CIE A Level Mathematics 9709 Trigonometry Comprehensive Trigonometry Study Guide

Core Trigonometric Identities and Elementary Simplifications

• The Pythagorean Identity and its Variants

  • The fundamental identity used across multiple problems is $\sin^2(x) + \cos^2(x) = 1$.

  • This identity allows for the conversion of terms within quadratic equations. For example, in paper 9709/11/M/J/16, the equation $3\sin^2(\theta) = 4\cos(\theta) - 1$ is solved by substituting $\sin^2(\theta) = 1 - \cos^2(\theta)$ to form the quadratic: $3(1 - \cos^2(\theta)) = 4\cos(\theta) - 1$, which simplifies to $3\cos^2(\theta) + 4\cos(\theta) - 4 = 0$.

• Tangent-Related Identities

  • Direct substitution of $\tan(x) = \frac{\sin(x)}{\cos(x)}$ is a standard first step.

  • Case Study (9709/1/M/J/02): To show $\sin(x)\tan(x) = \frac{1 - \cos^2(x)}{\cos(x)}$, substitute to get $\sin(x) \times \frac{\sin(x)}{\cos(x)}$. This leads to $\frac{\sin^2(x)}{\cos(x)}$, and using $\sin^2(x) = 1 - \cos^2(x)$, the result is verified.

  • Identity Proof (9709/01/M/J/09): $\frac{\sin(x)}{1 - \sin(x)} - \frac{\sin(x)}{1 + \sin(x)} \equiv 2\tan^2(x)$.

    • The process involves finding a common denominator: $\frac{\sin(x)(1 + \sin(x)) - \sin(x)(1 - \sin(x))}{(1 - \sin(x))(1 + \sin(x))} = \frac{\sin(x) + \sin^2(x) - \sin(x) + \sin^2(x)}{1 - \sin^2(x)}$.

    • This simplifies to $\frac{2\sin^2(x)}{\cos^2(x)}$, which is equivalent to $2\tan^2(x)$.

Advanced Identity Proofs

• Proof 1: Sum and Difference of Cubes Form (9709/11/M/J/18)

  • Identity: $(\sin(\theta) + \cos(\theta))(1 - \sin(\theta)\cos(\theta)) \equiv \sin^3(\theta) + \cos^3(\theta)$.

  • Expansion of Left-Hand Side (LHS):

    • $\sin(\theta) - \sin^2(\theta)\cos(\theta) + \cos(\theta) - \sin(\theta)\cos^2(\theta)$.

  • Factoring and Using Pythagorean Substitution:

    • $\sin(\theta)(1 - \cos^2(\theta)) + \cos(\theta)(1 - \sin^2(\theta))$

    • $\sin(\theta)\sin^2(\theta) + \cos(\theta)\cos^2(\theta) = \sin^3(\theta) + \cos^3(\theta)$.

• Proof 2: Tangent and Cosine Relationship (9709/13/M/J/12)

  • Identity: $\tan^2(\theta) - \sin^2(\theta) \equiv \tan^2(\theta)\sin^2(\theta)$.

  • LHS Manipulation: $\frac{\sin^2(\theta)}{\cos^2(\theta)} - \sin^2(\theta) = \frac{\sin^2(\theta) - \sin^2(\theta)\cos^2(\theta)}{\cos^2(\theta)}$.

  • Factoring: $\frac{\sin^2(\theta)(1 - \cos^2(\theta))}{\cos^2(\theta)}$.

  • Result: Since $(1 - \cos^2(\theta) = \sin^2(\theta))$, the expression becomes $\frac{\sin^2(\theta)\sin^2(\theta)}{\cos^2(\theta)} = \tan^2(\theta)\sin^2(\theta)$.

• Proof 3: Secant/Cosine Squared (9709/13/M/J/18)

  • Identity: $\frac{\tan^2(\theta) - 1}{\tan^2(\theta) + 1}$.

  • Substitution for Secant: $\frac{\tan^2(\theta) - 1}{\sec^2(\theta)} = (\tan^2(\theta) - 1)\cos^2(\theta)$.

  • Final form: $(\frac{\sin^2(\theta)}{\cos^2(\theta)} - 1)\cos^2(\theta) = \sin^2(\theta) - \cos^2(\theta)$, which can be further written as $2\sin^2(\theta) - 1$.

Solving Trigonometric Equations

• Linear and Single-Dimentional Equations

  • Solving $\sin(3x) + 2\cos(3x) = 0$ for $0^{\circ} \le x \le 180^{\circ}$ (9709/01/M/J/03):

    • Rewrite as $\sin(3x) = -2\cos(3x)$, then $\tan(3x) = -2$.

    • $3x = \tan^{-1}(-2) = 116.6^{\circ}$, then add multiples of $180^{\circ}$ for other solutions: $296.6^{\circ}$, $476.6^{\circ}$.

    • Dividing by $3$ gives $x = 38.9^{\circ}, 98.9^{\circ}, 158.9^{\circ}$.

• Quadratic Forms in Cosine or Sine

  • Equation $2\tan^2(\theta)\cos(\theta) = 3$ (9709/01/M/J/08):

    • Rewrite as $2\frac{\sin^2(\theta)}{\cos^2(\theta)}\cos(\theta) = 3 \rightarrow \frac{2\sin^2(\theta)}{\cos(\theta)} = 3$.

    • $2(1 - \cos^2(\theta)) = 3\cos(\theta) \rightarrow 2\cos^2(\theta) + 3\cos(\theta) - 2 = 0$.

    • Factoring the quadratic: $(2\cos(\theta) - 1)(\cos(\theta) + 2) = 0$.

    • Solutions: $\cos(\theta) = 0.5$ (since $\cos(\theta) = -2$ is impossible).

    • $\theta = 60^{\circ}, 300^{\circ}$.

• Multiple-Domain Solutions

  • Equation $\sin(2x) + 3\cos(2x) = 0$ for $0^{\circ} \le x \le 360^{\circ}$ (9709/13/M/J/12):

    • $\tan(2x) = -3$.

    • $2x = 108.4^{\circ}, 288.4^{\circ}, 468.4^{\circ}, 648.4^{\circ}$.

    • $x = 54.2^{\circ}, 144.2^{\circ}, 234.2^{\circ}, 324.2^{\circ}$.

    • Crucial Observation: For the range up to $1080^{\circ}$, there are $12$ solutions because the frequency doubles.

Geometric Applications

• Triangle BC with AC Perpendicular (9709/01/M/J/06)

  • In triangle $ABC$: $AB = 4\,cm$, $BC = 6\,cm$, $\angle ABC = 150^{\circ}$. Line $CX$ is perpendicular to the line extension $ABX$.

  • Calculations for $BX$: In triangle $BCX$, the angle $\angle CBX = 180^{\circ} - 150^{\circ} = 30^{\circ}$.

    • $BX = 6\cos(30^{\circ}) = 6\left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3}$.

    • $CX = 6\sin(30^{\circ}) = 6(0.5) = 3$.

  • Calculations for $\tan(CAB)$: $\tan(\angle CAB) = \frac{CX}{AX} = \frac{CX}{AB + BX} = \frac{3}{4 + 3\sqrt{3}}$.

  • Length of $AC$: $AC = \sqrt{CX^2 + AX^2} = \sqrt{3^2 + (4 + 3\sqrt{3})^2} = \sqrt{52 + 24\sqrt{3}}$.

• Triangle ABC with Sine Rule (9709/01/M/J/08)

  • $AB = 12\,cm$, $\angle BAC = 60^{\circ}$, $\angle ACB = 45^{\circ}$.

  • Sine Rule application: $\frac{BC}{\sin(60^{\circ})} = \frac{12}{\sin(45^{\circ})}$.

  • Exact Length: $BC = 12 \times \frac{\sin(60^{\circ})}{\sin(45^{\circ})} = 12 \times \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = 6\sqrt{6}\,cm$.

Real-World Modelling: The Big Vertical Wheel

• Formula and Constant Values

  • The height model (9709/12/M/J/15): $h = 60(1 - \cos(kt))$

    • $h$: height in meters above ground.

    • $t$: time in minutes.

    • $k$: constant.

  • Maximum Height Analysis: Max height occurs when $\cos(kt) = -1$, giving $h = 60(1 - (-1)) = 120\,m$.

• Solving for Constant $k$

  • Condition: One complete revolution takes $30$ minutes.

  • One full period of a cosine function is $2\pi$ radians.

  • $k(30) = 2\pi \rightarrow k = \frac{2\pi}{30} = \frac{\pi}{15}$.

• Threshold Height Calculation

  • Task: Find time for which the passenger is above $90\,m$.

  • $90 = 60(1 - \cos(\frac{\pi}{15}t)) \rightarrow 1.5 = 1 - \cos(\frac{\pi}{15}t)$.

  • $\cos(\frac{\pi}{15}t) = -0.5$.

  • The solutions within one cycle are $\frac{\pi}{15}t = \frac{2\pi}{3}$ and $\frac{\pi}{15}t = \frac{4\pi}{3}$.

  • $t = 10$ minutes and $t = 20$ minutes.

  • Duration: $20 - 10 = 10$ total minutes.

Graphs and Periodic Function Analysis

• Sketching Curves (9709/01/M/J/03)

  • Function: $y = 3\sin(x)$ for $-\pi \le x \le \pi$.

  • Maximum Point: $(\frac{\pi}{2}, 3)$.

  • Linear Intersection: For line $y = kx$ passing through the maximum point:

    • $3 = k(\frac{\pi}{2}) \rightarrow k = \frac{6}{\pi}$.

  • Symmetry Property: The line and curve also intersect at the origin $(0, 0)$ and the point $(-\frac{\pi}{2}, -3)$.

• Graphical Intersection Analysis (9709/13/M/J/18)

  • Functions: $y = \sin(x)$ and $y = 2\cos(x)$ for $-\pi \le x \le \pi$.

  • At intersections: $\sin(x) = 2\cos(x) \rightarrow \tan(x) = 2$.

  • $x$-coordinate of point $A$ (Positive region): $x = 1.11$ radians.

  • $y$-coordinate of point $B$ (Negative region): Intersection occurs also at $x = 1.11 - \pi \approx -2.03$. Substituting into $y = \sin(x)$ gives $y \approx -0.894$.

Radian Measure and obtuse/reflex Angle Expressing

• Expressing Constants (9709/11/M/J/15)

  • Given $x$ is obtuse and $\sin(x) = k$:

    • $\cos(x) = -\sqrt{1 - k^2}$ (negative because $\cos$ is negative in the 2nd quadrant).

    • $\tan(x) = \frac{k}{-\sqrt{1 - k^2}}$.

    • $\sin(x + \pi) = -k$ (quadrant shift property).

• Reflex Angles (9709/12/M/J/14)

  • Given reflex angle $\theta$ such that $\cos(\theta) = k$ where 0 < k < 1:

    • $\sin(\theta) = -\sqrt{1 - k^2}$ (negative because reflex/4th quadrant).

    • $\tan(\theta) = -\frac{\sqrt{1 - k^2}}{k}$.

    • Explanation for $\sin(2\theta)$: Since $\theta$ is in the 4th quadrant (270^{\circ} < \theta < 360^{\circ}), then $2\theta$ lies between $540^{\circ}$ and $720^{\circ}$. In both these regions (3rd and 4th quadrants of the second cycle), $\sin$ is negative.