos^2(x) = 3(1 - \cos^2(x))$

  • Substitute $1 - \cos^2(x) = \sin^2(x)$:     $= 3\sin^2(x)$

  • Rewrite the full algebraic fraction:     $\frac{3\sin^2(x)}{\sin(x)\cos(x)}$

  • Cancel out one factor of $\sin(x)$ from numerator and denominator:     $= \frac{3\sin(x)}{\cos(x)}$

  • Apply the quotient identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$:     $= 3\tan(x)$

• Worked Example 2: Show that $\frac{\sin^3(x) + \sin(x)\cos^2(x)}{\cos^3(x) + \cos(x)\sin^2(x)} = \tan(x)$.

  • Factor out common terms from the numerator:     $\sin^3(x) + \sin(x)\cos^2(x) = \sin(x)(\sin^2(x) + \cos^2(x)) = \sin(x)(1) = \sin(x)$

  • Factor out common terms from the denominator:     $\cos^3(x) + \cos(x)\sin^2(x) = \cos(x)(\cos^2(x) + \sin^2(x)) = \cos(x)(1) = \cos(x)$

  • Divide numerator by denominator:     $\frac{\sin(x)}{\cos(x)} = \tan(x)$

Deducing Trigonometric Ratios for Any Real Number

• Geometric Approach (Using Right Triangles & Quadrants):

  • Consider a right-angled triangle with hypotenuse $5$ and vertical side $3$.

  • By Pythagorean triple $(3, 4, 5)$, the adjacent horizontal side length is $4$.

  • Quadrant I (Angle $\theta$):

  • $\sin(\theta) = \frac{3}{5}$

  • $\cos(\theta) = \frac{4}{5}$

  • $\tan(\theta) = \frac{3}{4}$

  • Quadrant II (Angle $x = \pi - \theta$):

  • $\sin(x) = \frac{3}{5}$ (Positive in Quadrant II)

  • $\cos(x) = -\frac{4}{5}$ (Negative in Quadrant II)

  • $\tan(x) = -\frac{3}{4}$ (Negative in Quadrant II)

• Algebraic Approach (Using Identities):

  • Given $\sin(x) = \frac{3}{5}$ where $\frac{\pi}{2} \le x \le \pi$ (Quadrant II):

  • Calculate $\cos(x)$:       $\cos^2(x) = 1 - \sin^2(x) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$       $\cos(x) = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$       Since $x \in \text{Quadrant II}$, $\cos(x)$ must be negative:       $\cos(x) = -\frac{4}{5}$

  • Calculate $\tan(x)$:       $\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{3/5}{-4/5} = \frac{3}{5} \div \left(-\frac{4}{5}\right) = -\frac{3}{4}$

• DIY Practice Example:

  • Problem: Given $\cos(x) = -\frac{1}{3}$ where $\pi \le x \le \frac{3\pi}{2}$ (Quadrant III), deduce the exact values of $\sin(x)$ and $\tan(x)$.

• Step 1 (Find missing side length):     $a^2 + 1^2 = 3^2$     $a^2 + 1 = 9$     $a^2 = 8$     $a = \sqrt{8} = 2\sqrt{2}$

• Step 2 (Determine quadrant signs using ASTC):

  • Angle $x$ is in Quadrant III ($\pi \le x \le \frac{3\pi}{2}$).

  • In Quadrant III, sine is negative and tangent is positive.

• Step 3 (State final exact values):     $\sin(x) = -\frac{2\sqrt{2}}{3}$     $\tan(x) = \frac{-2\sqrt{2}/3}{-1/3} = 2\sqrt{2}$

Complementary and Symmetry Properties

• Complementary Properties (First Quadrant):

  • $\left(x, \frac{\pi}{2} - x\right)$ form a complementary pair of angles.

  • $\sin(x) = \cos\left(\frac{\pi}{2} - x\right)$

  • $\cos(x) = \sin\left(\frac{\pi}{2} - x\right)$

• Numerical Examples:

  • $\sin\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right)$

  • $\sin(30^\circ) = \cos(60^\circ)$

• Combining Complementary and Symmetry Properties:

  • Example: Evaluate $\sin\left(\frac{5\pi}{6}\right)$     $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) \quad \text{[symmetry property]}$     $= \cos\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right) \quad \text{[complementary property]}$     $= -\cos\left(\frac{2\pi}{3}\right)$

• Extending Complementary Properties Across Quadrants:

  $\cos\left(\frac{\pi}{2} + x\right) = \cos\left(\pi - \left(\frac{\pi}{2} - x\right)\right) = -\cos\left(\frac{\pi}{2} - x\right) = -\sin(x)$

• Key Distinction Rule:

  • Complementary properties are measured relative to the vertical axis ($y$-axis).

  • Symmetry properties are always measured relative to the horizontal axis ($x$-axis).

• Worked Examples:

  • Example A: Simplify $\sin\left(\frac{3\pi}{2} + x\right)$.

  • The angle $\frac{3\pi}{2} + x$ is located in Quadrant IV.

  • Using vertical axis reference ($\frac{3\pi}{2}$), sine changes to cosine, and sine is negative in Quadrant IV:       $\sin\left(\frac{3\pi}{2} + x\right) = \sin\left(2\pi - \left(\frac{\pi}{2} - x\right)\right) = -\sin\left(\frac{\pi}{2} - x\right) = -\cos(x)$

  • Example B: Find a value for $\theta$ if $\cos\left(\frac{2\pi}{5}\right) = \sin(\theta)$.

  • The angle $\frac{2\pi}{5}$ exists in Quadrant I.

  • Apply complementary identity $\cos(x) = \sin\left(\frac{\pi}{2} - x\right)$:       $\cos\left(\frac{2\pi}{5}\right) = \sin\left(\frac{\pi}{2} - \frac{2\pi}{5}\right) = \sin\left(\frac{5\pi - 4\pi}{10}\right) = \sin\left(\frac{\pi}{10}\right)$

  • Equating $\sin\left(\frac{\pi}{10}\right) = \sin(\theta)$ yields:       $\theta = \frac{\pi}{10}$

• Additional Coursework Practice:

  • Problem A: Give a value for $\theta$ if $\sin\left(\frac{\pi}{12}\right) = \cos(\theta)$.

  • Apply complementary property:       $\sin\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \cos\left(\frac{5\pi}{12}\right)$

  • Therefore, $\theta = \frac{5\pi}{12}$.

  • Problem B: Simplify $\cos\left(\frac{3\pi}{2} - x\right)$.

  • The angle $\frac{3\pi}{2} - x$ is in Quadrant III (where cosine is negative).

  • Measuring from the vertical axis ($\frac{3\pi}{2}$) changes cosine to sine:       $\cos\left(\frac{3\pi}{2} - x\right) = -\sin(x)$

Coursework & Independent Practice

• Textbook Reference: Jacaranda Maths Quest Mathematical Methods 11 VCE Units 1 and 2 (Third Edition).

• Topic: Exercise 9.6 – Trigonometric Relationships.

• Assigned Tasks:

  • Tech-Free Questions: 1–10, 12, 15

  • Tech-Active Questions: 16, 18, 20

  • Homework: Exam Questions located at the end of the exercise.