IB Syllabus Alignment

Standard Level Topic 3.6: The Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$; the quotient identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$. Double angle identities for sine and cosine: $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$. Simplifying trigonometric expressions, proving identities from LHS to RHS, and evaluating exact trigonometric values without a calculator.

1. The Fundamental Pythagorean & Quotient Identities

An identity is an equation that is true for all values of the variable in its domain. Unlike conditional equations (which are true only for specific solutions), identities can be substituted into expressions at any stage of algebraic working.

  • The Fundamental Pythagorean Identity: $$\cos^2\theta + \sin^2\theta = 1$$ Rearranged variations: $$\sin^2\theta = 1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$$ $$\cos^2\theta = 1 - \sin^2\theta = (1 - \sin\theta)(1 + \sin\theta)$$
  • The Quotient Identity: $$\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad (\cos\theta \neq 0)$$
  • Square of a Binomial Identity: Expanding $(\sin\theta \pm \cos\theta)^2$: $$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta = 1 + \sin(2\theta)$$ $$(\sin\theta - \cos\theta)^2 = 1 - \sin(2\theta)$$
Worked Example 1.1 — Simplifying Rational Trigonometric Expressions

Simplify the expression $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}$ for all values of $x$ where the expression is defined.

  1. Step 1: Find a common denominator: Common denominator is $\sin x (1 + \cos x)$: $$\frac{\sin^2 x + (1 + \cos x)^2}{\sin x (1 + \cos x)}$$
  2. Step 2: Expand the numerator: $$(1 + \cos x)^2 = 1 + 2\cos x + \cos^2 x$$ $$\text{Numerator} = \sin^2 x + 1 + 2\cos x + \cos^2 x$$
  3. Step 3: Apply the Pythagorean identity: Group $\sin^2 x + \cos^2 x = 1$: $$\text{Numerator} = (\sin^2 x + \cos^2 x) + 1 + 2\cos x = 1 + 1 + 2\cos x = 2 + 2\cos x = 2(1 + \cos x)$$
  4. Step 4: Cancel common factor: $$\frac{2(1 + \cos x)}{\sin x (1 + \cos x)} = \frac{2}{\sin x}$$
Answer: $\frac{2}{\sin x}$ (or $2\csc x$ where defined).

2. The Double Angle Identities

The double angle identities relate trigonometric functions of $2\theta$ to functions of $\theta$. In Paper 1, they are indispensable for solving equations, simplifying expressions, and finding exact non-calculator values.

  • Sine Double Angle: $$\sin(2\theta) = 2\sin\theta\cos\theta$$ Key Application: Whenever you see the product $\sin\theta\cos\theta$, rewrite it as $\frac{1}{2}\sin(2\theta)$.
  • Cosine Double Angle (Three Equivalent Forms): $$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$ $$\cos(2\theta) = 2\cos^2\theta - 1$$ $$\cos(2\theta) = 1 - 2\sin^2\theta$$
  • Strategic Choice of Cosine Form:
    • To eliminate $+1$: Use $\cos(2\theta) = 2\cos^2\theta - 1$, so $1 + \cos(2\theta) = 2\cos^2\theta$.
    • To eliminate $-1$: Use $\cos(2\theta) = 1 - 2\sin^2\theta$, so $1 - \cos(2\theta) = 2\sin^2\theta$.
Worked Example 1.2 — Exact Evaluation Using Double Angles

Given that $\cos\theta = -\frac{3}{5}$ and $\frac{\pi}{2} < \theta < \pi$, find the exact values of:
(a) $\sin\theta$
(b) $\sin(2\theta)$
(c) $\cos(2\theta)$

  1. Step 1: Find $\sin\theta$: Since $\theta$ is in Quadrant II, $\sin\theta > 0$: $$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
  2. Step 2: Evaluate $\sin(2\theta)$: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25}$$
  3. Step 3: Evaluate $\cos(2\theta)$: $$\cos(2\theta) = 2\cos^2\theta - 1 = 2\left(-\frac{3}{5}\right)^2 - 1 = 2\left(\frac{9}{25}\right) - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25}$$ Check with alternative form: $\cos^2\theta - \sin^2\theta = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}$. Perfect agreement!
Answers: (a) $\sin\theta = \frac{4}{5}$; (b) $\sin(2\theta) = -\frac{24}{25}$; (c) $\cos(2\theta) = -\frac{7}{25}$.

Visual Proof of the Double Angle Identity

The diagram below provides a geometric intuition for the relationship between angle $\theta$ and double angle $2\theta$ inside the unit circle.

x y O cos θ sin θ r = 1 sin(2θ) = 2 sin θ cos θ cos(2θ) θ 2θ Pythagorean Identity cos²θ + sin²θ = 1 Double Angle Formulas cos(2θ) = 2cos²θ - 1 = 1 - 2sin²θ
Figure 3.5 — Geometric relationships on the unit circle: $\cos^2\theta + \sin^2\theta = 1$ in right triangle, and the double angle projections $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = \cos^2\theta - \sin^2\theta$.

3. Proving Trigonometric Identities (LHS to RHS)

In IB examination papers, identity proofs are marked strictly on logical progression. You must follow the standard formal proof protocol:

  1. Start with One Side Only: Choose the more complicated side (usually the Left-Hand Side, LHS) and write down: $$\text{LHS} = \dots$$
  2. Do NOT Move Terms Across the Equals Sign: An identity proof is not an equation to be solved! You cannot add, subtract, multiply, or divide both sides by an expression.
  3. Standard Strategic Moves:
    • Convert $\tan\theta$ into $\frac{\sin\theta}{\cos\theta}$.
    • Find common denominators when adding algebraic fractions.
    • Factorise quadratic forms (e.g., $1 - \sin^2\theta = (1 - \sin\theta)(1 + \sin\theta)$ or difference of squares $\cos^4\theta - \sin^4\theta$).
    • Replace double angles with single angle identities.
  4. Conclude Clearly: Conclude your proof with: $$\dots = \text{RHS} \quad \text{(Hence proved)}$$
Worked Example 1.3 — Formal Identity Proof

Prove the identity $\frac{1 - \cos(2\theta)}{\sin(2\theta)} = \tan\theta$ for all values of $\theta$ where $\sin(2\theta) \neq 0$.

  1. Step 1: State the Left-Hand Side: $$\text{LHS} = \frac{1 - \cos(2\theta)}{\sin(2\theta)}$$
  2. Step 2: Choose double angle substitutions: In the numerator, we want to cancel the $1$. Use $\cos(2\theta) = 1 - 2\sin^2\theta$: $$1 - \cos(2\theta) = 1 - (1 - 2\sin^2\theta) = 2\sin^2\theta$$ In the denominator, substitute $\sin(2\theta) = 2\sin\theta\cos\theta$: $$\text{LHS} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta}$$
  3. Step 3: Simplify the fraction: Cancel common factors of $2$ and $\sin\theta$: $$\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta$$
  4. Step 4: Conclude: $$\text{LHS} = \text{RHS}$$
Conclusion: Identity established algebraically through direct double angle substitution.

4. Identity Proofs & Exact Double Angles

Worked Example 1.4 — Rigorous LHS to RHS Identity Proof

Prove the identity $\frac{\sin(2x)}{1 + \cos(2x)} = \tan x$ for all $x$ where both sides are defined. [4 marks]

  1. Step 1: Start strictly from the Left-Hand Side (LHS): $$\text{LHS} = \frac{\sin(2x)}{1 + \cos(2x)}$$
  2. Step 2: Substitute double angle formulas: $$\sin(2x) = 2\sin x\cos x \quad [A1]$$

    Choose the form of $\cos(2x)$ that eliminates $+1$, namely $\cos(2x) = 2\cos^2 x - 1$ [M1]:

    $$\text{LHS} = \frac{2\sin x\cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x\cos x}{2\cos^2 x} \quad [A1]$$
  3. Step 3: Cancel common factors of $2$ and $\cos x$: $$\text{LHS} = \frac{\sin x}{\cos x} = \tan x = \text{RHS} \quad [AG]$$
Conclusion: Identity proved rigorously by manipulating LHS into RHS.
Worked Example 1.5 — Exact Value Calculation Using Double Angles

Given that $\tan\theta = -\frac{3}{4}$ and $\frac{\pi}{2} < \theta < \pi$, find the exact value of:
(a) $\sin\theta$ and $\cos\theta$ [3 marks]
(b) $\cos(2\theta)$ [2 marks]
(c) $\sin(2\theta)$ [2 marks]

  1. Part (a): Find $\sin\theta$ and $\cos\theta$:

    In Quadrant II ($\frac{\pi}{2} < \theta < \pi$), $\sin\theta > 0$ and $\cos\theta < 0$ [R1].

    Using reference right triangle with legs $3$ and $4$, hypotenuse is $\sqrt{3^2 + 4^2} = 5$:

    $$\sin\theta = \frac{3}{5}, \quad \cos\theta = -\frac{4}{5} \quad [A1][A1]$$
  2. Part (b): Evaluate $\cos(2\theta)$: $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = \left(-\frac{4}{5} \right)^2 - \left(\frac{3}{5} \right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \quad [M1][A1]$$
  3. Part (c): Evaluate $\sin(2\theta)$: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{3}{5} \right)\left(-\frac{4}{5} \right) = -\frac{24}{25} \quad [M1][A1]$$
Answers: (a) $\sin\theta = \frac{3}{5}, \cos\theta = -\frac{4}{5}$; (b) $\cos(2\theta) = \frac{7}{25}$; (c) $\sin(2\theta) = -\frac{24}{25}$.

10 Tiered Practice Exercises

Progress from procedural algebraic expansions in Tier 1 to multi-step identity proofs and exact value derivations in Tier 2, concluding with higher-order synthesis in Tier 3.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Pythagorean Quadrant Evaluation

Given that $\cos\theta = \frac{5}{13}$ and $\frac{3\pi}{2} < \theta < 2\pi$, find the exact values of:
(a) $\sin\theta$
(b) $\tan\theta$

View Full Solution

(a) Find $\sin\theta$:
In Quadrant IV, sine is negative: $$\sin\theta = -\sqrt{1 - \cos^2\theta} = -\sqrt{1 - \left(\frac{5}{13}\right)^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$$

(b) Find $\tan\theta$:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{5/13} = -\frac{12}{5}$$

Final Answers: (a) $-\frac{12}{13}$; (b) $-\frac{12}{5}$.

Exercise 2 Double Angle Direct Calculation

If $\sin x = \frac{3}{5}$ and $x$ is acute, calculate the exact values of:
(a) $\sin(2x)$
(b) $\cos(2x)$

View Full Solution

Since $x$ is acute, $\cos x = \sqrt{1 - (3/5)^2} = \frac{4}{5}$.

(a) $\sin(2x)$:
$$\sin(2x) = 2\sin x\cos x = 2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right) = \frac{24}{25}$$

(b) $\cos(2x)$:
$$\cos(2x) = 1 - 2\sin^2 x = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$$

Final Answers: (a) $\frac{24}{25}$; (b) $\frac{7}{25}$.

Exercise 3 Algebraic Simplification

Simplify the expression $\frac{\sin(2\theta)}{2 - 2\cos^2\theta}$ for all valid $\theta$.

View Full Solution

Step 1: Factor denominator:
$$2 - 2\cos^2\theta = 2(1 - \cos^2\theta) = 2\sin^2\theta$$

Step 2: Expand numerator:
$$\sin(2\theta) = 2\sin\theta\cos\theta$$

Step 3: Simplify:
$$\frac{2\sin\theta\cos\theta}{2\sin^2\theta} = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta}\text{ (or }\cot\theta\text{)}$$

Final Answer: $\frac{\cos\theta}{\sin\theta}$ (or $\frac{1}{\tan\theta}$).

Exercise 4 Binomial Expansion & Identity

Show that $(\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 = 2$ for all $\theta \in \mathbb{R}$.

View Full Solution

Step 1: Expand both squared terms:
$$(\cos\theta + \sin\theta)^2 = \cos^2\theta + 2\cos\theta\sin\theta + \sin^2\theta = 1 + \sin(2\theta)$$ $$(\cos\theta - \sin\theta)^2 = \cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta = 1 - \sin(2\theta)$$

Step 2: Add expressions:
$$(1 + \sin(2\theta)) + (1 - \sin(2\theta)) = 1 + 1 = 2$$

Final Answer: Result established ($1 + 1 = 2$).

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Identity Proof with Double Angle

Prove that $\frac{\sin(2x)}{1 + \cos(2x)} = \tan x$ for all $x$ where the expression is defined.

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Proof from LHS:
$$\text{LHS} = \frac{\sin(2x)}{1 + \cos(2x)}$$ Substitute $\sin(2x) = 2\sin x\cos x$ and $\cos(2x) = 2\cos^2 x - 1$: $$\text{LHS} = \frac{2\sin x\cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x\cos x}{2\cos^2 x}$$ Cancel $2\cos x$: $$\text{LHS} = \frac{\sin x}{\cos x} = \tan x = \text{RHS}$$

Final Answer: Identity verified.

Exercise 6 Rational Trigonometric Proof

Prove that $\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2\tan x$.

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Proof from LHS:
Combine over common denominator $(1 - \sin x)(1 + \sin x) = 1 - \sin^2 x$: $$\text{LHS} = \frac{\cos x(1 + \sin x) - \cos x(1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$$ $$\text{Numerator} = \cos x + \cos x\sin x - \cos x + \cos x\sin x = 2\sin x\cos x$$ $$\text{Denominator} = 1 - \sin^2 x = \cos^2 x$$ $$\text{LHS} = \frac{2\sin x\cos x}{\cos^2 x} = \frac{2\sin x}{\cos x} = 2\tan x = \text{RHS}$$

Final Answer: Identity verified.

Exercise 7 Exact Value via Double Angle Inversion

Given that $\cos(2\theta) = \frac{7}{25}$ and $0 < \theta < \frac{\pi}{2}$, find the exact values of $\cos\theta$ and $\sin\theta$.

View Full Solution

Step 1: Find $\cos\theta$:
$$\cos(2\theta) = 2\cos^2\theta - 1 \implies \frac{7}{25} = 2\cos^2\theta - 1$$ $$2\cos^2\theta = 1 + \frac{7}{25} = \frac{32}{25} \implies \cos^2\theta = \frac{16}{25}$$ Since $\theta$ is acute, $\cos\theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Step 2: Find $\sin\theta$:
$$\cos(2\theta) = 1 - 2\sin^2\theta \implies \frac{7}{25} = 1 - 2\sin^2\theta$$ $$2\sin^2\theta = 1 - \frac{7}{25} = \frac{18}{25} \implies \sin^2\theta = \frac{9}{25} \implies \sin\theta = \frac{3}{5}$$

Final Answers: $\cos\theta = \frac{4}{5}, \sin\theta = \frac{3}{5}$.

Exercise 8 Difference of Powers Proof

Show that $\cos^4 x - \sin^4 x = \cos(2x)$ for all $x \in \mathbb{R}$.

View Full Solution

Step 1: Factor difference of squares:
$$\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$$

Step 2: Apply identities:
$$\cos^2 x + \sin^2 x = 1$$ $$\cos^2 x - \sin^2 x = \cos(2x)$$ $$\text{LHS} = \cos(2x) \times 1 = \cos(2x) = \text{RHS}$$

Final Answer: Identity verified.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Exact Half-Angle Derivation

By using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$ with $\theta = \frac{\pi}{8}$ ($22.5^\circ$), show that $\sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ and find the exact value of $\cos\left(\frac{\pi}{8}\right)$.

View Full Solution

Step 1: Find $\sin(\pi/8)$:
Let $\theta = \frac{\pi}{8}$. Then $2\theta = \frac{\pi}{4}$: $$\cos\left(\frac{\pi}{4}\right) = 1 - 2\sin^2\left(\frac{\pi}{8}\right) \implies \frac{\sqrt{2}}{2} = 1 - 2\sin^2\left(\frac{\pi}{8}\right)$$ $$2\sin^2\left(\frac{\pi}{8}\right) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$$ $$\sin^2\left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4} \implies \sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$$

Step 2: Find $\cos(\pi/8)$:
Using $\cos(2\theta) = 2\cos^2\theta - 1$: $$\frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \implies 2\cos^2\left(\frac{\pi}{8}\right) = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$$ $$\cos^2\left(\frac{\pi}{8}\right) = \frac{2 + \sqrt{2}}{4} \implies \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

Final Answers: $\sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$, $\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$.

Exercise 10 Multi-Angle Identity & Domain Restrictions

Given the identity $\frac{\sin(3x)}{\sin x} - \frac{\cos(3x)}{\cos x} = 2$ for all $x \in \mathbb{R}$ where $\sin x \neq 0$ and $\cos x \neq 0$:
(a) Prove the identity using compound angle expansion on $3x = 2x + x$.
(b) State the exact set of excluded values in $[0, 2\pi]$ where the expression is undefined.

View Full Solution

(a) Proof from LHS:
Combine fractions: $$\text{LHS} = \frac{\sin(3x)\cos x - \cos(3x)\sin x}{\sin x\cos x}$$ By the sine difference identity $\sin(A - B) = \sin A\cos B - \cos A\sin B$: $$\text{Numerator} = \sin(3x - x) = \sin(2x)$$ Substitute double angle in numerator: $$\text{LHS} = \frac{\sin(2x)}{\sin x\cos x} = \frac{2\sin x\cos x}{\sin x\cos x} = 2 = \text{RHS}$$

(b) Excluded values in $[0, 2\pi]$:
Undefined when $\sin x = 0$ or $\cos x = 0$: $$\sin x = 0 \implies x = 0, \pi, 2\pi$$ $$\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$$ Excluded values: $\left\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\right\}$.

Final Answers: (a) Proof complete; (b) $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$.

Trigonometric Identities Reference Sheet

In the IB Formula Booklet (Topic 3.6)

The official IB Mathematics AA SL formula booklet lists the following fundamental identities:

Pythagorean Identity $$\cos^2\theta + \sin^2\theta = 1$$
Quotient Identity $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
Double Angle: Sine $$\sin(2\theta) = 2\sin\theta\cos\theta$$
Double Angle: Cosine $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$
Must Memorize & Identity Techniques
  • Constant Elimination Rules: $$1 + \cos(2\theta) = 2\cos^2\theta \qquad \text{and} \qquad 1 - \cos(2\theta) = 2\sin^2\theta$$
  • Binomial Square Shortcuts: $$(\sin\theta \pm \cos\theta)^2 = 1 \pm \sin(2\theta)$$
  • Difference of 4th Powers: $$\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) = \cos(2\theta)$$
  • Half-Product Substitution: $$\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$$
Critical IB Examination Pitfalls
  • Pitfall 1: Distributing Over Functions: Writing $\sin(2x) = 2\sin x$. This is a severe conceptual error. The correct identity is $\sin(2x) = 2\sin x\cos x$.
  • Pitfall 2: Treating Proofs as Equations: Moving terms across the equals sign during an identity verification question ("Show that LHS = RHS"). You must work strictly on one side until it matches the other.
  • Pitfall 3: Sign Error in Subtracting Cosine Double Angle: Writing $1 - \cos(2x) = 1 - 1 - 2\sin^2 x = -2\sin^2 x$. The minus sign distributes: $1 - (1 - 2\sin^2 x) = +2\sin^2 x$.
  • Pitfall 4: Dividing by Zero: Canceling $\sin x$ or $\cos x$ without checking if it can equal zero.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: Simplify $(\sin x + \cos x)^2 - \sin(2x)$.

Question 2: If $\sin\theta = \frac{3}{5}$ and $\theta$ is acute, what is the exact value of $\sin(2\theta)$?

Question 3: Which expression is identically equal to $\frac{1 - \cos(2\theta)}{2}$?

Question 4: What is the exact value of $\cos^2(15^\circ) - \sin^2(15^\circ)$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

Let $\cos\theta = \frac{1}{3}$, where $0 < \theta < \frac{\pi}{2}$.
(a) Find the exact value of $\sin\theta$. [2]
(b) Find the exact value of $\sin(2\theta)$. [2]
(c) Find the exact value of $\cos(2\theta)$. [2]

View Full Markscheme

(a) Exact value of $\sin\theta$: [2 marks]
$$\sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \quad (M1)$$ Since $\theta$ is in Quadrant I: $$\sin\theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \quad (A1)$$

(b) Exact value of $\sin(2\theta)$: [2 marks]
$$\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{2\sqrt{2}}{3}\right)\left(\frac{1}{3}\right) \quad (M1)$$ $$\sin(2\theta) = \frac{4\sqrt{2}}{9} \quad (A1)$$

(c) Exact value of $\cos(2\theta)$: [2 marks]
$$\cos(2\theta) = 2\cos^2\theta - 1 = 2\left(\frac{1}{3}\right)^2 - 1 = 2\left(\frac{1}{9}\right) - 1 \quad (M1)$$ $$\cos(2\theta) = \frac{2}{9} - \frac{9}{9} = -\frac{7}{9} \quad (A1)$$

Question 6 (Paper 2 Style) 8 Marks

(a) Prove the identity $\frac{1 - \cos(2x)}{1 + \cos(2x)} = \tan^2 x$ for all $x$ where defined. [3]
(b) Hence solve the equation $\frac{1 - \cos(2x)}{1 + \cos(2x)} = 3$ for $0 \le x \le 2\pi$. [3]
(c) Show that the curve $y = \frac{1 - \cos(2x)}{1 + \cos(2x)}$ has no real roots where $y < 0$. [2]

View Full Markscheme

(a) Proof of identity: [3 marks]
$$\text{Numerator: } 1 - \cos(2x) = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \quad (M1)$$ $$\text{Denominator: } 1 + \cos(2x) = 1 + (2\cos^2 x - 1) = 2\cos^2 x \quad (M1)$$ $$\text{LHS} = \frac{2\sin^2 x}{2\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \left(\frac{\sin x}{\cos x}\right)^2 = \tan^2 x = \text{RHS} \quad (A1)$$

(b) Solving $\tan^2 x = 3$: [3 marks]
$$\tan x = \pm\sqrt{3} \quad (M1)$$ Reference angle $\alpha = \frac{\pi}{3}$.
Quadrant I: $x = \frac{\pi}{3}$
Quadrant II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
Quadrant IV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ (A2)

(c) No negative values: [2 marks]
Since $\frac{1 - \cos(2x)}{1 + \cos(2x)} = \tan^2 x$ and $\tan^2 x \ge 0$ for all real $x$ in its domain, $y$ cannot be strictly negative ($y \not< 0$). (R2)