6.8.1 Trigonometric Functions Long Questions (Question 1 & 2)

Question 1:

(a) Sketch the graph of y = cos 2x for 0° ≤ x ≤ 180°.

(b) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation

$2 {sin}^{2} x = 2 - \frac{x}{180}$ for 0° ≤ x ≤ 180°.

Solution:
(a)(b)

$\begin{array}{l} 2 {sin}^{2} x = 2 - \frac{x}{180} \\ 1 - 2 {sin}^{2} x = 1 - (2 - \frac{x}{180}) \\ \cos 2 x = \frac{x}{180} - 1 \\ y = \frac{x}{180} - 1 \\ x = 0, y = - 1 \\ x = 180, y = 0 \\ Number of solutions = 2 \end{array}$

Question 2:

(a) Prove that

$\frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x .$

(b)(i) Sketch the graph of y = – tan 2x for 0 ≤ x ≤ π .

(b)(ii) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation

$\frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0$ for 0 ≤ x ≤ π .

State the number of solutions.

Solution:

(a)

$\begin{array}{l} \frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x \\ L H S : \\ \frac{2 \tan x}{2 - \sec^{2} x} = \frac{2 \tan x}{2 - (1 + \tan^{2} x)} \\ = \frac{2 \tan x}{2 - \tan^{2} x} \\ = \tan 2 x (RHS) \end{array}$

(b)(i)

(b)(ii)

$\begin{array}{l} \frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0 \\ \frac{3 x}{π} + \tan 2 x = 0 \leftarrow from part (a) \\ - \tan 2 x = \frac{3 x}{π} \\ y = \frac{3 x}{π} \\ The suitable straight line to sketch is y = \frac{3 x}{π} . \end{array}$

When x = 0, y = 0.

When x = π, y = 3.

Number of solutions = 3

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