Question 3(xi, xii & xiii) Exercise 8.3

Solutions of Question 3(xi, xii & xiii) of Exercise 8.3 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 the identity $2\cos2u \cos u-\sin 2u \sin u=2\cos^3 u$

Solution.

$$\begin{align*} LHS & = 2\cos 2u \cos u - \sin 2u \sin u \\ & = 2\left(\cos^2 u - \sin^2 u\right) \cos u - 2\sin u \cos u \sin u \\ & = 2\cos^3 u - 2\sin^2 u \cos u \\ & = 2\cos^3 u\\ &=RHS \end{align*}$$

Prove the identity $2 \sin 2y \sin 3y=\cos y-\cos 5y$

Solution.

$$\begin{align*} LHS & = 2 \sin 2y \sin 3y \\ & = \cos(3y - 2y) - \cos(3y + 2y) \\ & = \cos y - \cos 5y\\ &=RHS \end{align*}$$

Prove the identity $\dfrac{\cos 10x + \cos 6x}{\cos 6x - \cos 10x}=\cot 2x \cot 8x$

Solution.

$$\begin{align*} LHS & = \dfrac{\cos 10x + \cos 6x}{\cos 6x - \cos 10x} \\ & = \dfrac{2 \cos \left( \frac{10x + 6x}{2} \right) \cos \left( \frac{10x - 6x}{2} \right)}{-2 \sin \left( \frac{10x + 6x}{2} \right) \sin \left( \frac{10x - 6x}{2} \right)} \\ & = -\frac{\cos 8x \cos 2x}{\sin 8x \sin 2x} \\ & = \cot 2x \cot 8x\\ &=RHS \end{align*}$$

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