6.8.2 Trigonometric Functions Long Questions (Question 3 & 4)

Question 3:
(a) Sketch the graph of

y = \frac{3}{2} cos 2 x for 0 \leq x \leq \frac{3}{2} π .

$y = \frac{3}{2} \cos 2 x for 0 \leq x \leq \frac{3}{2} π .$
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation

\frac{4}{3 π} x - cos 2 x = \frac{3}{2} for 0 \leq x \leq \frac{3}{2} π

$\frac{4}{3 π} x - \cos 2 x = \frac{3}{2} for 0 \leq x \leq \frac{3}{2} π$
State the number of solutions.

Solution:
(a)(b)

\begin{matrix} \frac{4}{3 π} x - cos 2 x = \frac{3}{2} cos 2 x = \frac{4}{3 π} x - \frac{3}{2} \frac{3}{2} cos 2 x = \frac{3}{2} (\frac{4}{3 π} x - \frac{3}{2}) y = \frac{2}{π} x - \frac{9}{4} To sketch the graph of y = \frac{2}{π} x - \frac{9}{4} x = 0, y = - \frac{9}{4} x = \frac{3 π}{2}, y = \frac{3}{4} Number of solutions = Number of intersection points = 3 \end{matrix}

$\begin{array}{l} \frac{4}{3 π} x - \cos 2 x = \frac{3}{2} \\ \cos 2 x = \frac{4}{3 π} x - \frac{3}{2} \\ \frac{3}{2} \cos 2 x = \frac{3}{2} (\frac{4}{3 π} x - \frac{3}{2}) \\ y = \frac{2}{π} x - \frac{9}{4} \\ To sketch the graph of y = \frac{2}{π} x - \frac{9}{4} \\ x = 0, y = - \frac{9}{4} \\ x = \frac{3 π}{2}, y = \frac{3}{4} \\ Number of solutions \\ = Number of intersection points \\ = 3 \end{array}$

Question 4:

(a)

Prove that {(\frac{cos e c x - sec x}{sec x cos e c x})}^{2} = 1 - sin 2 x

$Prove that {(\frac{\cos e c x - \sec x}{\sec x \cos e c x})}^{2} = 1 - \sin 2 x$

[3 marks]

(b)(i) Sketch the graph of y = 1 – sin2x for 0 ≤ x ≤ 2π.

(b)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation

2 - {(\frac{cos e c x - sec x}{sec x cos e c x})}^{2} = \frac{x}{π}

$2 - {(\frac{\cos e c x - \sec x}{\sec x \cos e c x})}^{2} = \frac{x}{π}$ for 0 ≤ x ≤ 2π.

State the number of soultions.
[7 marks]

Solution:

(a)

\begin{matrix} L H S = {(\frac{cos e c x - sec x}{sec x cos e c x})}^{2} = {(\frac{cos e c x}{sec x cos e c x} - \frac{sec x}{sec x cos e c x})}^{2} = {(\frac{1}{sec x} - \frac{1}{cos e c x})}^{2} = {(cos x - sin x)}^{2} = {cos}^{2} x - 2 cos x sin x + {sin}^{2} x = 1 - sin 2 x (RHS) \end{matrix}

$\begin{array}{l} L H S \\ = {(\frac{\cos e c x - \sec x}{\sec x \cos e c x})}^{2} \\ = {(\frac{\cos e c x}{\sec x \cos e c x} - \frac{\sec x}{\sec x \cos e c x})}^{2} \\ = {(\frac{1}{\sec x} - \frac{1}{\cos e c x})}^{2} \\ = {(\cos x - \sin x)}^{2} \\ = \cos^{2} x - 2 \cos x \sin x + \sin^{2} x \\ = 1 - \sin 2 x (RHS) \end{array}$

(b)(i)

(b)(ii)

\begin{matrix} 2 - {(\frac{cos e c x - sec x}{sec x cos e c x})}^{2} = \frac{x}{π} 2 - (1 - sin 2 x) = \frac{x}{π} \leftarrow From 4(a) 2 - y = \frac{x}{π} \leftarrow From 4(b)(i) y = 2 - \frac{x}{π} (suitable straight line) \end{matrix}

$\begin{array}{l} 2 - {(\frac{\cos e c x - \sec x}{\sec x \cos e c x})}^{2} = \frac{x}{π} \\ 2 - (1 - \sin 2 x) = \frac{x}{π} \leftarrow From 4(a) \\ 2 - y = \frac{x}{π} \leftarrow From 4(b)(i) \\ y = 2 - \frac{x}{π} (suitable straight line) \end{array}$

x	0	π	2π
y	2	1	0

From the graph, there is 3 number of solutions.

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