Question 3(i, ii, iii, iv & v) Exercise 8.3

Solutions of Question 3(i, ii, iii, iv & v) 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 $\dfrac{\cos (\alpha + \beta)}{\cos(\alpha - \beta)}=\dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta}$

Solution.

\begin{align*} RHS & = \dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta} \\ & = \dfrac{1- \frac{\sin\alpha}{\cos\alpha} \frac{\sin\beta}{\cos\beta}}{1+ \frac{\sin\alpha}{\cos\alpha} \frac{\sin\beta}{\cos\beta}} \\ & = \dfrac{\dfrac{\cos\alpha \cos\beta - \sin\alpha \sin\beta}{\cos\alpha \cos\beta}}{\dfrac{\cos\alpha \cos\beta + \sin\alpha \sin\beta}{\cos\alpha \cos\beta}} \\ & = \dfrac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)} \\ & = LHS \end{align*}

Prove the identity $\dfrac{6 \cos 8u \sin 2u}{\sin (-6u)}=-\dfrac{\sin 10 u}{\sin 6u}+3$

Solution.

\begin{align*} RHS & = \dfrac{6 \cos 8u \sin 2u}{\sin (-6u)} \\ & = \dfrac{6 \cos 8u \sin 2u}{-\sin 6u} \\ & = -\dfrac{6 \cos 8u \sin 2u}{\sin 6u} \\ & = -\dfrac{3(\sin(10u) - \sin(2u))}{\sin 6u} \\ & = -3\dfrac{\sin 10u}{\sin 6u} + 3\\ & = LHS \end{align*}

Prove the identity $4 \cos 4v \sin 3v=2(\sin 7v - \sin v)$

Solution.

\begin{align*} RHS & = 4 \cos 4v \sin 3v \\ & = 2 \cdot 2 \cos 4v \sin 3v \\ & = 2 \cdot \left( \sin(3v + 4v) - \sin(3v - 4v) \right) \\ & = 2 \cdot \left( \sin 7v - \sin(-v) \right) \\ & = 2(\sin 7v + \sin v) \\ & = 2(\sin 7v - \sin v) \quad \text{(since \sin(-v) = -\sin v)}\\ & = LHS \end{align*}

Prove the identity $\sin 3 \theta + \sin \theta = 4\cos^2 \theta \sin \theta$

Solution.

\begin{align*} LHS & = \sin 3 \theta + \sin \theta \\ & = 2 \sin \left( \frac{3\theta + \theta}{2} \right) \cos \left( \frac{3\theta - \theta}{2} \right) \\ & = 2 \sin \left( \frac{4\theta}{2} \right) \cos \left( \frac{2\theta}{2} \right) \\ & = 2 \sin 2\theta \cos \theta \\ & = 2 (2 \sin \theta \cos \theta) \sin \theta \\ & = 4 \cos^2 \theta \sin \theta\\ &=RHS \end{align*}

Prove the identity $\cos 3x + \cos x = 2 \cos x (\cos 2x)$.

Solution.

\begin{align*} LHS & = \cos 3x + \cos x \\ & = 2 \cos \left( \frac{3x + x}{2} \right) \cos \left( \frac{3x - x}{2} \right) \\ & = 2 \cos \left( \frac{4x}{2} \right) \cos \left( \frac{2x}{2} \right) \\ & = 2 \cos 2x \cos x \\ & = 2 \cos x (\cos 2x) \\ &=RHS \end{align*}

< Question 2 Question 3(vi, vii, viii, ix & x) >