Question 10, Review Exercise

Solutions of Question 10 of Review Exercise of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Prove that: $\sin (16 x)=16 \sin (x) \cos (x) \cos (2 x) \cos (4 x) \cos (8 x)$

Solution.

$$\begin{align*} RHS&=16 \sin (x) \cos (x) \cos (2 x) \cos (4 x) \cos (8 x) \\ &= 8(2 \sin (x) \cos (x) )\cos (2 x) \cos (4 x) \cos (8 x) \\ &= 8 \sin2 (x) \cos (2 x) \cos (4 x) \cos (8 x) \\ &= 4(2 \sin2 (x) \cos (2 x)) \cos (4 x) \cos (8 x) \\ &= 4 \sin4 (x) \cos (4 x) \cos (8 x) \\ &= 2(2 \sin4 (x) \cos (4 x)) \cos (8 x) \\ &=2 \sin8 (x) \cos (8 x) \\ &= \sin16 (x) \\ &= LHS \end{align*}$$

Prove that: $$\frac{1+\cos 2 \theta}{\sin 2 \theta-\cos \theta}=\frac{2 \cos \theta}{2 \sin \theta-1}.$$

Solution.

$$\begin{align*} LHS &=\frac{1+\cos 2 \theta}{\sin 2 \theta-\cos \theta}\\ &=\frac{1+\cos ^2 \theta-\sin ^2 \theta}{2\sin \theta\cos\theta-\cos \theta}\\ &=\frac{1-\sin ^2 \theta+\cos ^2 \theta}{\cos \theta(2\sin \theta-1)}\\ &=\frac{2\cos ^2 \theta}{\cos \theta(2\sin \theta-1)}\\ &=\frac{2\cos \theta}{2\sin \theta-1}\\ &= RHS \end{align*}$$

Prove that: $$\frac{\cos 3 \theta-\cos \theta}{\cos 3 \theta+\cos \theta}=\frac{2 \tan ^{2} \theta}{\tan ^{2} \theta-1}=\frac{-2 \tan ^{2} \theta}{\sec ^{2} \theta-2 \tan ^{2} \theta}$$

Solution. (statement is wrong) $$\begin{align*} LHS &= \frac{\cos 3 \theta - \cos \theta}{\cos 3 \theta + \cos \theta} \\ &= \frac{4 \cos^ 3 \theta -3\cos \theta-\cos \theta}{4 \cos^ 3 \theta -3\cos \theta+\cos \theta} \\ &= \frac{4 \cos^ 3 \theta -4\cos \theta}{4 \cos^ 3 \theta -2\cos \theta} \\ &= \frac{-4 \cos \theta(1- \cos^ 2 \theta)}{2 \cos \theta(2 \cos^2 \theta -1)} \\ &= \frac{2 \sin^ 2 \theta}{1-2 \cos^2 \theta } \\ &= \frac{2 \tan^2 \theta}{\sec^2 \theta - 2 } \\ &= RHS \end{align*}$$

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