Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}} = \cot \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right)$, $\;\;\;$ $\pi < \alpha < 2 \pi$

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

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

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $3 - 4 \cos 2 \alpha + \cos 4 \alpha = 8 \sin^4 \alpha$

$\begin{aligned} LHS & = 3 - 4 \cos 2 \alpha + \cos 4 \alpha \\\\ & = 3 - 4 \left(1 - 2 \sin^2 \alpha\right) + \cos \left[2 \left(2 \alpha\right)\right] \\\\ & = 3 - 4 + 8 \sin^2 \alpha + 1 - 2 \sin^2 2 \alpha \\\\ & = 8 \sin^2 \alpha - 2 \sin^2 2 \alpha \\\\ & = 8 \sin^2 \alpha - 2 \left(2 \sin \alpha \cos \alpha\right)^2 \\\\ & = 8 \sin^2 \alpha - 2 \times 4 \sin^2 \alpha \cos^2 \alpha \\\\ & = 8 \sin^2 \alpha \left(1 - \cos^2 \alpha\right) \\\\ & = 8 \sin^2 \alpha \times \sin^2 \alpha \\\\ & = 8 \sin^4 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

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

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

Hence proved.