Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression:
$\dfrac{\cos 2 \alpha}{\sin^2 2 \alpha \left(\cos^2 \alpha - \tan^2 \alpha\right)}$

$\dfrac{\cos 2 \alpha}{\sin^2 2 \alpha \left(\cos^2 \alpha - \tan^2 \alpha\right)}$

$= \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\left(2 \sin \alpha \cos \alpha\right)^2 \times \left(\dfrac{\cos^2 \alpha}{\sin^2 \alpha} - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}\right)}$

$= \dfrac{\left(\cos^2 \alpha - \sin^2 \alpha\right) \times \sin^2 \alpha \cos^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha \times \left(\cos^4 \alpha - \sin^4 \alpha\right)}$

$= \dfrac{\cos^2 \alpha - \sin^2 \alpha}{4 \times \left[\left(\cos^2 \alpha\right)^2 - \left(\sin^2 \alpha\right)^2\right]}$

$= \dfrac{\cos^2 \alpha - \sin^2 \alpha}{4 \times \left(\cos^2 \alpha + \sin^2 \alpha\right) \times \left(\cos^2 \alpha - \sin^2 \alpha\right)}$

$= \dfrac{1}{4 \times 1} = \dfrac{1}{4}$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\tan \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right) \left(\dfrac{1 - \sin \alpha}{\cos \alpha}\right)$

$\tan \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right) \left(\dfrac{1 - \sin \alpha}{\cos \alpha}\right)$

$= \left(\dfrac{\tan \dfrac{\pi}{4} + \tan \dfrac{\alpha}{2}}{1 - \tan \dfrac{\pi}{4} \times \tan \dfrac{\alpha}{2}}\right) \left(\dfrac{\sin^2 \dfrac{\alpha}{2} + \cos^2 \dfrac{\alpha}{2} - 2 \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}}{\cos^2 \dfrac{\alpha}{2} - \sin^2 \dfrac{\alpha}{2}}\right)$

$= \left(\dfrac{1 + \tan \dfrac{\alpha}{2}}{1 - \tan \dfrac{\alpha}{2}}\right) \times \dfrac{\left(\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2}\right)^2}{\left(\cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2}\right) \left(\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2}\right)}$

$= \left(\dfrac{1 + \dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}}{1 - \dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}}\right) \times \left(\dfrac{\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2}}\right)$

$= \left(\dfrac{\cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2}}\right) \times \left(\dfrac{\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2}}\right)$

$= 1$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\dfrac{2 \cos^2 \alpha - 1}{4 \tan \left(\dfrac{\pi}{4} - \alpha\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)}$

$\dfrac{2 \cos^2 \alpha - 1}{4 \tan \left(\dfrac{\pi}{4} - \alpha\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)}$

$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{\tan \dfrac{\pi}{4} - \tan \alpha}{1 + \tan \dfrac{\pi}{4} \tan \alpha}\right] \left[\sin \dfrac{\pi}{4} \cos \alpha + \cos \dfrac{\pi}{4} \sin \alpha\right]^2}$

$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{1 - \tan \alpha}{1 + \tan \alpha}\right] \left[\dfrac{1}{\sqrt{2}} \cos \alpha + \dfrac{1}{\sqrt{2}} \sin \alpha\right]^2}$

$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{1 - \dfrac{\sin \alpha}{\cos \alpha}}{1 + \dfrac{\sin \alpha}{\cos \alpha}}\right] \left[\dfrac{\cos^2 \alpha}{2} + \dfrac{\sin^2 \alpha}{2} + 2 \times \dfrac{1}{\sqrt{2}} \cos \alpha \times \dfrac{1}{\sqrt{2}} \sin \alpha\right]}$

$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right] \left[\dfrac{1}{2} + \sin \alpha \cos \alpha\right]}$

$= \dfrac{\cos 2 \alpha}{4 \times \dfrac{\left(\cos \alpha - \sin \alpha\right) \left(\cos \alpha + \sin \alpha\right)}{\left(\cos \alpha + \sin \alpha\right)^2} \times \left(\dfrac{1 + 2 \sin \alpha \cos \alpha}{2}\right)}$

$= \dfrac{\cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2 \times 2}{4 \times \left(\cos^2 \alpha - \sin^2 \alpha\right) \times \left(\sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha\right)}$

$= \dfrac{\cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2}{2 \times \cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2}$

$= \dfrac{1}{2}$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\cos 0 + \cos \dfrac{\pi}{7} + \cos \dfrac{2 \pi}{7} + \cos \dfrac{3 \pi}{7} + \cos \dfrac{4 \pi}{7} + \cos \dfrac{5 \pi}{7} + \cos \dfrac{6 \pi}{7}$

$\cos 0 + \cos \dfrac{\pi}{7} + \cos \dfrac{2 \pi}{7} + \cos \dfrac{3 \pi}{7} + \cos \dfrac{4 \pi}{7} + \cos \dfrac{5 \pi}{7} + \cos \dfrac{6 \pi}{7}$

$= 1 + \left(\cos \dfrac{6 \pi}{7} + \cos \dfrac{\pi}{7}\right) + \left(\cos \dfrac{5 \pi}{7} + \cos \dfrac{2 \pi}{7}\right) + \left(\cos \dfrac{4 \pi}{7} + \cos \dfrac{3 \pi}{7}\right)$

$= 1 + 2 \cos \left(\dfrac{\dfrac{6 \pi}{7} + \dfrac{\pi}{7}}{2}\right) \cos \left(\dfrac{\dfrac{6 \pi}{7} - \dfrac{\pi}{7}}{2}\right)$
$\hspace{1.5cm} + 2 \cos \left(\dfrac{\dfrac{5 \pi}{7} + \dfrac{2\pi}{7}}{2}\right) \cos \left(\dfrac{\dfrac{5 \pi}{7} - \dfrac{2\pi}{7}}{2}\right)$
$\hspace{2.5cm} + 2 \cos \left(\dfrac{\dfrac{4\pi}{7} + \dfrac{3\pi}{7}}{2}\right) \cos \left(\dfrac{\dfrac{4\pi}{7} - \dfrac{3\pi}{7}}{2}\right)$

$= 1 + 2 \cos \dfrac{\pi}{2} \cos \dfrac{5 \pi}{14} + 2 \cos \dfrac{\pi}{2} \cos \dfrac{3 \pi}{14} + 2 \cos \dfrac{\pi}{2} \cos \dfrac{\pi}{14}$

$= 1 + 2 \times 0 \times \cos \dfrac{5 \pi}{14} + 2 \times 0 \times \cos \dfrac{3\pi}{14} + 2 \times 0 \times \cos \dfrac{\pi}{14}$

$= 1$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\dfrac{\sin 3 \alpha + \sin 5 \alpha + \sin 7 \alpha}{\cos 3 \alpha + \cos 5 \alpha + \cos 7 \alpha}$

$\dfrac{\sin 3 \alpha + \sin 5 \alpha + \sin 7 \alpha}{\cos 3 \alpha + \cos 5 \alpha + \cos 7 \alpha}$

$= \dfrac{\left(\sin 3 \alpha + \sin 7 \alpha\right) + \sin 5 \alpha}{\left(\cos 3 \alpha + \cos 7 \alpha\right) + \cos 5 \alpha}$

$= \dfrac{2 \sin \left(\dfrac{7 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{7 \alpha - 3 \alpha}{2}\right) + \sin 5 \alpha}{2 \cos \left(\dfrac{7 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{7 \alpha - 3 \alpha}{2}\right) + \cos 5 \alpha}$

$= \dfrac{2 \sin 5 \alpha \cos 2 \alpha + \sin 5 \alpha}{2 \cos 5 \alpha \cos 2 \alpha + \cos 5 \alpha}$

$= \dfrac{\sin 5 \alpha \left(2 \cos 2 \alpha + 1\right)}{\cos 5 \alpha \left(2 \cos 2 \alpha + 1\right)}$

$= \dfrac{\sin 5 \alpha}{\cos 5 \alpha}$

$= \tan 5 \alpha$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\dfrac{1 - \cos 4 \alpha}{\sec^2 2 \alpha - 1} + \dfrac{1 + \cos 4 \alpha}{\text{cosec}^2 2 \alpha - 1}$

$\dfrac{1 - \cos 4 \alpha}{\sec^2 2 \alpha - 1} + \dfrac{1 + \cos 4 \alpha}{\text{cosec}^2 2 \alpha - 1}$

$= \dfrac{2 \sin^2 2 \alpha}{\dfrac{1}{\cos^2 2 \alpha} - 1} + \dfrac{2 \cos^2 2 \alpha}{\dfrac{1}{\sin^2 2 \alpha} - 1}$

$\left[\because \; 1 - \cos 2 \theta = 2 \sin^2 \theta \implies 1 - \cos 4 \theta = 2 \sin^2 2 \theta\right] \;\;\;$ and

$\left[1 + \cos 2 \theta = 2 \cos^2 \theta \implies 1 + \cos 4 \theta = 2 \cos^2 2 \theta\right]$

$= \dfrac{2 \sin^2 2 \alpha \cos^2 2 \alpha}{1 - \cos^2 2 \alpha} + \dfrac{2 \cos^2 2 \alpha \sin^2 2 \alpha}{1 - \sin^2 2 \alpha}$

$= \dfrac{2 \sin^2 2 \alpha \cos^2 2 \alpha}{\sin^2 2 \alpha} + \dfrac{2 \cos^2 2 \alpha \sin^2 2 \alpha}{\cos^2 2 \alpha}$

$= 2 \sin^2 2 \alpha \cos^2 2 \alpha \left(\dfrac{1}{\sin^2 2 \alpha} + \dfrac{1}{\cos^2 2 \alpha}\right)$

$= 2 \sin^2 2 \alpha \cos^2 2 \alpha \left(\dfrac{\cos^2 2 \alpha + \sin^2 2 \alpha}{\sin^2 2 \alpha \cos^2 2 \alpha}\right)$

$= 2 \times 1 = 2$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\dfrac{\tan \alpha + \sin \alpha}{2 \cos^2 \left(\dfrac{\alpha}{2}\right)}$

$\dfrac{\tan \alpha + \sin \alpha}{2 \cos^2 \left(\dfrac{\alpha}{2}\right)}$

$= \dfrac{\tan \alpha + \sin \alpha}{1 + \cos \alpha}$

$= \dfrac{\dfrac{\sin \alpha}{\cos \alpha} + \sin \alpha}{1 + \cos \alpha}$

$= \dfrac{\sin \alpha \left(1 + \cos \alpha\right)}{\cos \alpha \left(1 + \cos \alpha\right)}$

$= \dfrac{\sin \alpha}{\cos \alpha}$

$= \tan \alpha$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $3 \cos^2 x + 4 \sin x \cos x - \sin^2 x - 1$

$3 \cos^2 x + 4 \sin x \cos x - \sin^2 x - 1$

$= 2 \cos^2 x + 2 \times 2 \sin x \cos x - \sin^2 x - \left(1 - \cos^2 x\right)$

$= 2 \cos^2 x + 2 \sin 2 x - \sin^2 x - \sin^2 x$

$= 2 \cos^2 x + 2 \sin 2 x - 2 \sin^2 x$

$= 2 \left(\cos^2 x - \sin^2 x\right) + 2 \sin 2 x$

$= 2 \left(\cos 2 x + \sin 2 x\right)$

$= 2 \sqrt{2} \left(\cos 2 x \times \dfrac{1}{\sqrt{2}} + \sin 2 x \times \dfrac{1}{\sqrt{2}}\right)$

$= 2 \sqrt{2} \left(\sin 2 x \times \cos \dfrac{\pi}{4} + \cos 2 x \times \sin \dfrac{\pi}{4}\right)$

$= 2 \sqrt{2} \sin \left(2x + \dfrac{\pi}{4}\right)$

Trigonometry - Simplification of Trigonometric Expressions

Simplify the expression: $\dfrac{2 \left(\sin 2 \alpha + 2 \cos^2 \alpha - 1\right)}{\cos \alpha - \sin \alpha - \cos 3 \alpha + \sin 3 \alpha}$

$\dfrac{2 \left(\sin 2 \alpha + 2 \cos^2 \alpha - 1\right)}{\cos \alpha - \sin \alpha - \cos 3 \alpha + \sin 3 \alpha}$

$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{\left(\cos \alpha - \cos 3 \alpha \right) + \left(\sin 3 \alpha - \sin \alpha\right)}$

$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{2 \sin \left(\dfrac{3 \alpha - \alpha}{2}\right) \sin \left(\dfrac{\alpha + 3 \alpha}{2}\right) + 2 \sin \left(\dfrac{3 \alpha - \alpha}{2}\right) \cos \left(\dfrac{3 \alpha + \alpha}{2}\right)}$

$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{2 \sin \alpha \sin 2 \alpha + 2 \sin \alpha \cos 2 \alpha}$

$= \dfrac{\sin 2 \alpha + \cos 2 \alpha}{\sin \alpha \left(\sin 2 \alpha + \cos 2 \alpha\right)}$

$= \dfrac{1}{\sin \alpha}$

$= \text{cosec } \alpha$

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) = 2 \cos 2 \alpha$

$\begin{aligned} LHS & = \dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) \\\\ & = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + 2 \cos \left(\dfrac{270^\circ + 2 \alpha}{2}\right) \cos \left(\dfrac{270^\circ - 2 \alpha}{2}\right) \\\\ & = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + \left[\cos \left(270^\circ\right) + \cos 2 \alpha\right] \\\\ & = \dfrac{\dfrac{1}{\tan \alpha} - \tan \alpha}{\dfrac{1}{\tan \alpha} + \tan \alpha} + 0 + \cos 2 \alpha \\\\ & = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} + \cos 2 \alpha \\\\ & = \dfrac{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} + \cos 2 \alpha \\\\ & = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha + \sin^2 \alpha} + \cos 2 \alpha \\\\ & = \dfrac{\cos 2 \alpha}{1} + \cos 2 \alpha \\\\ & = 2 \cos 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 = 0$

$\begin{aligned} LHS & = 2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 \\\\ & = 2 \left[\left(\sin^2 \alpha\right)^3 + \left(\cos^2 \alpha\right)^3\right] - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = 2 \left[\left(\sin^2 \alpha + \cos^2 \alpha\right) \left(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right)\right] \\ & \hspace{6cm} - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = 2 \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha + 2 \cos^4 \alpha - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = - \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha - \cos^4 \alpha + 1 \\\\ & = - \left[\sin^4 \alpha + 2 \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right] + 1 \\\\ & = - \left[\sin^2 \alpha + \cos^2 \alpha\right]^2 + 1 \\\\ & = - 1 + 1 \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cos^6 \alpha - \sin^6 \alpha = \dfrac{\left(3 + \cos^2 2 \alpha\right) \cos 2 \alpha}{4}$

$\begin{aligned} LHS & = \cos^6 \alpha - \sin^6 \alpha \\\\ & = \left(\cos^2 \alpha\right)^3 - \left(\sin^2 \alpha\right)^3 \\\\ & = \left[\cos^2 \alpha - \sin^2 \alpha\right] \left[\cos^4 \alpha + \cos^2 \alpha \sin^2 \alpha + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha \left(\cos^2 \alpha + \sin^2 \alpha\right) + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \left(\sin^2 \alpha\right)^2 \right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \left(\dfrac{1 - \cos 2 \alpha}{2}\right)^2\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \dfrac{1 - 2 \cos 2 \alpha + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{4 \cos^2 \alpha + 1 - 2 \left(2 \cos^2 \alpha - 1\right) + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{4 \cos^2 \alpha + 1 - 4 \cos^2 \alpha + 2 + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{3 + \cos^2 2 \alpha}{4}\right] = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} = {\tan^4 \alpha}$

$\begin{aligned} LHS & = \dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} \\\\ & = \dfrac{3 - 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha}{3 + 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha} \\\\ & = \dfrac{4 - 4 \cos 2 \alpha - 2 \sin^2 2 \alpha}{4 + 4 \cos 2 \alpha - 2 \sin^2 2 \alpha} \\\\ & = \dfrac{2 \left(1 - \cos 2 \alpha\right) - \sin^2 2 \alpha}{2 \left(1 + \cos 2 \alpha \right) - \sin^2 2 \alpha} \\\\ & = \dfrac{2 \times 2 \sin^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2}{2 \times 2 \cos^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2} \\\\ & = \dfrac{4 \sin^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha}{4 \cos^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^2 \alpha \left(1 - \cos^2 \alpha\right)}{\cos^2 \alpha \left(1 - \sin^2 \alpha\right)} \\\\ & = \dfrac{\sin^2 \alpha \times \sin^2 \alpha}{\cos^2 \alpha \times \cos^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^4 \alpha} \\\\ & = \tan^4 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right)$

$\begin{aligned} LHS & = \cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha \\\\ & = \left(\cos 5 \alpha + \cos 2 \alpha\right) - \left(\cos 4 \alpha + \cos 3 \alpha\right) \\\\ & = 2 \cos \left(\dfrac{5 \alpha + 2 \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - 2 \alpha}{2}\right) - 2 \cos \left(\dfrac{4 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{4 \alpha - 3 \alpha}{2}\right) \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{3 \alpha}{2}\right) - 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[\cos \left(\dfrac{3 \alpha}{2}\right) - \cos \left(\dfrac{\alpha}{2}\right)\right] \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[- 2 \sin \left(\dfrac{\dfrac{3 \alpha}{2} - \dfrac{\alpha}{2}}{2}\right) \sin \left(\dfrac{\dfrac{3 \alpha}{2} + \dfrac{\alpha}{2}}{2}\right) \right] \\\\ & = - 4 \cos \left(\dfrac{7 \alpha}{2}\right) \sin \left(\dfrac{\alpha}{2}\right) \sin \left(\dfrac{2 \alpha}{2}\right) \\\\ & = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right) = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin x + \cos x}{\sin x - \cos x} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} = \dfrac{2}{\tan x + 1}$

$\begin{aligned} LHS & = \dfrac{\sin x + \cos x}{\sin x - \cos x} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\dfrac{\sin x}{\cos x} + 1}{\dfrac{\sin x}{\cos x} - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\tan x + 1}{\tan x - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\left(\tan x + 1\right)^2}{\left(\tan x - 1\right) \left(\tan x + 1\right)} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\tan^2 x + 1 + 2 \tan x}{\tan^2 x - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\sec^2 x + 2 \tan x - sec^2 x - 2}{\tan^2 x - 1} \\\\ & = \dfrac{2 \left(\tan x - 1\right)}{\left(\tan x + 1\right) \left(\tan x - 1\right)} \\\\ & = \dfrac{2}{\tan x + 1} = RHS \end{aligned}$

Hence proved.