Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha = 8 \cot 8 \alpha$

$\begin{aligned} LHS & = \cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha \\\\ & = \dfrac{\cos \alpha}{\sin \alpha} - \dfrac{\sin \alpha}{\cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{\cos 2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 2 \alpha}{2 \sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 2 \alpha}{\sin 2 \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \left(\cos^2 2 \alpha - \sin^2 2 \alpha\right)}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 4 \alpha}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \cos 4 \alpha}{2 \sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \cos 4 \alpha}{\sin 4 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \left(\cos^2 4 \alpha - \sin^2 4 \alpha\right)}{\sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{4 \cos 8 \alpha}{\sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{8 \cos 8 \alpha}{2 \sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{8 \cos 8 \alpha}{\sin 8 \alpha} \\\\ & = 8 \cot 8 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\left(\sin \alpha - \sin \beta\right) \left(\sin \alpha + \sin \beta\right) = \sin \left(\alpha - \beta\right) \sin \left(\alpha + \beta\right)$

$\begin{aligned} RHS & = \sin \left(\alpha - \beta\right) \sin \left(\alpha + \beta\right) \\\\ & = \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \\\\ & = \sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta \\\\ & = \sin^2 \alpha \left(1 - \sin^2 \beta\right) - \sin^2 \beta \left(1 - \sin^2 \alpha\right) \\\\ & = \sin^2 \alpha - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta + \sin^2 \alpha \sin^2 \beta \\\\ & = \sin^2 \alpha - \sin^2 \beta \\\\ & = \left(\sin \alpha - \sin \beta\right) \left(\sin \alpha + \sin \beta\right) = LHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\tan 3 \alpha}{\tan \alpha} = \dfrac{3 - \tan^2 \alpha}{1 - 3 \tan^2 \alpha}$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} = 2 \cot 2 \alpha$

$\begin{aligned} \cot \alpha + \tan \alpha & = \dfrac{\cos \alpha}{\sin \alpha} + \dfrac{\sin \alpha}{\cos \alpha} \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\sin \alpha \cos \alpha} \\\\ & = \dfrac{1}{\sin \alpha \cos \alpha} \\\\ & = \dfrac{2}{2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{2}{\sin 2 \alpha} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} 1 + \tan 2 \alpha \tan \alpha & = 1 + \left(\dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\right) \tan \alpha \\\\ & = \dfrac{1 - \tan^2 \alpha + 2 \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\ & = \dfrac{1 + \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\ & = \dfrac{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha} \\\\ & = \dfrac{1}{\cos^2 \alpha - \sin^2 \alpha} \\\\ & = \dfrac{1}{\cos 2 \alpha} \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} \\\\ & = \dfrac{\dfrac{2}{\sin 2 \alpha}}{\dfrac{1}{\cos 2 \alpha}} \;\;\; \left[\text{By equations (1) and (2)}\right] \\\\ & = 2 \times \dfrac{\cos 2 \alpha}{\sin 2 \alpha} \\\\ & = 2 \cot 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

$\begin{aligned} \tan \left(\alpha - \dfrac{\pi}{4}\right) & = \dfrac{\tan \alpha - \tan \left(\dfrac{\pi}{4}\right)}{1 + \tan \alpha \tan \left(\dfrac{\pi}{4}\right)} \\\\ & = \dfrac{\tan \alpha - 1}{1 + \tan \alpha} \\\\ & = \dfrac{\dfrac{\sin \alpha}{\cos \alpha} - 1}{\dfrac{\sin \alpha}{\cos \alpha} + 1} \\\\ & = \dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{\pi}{4} + \alpha\right) & = \sin \left(\dfrac{\pi}{4}\right) \cos \alpha + \cos \left(\dfrac{\pi}{4}\right) \sin \alpha \\\\ & = \dfrac{1}{\sqrt{2}} \times \cos \alpha + \dfrac{1}{\sqrt{2}} \times \sin \alpha \\\\ & = \dfrac{1}{\sqrt{2}} \left(\sin \alpha + \cos \alpha\right) \end{aligned}$

$\begin{aligned} \therefore \; \sin^2 \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right) ^2 \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{1 - 2 \cos^2 \alpha}{2 \tan \left(\alpha - \dfrac{\pi}{4}\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)} \\\\ & = \dfrac{- \left(2 \cos^2 \alpha - 1\right)}{2 \left(\dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}\right) \times \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right)^2} \\ & \hspace{2cm} \;\;\; \left[\text{by equations (1) and (2)}\right] \\\\ & = \dfrac{- \cos 2 \alpha}{- \left(\cos \alpha - \sin \alpha\right) \left(\sin \alpha + \cos \alpha\right)} \\\\ & = \dfrac{- \cos 2 \alpha}{- \left(\cos^2 \alpha - \sin^2 \alpha\right)} \\\\ & = \dfrac{- \cos 2 \alpha}{- \cos 2 \alpha} \\\\ & = 1 = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} = \sin^2 x$

$\begin{aligned} \tan \left(x - \dfrac{\pi}{2}\right) & = \tan \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\ & = - \tan \left(\dfrac{\pi}{2} - x\right) \\\\ & = - \cot x \\\\ & = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \cos \left(\dfrac{3 \pi}{2} + x\right) & = \cos \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\ & = -\cos \left(\dfrac{\pi}{2} + x\right) \\\\ & = - \left(- \sin x\right) \\\\ & = \sin x \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{7 \pi}{2} - x\right) & = \sin \left[\dfrac{\pi}{2} + \left(3 \pi - x\right)\right] \\\\ & = \cos \left(3 \pi - x\right) \\\\ & = - \cos x \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \cos \left(x - \dfrac{\pi}{2}\right) & = \cos \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\ & = \cos \left(\dfrac{\pi}{2} - x\right) \\\\ & = \sin x \;\;\; \cdots \; (4) \end{aligned}$

$\begin{aligned} \tan \left(\dfrac{3 \pi}{2} + x\right) & = \tan \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\ & = \tan \left(\dfrac{\pi}{2} + x\right) \\\\ & = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (5) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} \\\\ & = \dfrac{\dfrac{- \cos x}{\sin x} \times \sin x - \left(- \cos x\right)^3}{\sin x \times \left(\dfrac{- \cos x}{\sin x}\right)} \;\;\; \left[\text{In view of equations } (1) \cdots (5) \right] \\\\ & = \dfrac{- \cos x + \cos^3 x}{- \cos x} \\\\ & = 1 - \cos^2 x \\\\ & = \sin^2 x = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right)$

$\begin{aligned} \cot \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{\tan \left(\dfrac{\pi}{4} + \alpha\right)} \\\\ & = \dfrac{1 - \tan \left(\dfrac{\pi}{4}\right) \tan \alpha}{\tan \dfrac{\pi}{4} + \tan \alpha} \\\\ & = \dfrac{1 - \tan \alpha}{1 + \tan \alpha} \\\\ & = \dfrac{1 - \dfrac{\sin \alpha}{\cos \alpha}}{1 + \dfrac{\sin \alpha}{\cos \alpha}} \\\\ & = \dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha} \end{aligned}$

$\begin{aligned} \therefore \; RHS & = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right) \\\\ & = \left(\dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2 \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha}{\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = LHS \end{aligned}$

Hence proved.

Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$

To Prove That (TPT) $\;$ $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$

i.e. $\;$ TPT $\;$ $1 + \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = \dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha}$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin \left(\pi - \alpha\right)}{\sin \alpha - \cos \alpha \tan \dfrac{\alpha}{2}} + \cos \left(\pi - \alpha\right) = 1$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\tan 2x \tan x}{\tan 2x - \tan x} = \sin 2x$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

$\begin{aligned} LHS & = \dfrac{\left(\sin 2 \alpha - \sin 6 \alpha\right) + \left(\cos 2 \alpha - \cos 6 \alpha\right)}{\sin 4 \alpha - \cos 4 \alpha} \\\\ & = \dfrac{2 \sin \left(\dfrac{2 \alpha - 6 \alpha}{2}\right) \cos \left(\dfrac{2 \alpha + 6 \alpha}{2}\right) }{\sin 4 \alpha - \cos 4 \alpha} \\ & \hspace{2cm} + \dfrac{2 \sin \left(\dfrac{2 \alpha + 6 \alpha}{2}\right) \sin \left(\dfrac{6 \alpha - 2 \alpha}{2}\right)}{\sin 4 \alpha - \cos 4 \alpha} \\\\ & = \dfrac{2 \sin \left(-2 \alpha\right) \cos 4 \alpha + 2 \sin 4 \alpha \sin 2 \alpha}{\sin 4 \alpha - \cos 4 \alpha} \\\\ & = \dfrac{2 \sin 2 \alpha \sin 4 \alpha - 2 \sin 2 \alpha \cos 4 \alpha}{\sin 4 \alpha - \cos 4 \alpha} \\\\ & = \dfrac{2 \sin 2 \alpha \left(\sin 4 \alpha - \cos 4 \alpha\right)}{\sin 4 \alpha - \cos 4 \alpha} \\\\ & = 2 \sin 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) = \sin 2 \alpha$

$\begin{aligned} \sin^2 A - \sin^2 B & = \left(\sin A + \sin B\right) \left(\sin A - \sin B\right) \\\\ & = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \\ & \hspace{2cm} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \\\\ & = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \\ & \hspace{2cm} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \\\\ & = \sin \left(A + B\right) \times \sin \left(A - B\right) \end{aligned}$

$\begin{aligned} \therefore \; \sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) & = \sin \left(45^\circ + \alpha + 30^\circ - \alpha\right) \\ & \hspace{2cm} \sin \left(45^\circ + \alpha - 30^\circ + \alpha\right) \\\\ & = \sin 75^\circ \sin \left(15^\circ + 2 \alpha\right) \\\\ & = \sin \left(90^\circ - 15^\circ\right) \times \sin \left(15^\circ + 2 \alpha\right) \\\\ & = \cos 15^\circ \sin \left(15^\circ + 2 \alpha\right) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) \\\\ & = \cos 15^\circ \sin \left(15^\circ + 2 \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) \\\\ & = \sin \left(15^\circ + 2 \alpha - 15^\circ\right) \\\\ & = \sin 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha}{\cos \alpha + \cos 3 \alpha + \cos 5 \alpha} = \tan 3 \alpha$

$\begin{aligned} \sin \alpha + \sin 3 \alpha + \sin 5 \alpha & = \left(\sin \alpha + \sin 5 \alpha\right) + \sin 3 \alpha \\\\ & = 2 \sin \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) + \sin 3 \alpha \\\\ & = 2 \sin 3 \alpha \cos 2 \alpha + \sin 3 \alpha \\\\ & = \sin 3 \alpha \left(2 \cos 2 \alpha + 1\right) \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \cos \alpha + \cos 3 \alpha + \cos 5 \alpha & = \left(\cos \alpha + \cos 5 \alpha\right) + \cos 3 \alpha \\\\ & = 2 \cos \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) + \cos 3 \alpha \\\\ & = 2 \cos 3 \alpha \cos 2 \alpha + \cos 3 \alpha \\\\ & = \cos 3 \alpha \left(2 \cos 2 \alpha + 1\right) \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha}{\cos \alpha + \cos 3 \alpha + \cos 5 \alpha} \\\\ & = \dfrac{\sin 3 \alpha \left(2 \cos 2 \alpha + 1\right)}{\cos 3 \alpha \left(2 \cos 2 \alpha + 1\right)} \;\;\; \left[\text{in view of equations (1) and (2)}\right] \\\\ & = \dfrac{\sin 3 \alpha}{\cos 3 \alpha} \\\\ & = \tan 3 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\sin 2 \alpha + \sin 4 \alpha - \sin 6 \alpha = 4 \sin \alpha \sin 2 \alpha \sin 3 \alpha$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\sin \alpha + 2 \sin 3 \alpha + \sin 5 \alpha = 4 \sin 3 \alpha \cos^2 \alpha$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin^2 4 \alpha}{2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha} = 2 \sin \alpha \sin 2 \alpha$

$\begin{aligned} 2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha & = \left(\cos \alpha + \cos 3 \alpha\right) + \left(\cos \alpha + \cos 5 \alpha\right) \\\\ & = 2 \cos \left(\dfrac{3 \alpha + \alpha}{2}\right) \cos \left(\dfrac{3 \alpha - \alpha}{2}\right) \\ & \hspace{2cm} + 2 \cos \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) \\\\ & = 2 \cos 2 \alpha \cos \alpha + 2 \cos 3 \alpha \cos 2 \alpha \\\\ & = 2 \cos 2 \alpha \left(\cos 3 \alpha + \cos \alpha\right) \\\\ & = 2 \cos 2 \alpha \times 2 \cos 2 \alpha \cos \alpha \\\\ & = 4 \cos^2 2 \alpha \cos \alpha \end{aligned}$

$\begin{aligned} \sin^2 4 \alpha & = \left(2 \sin 2 \alpha \cos 2 \alpha\right)^2 \\\\ & = 4 \times \left(2 \sin \alpha \cos \alpha\right)^2 \times \cos^2 2 \alpha \\\\ & = 16 \sin^2 \alpha \cos^2 \alpha \cos^2 2 \alpha \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\sin^2 4 \alpha}{2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha} \\\\ & = \dfrac{16 \sin^2 \alpha \cos^2 \alpha \cos^2 2 \alpha}{4 \cos^2 2 \alpha \cos \alpha} \\\\ & = 4 \sin^2 \alpha \cos \alpha \\\\ & = 2 \sin \alpha \times 2 \sin \alpha \cos \alpha \\\\ & = 2 \sin \alpha \sin 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

$\begin{aligned} LHS & = \sin \alpha - \sin 2 \alpha + \sin 3 \alpha \\\\ & = 2 \sin \left(\dfrac{3 \alpha - 2 \alpha}{2}\right) \cos \left(\dfrac{3 \alpha + 2 \alpha}{2}\right) + \sin \alpha \\\\ & = 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{5 \alpha}{2}\right) + 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) \\\\ & = 2 \sin \left(\dfrac{\alpha}{2}\right) \left[\cos \left(\dfrac{5 \alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right] \\\\ & = 2 \sin \left(\dfrac{\alpha}{2}\right) \times 2 \cos \left(\dfrac{\dfrac{5 \alpha}{2} + \dfrac{\alpha}{2}}{2}\right) \cos \left(\dfrac{\dfrac{5 \alpha}{2} - \dfrac{\alpha}{2}}{2}\right) \\\\ & = 4 \cos \left(\dfrac{6 \alpha}{4}\right) \cos \left(\dfrac{4 \alpha}{4}\right) \sin \left(\dfrac{\alpha}{2}\right) \\\\ & = 4 \cos \left(\dfrac{3 \alpha}{2}\right) \cos \alpha \sin \left(\dfrac{\alpha}{2}\right) = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\left(1 + \tan \beta \tan 2 \beta\right) \sin 2 \beta = \tan 2 \beta$

To prove that (TPT): $\;\;\;$ $\left(1 + \tan \beta \tan 2 \beta\right) \sin 2 \beta = \tan 2 \beta$

i.e. $\;$ TPT $\;\;\;$ $1 + \tan \beta \tan 2 \beta = \dfrac{\tan 2 \beta}{\sin 2 \beta}$

i.e. $\;$ TPT $\;\;\;$ $1 + \tan \beta \tan 2 \beta = \dfrac{\sin 2 \beta}{\cos 2 \beta} \times \dfrac{1}{\sin 2\beta}$

i.e. $\;$ TPT $\;\;\;$ $\left(1 + \tan \beta \tan 2 \beta\right) \cos 2 \beta = 1$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\sin^2 \left(\alpha + \beta\right) - \sin^2 \alpha - \sin^2 \beta = 2 \sin \alpha \sin \beta \cos \left(\alpha + \beta\right)$

$\begin{aligned} LHS & = \sin^2 \left(\alpha + \beta\right) - \sin^2 \alpha - \sin^2 \beta \\\\ & = \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right)^2 - \sin^2 \alpha - \sin^2 \beta \\\\ & = \sin^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - \sin^2 \alpha - \sin^2 \beta \\\\ & = \sin^2 \alpha \left(\cos^2 \beta - 1\right) + \sin^2 \beta \left(\cos^2 \alpha - 1\right) + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \\\\ & = - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta \sin^2 \alpha + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \\\\ & = 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - 2 \sin^2 \alpha \sin^2 \beta \\\\ & = 2 \sin \alpha \sin \beta \left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) \\\\ & = 2 \sin \alpha \sin \beta \cos \left(\alpha + \beta\right) = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\tan \left(\dfrac{\pi}{4} + \alpha\right) = \dfrac{1 + \sin 2 \alpha}{\cos 2 \alpha}$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\cos^2 \beta \left(\tan^2 \alpha - \tan^2 \beta\right)}{\sec^2 \alpha} = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right)$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin^4 \alpha + 2 \sin \alpha \cos \alpha - \cos^4 \alpha}{\tan 2 \alpha - 1} = \cos 2 \alpha$

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

Hence proved.