5.5.1 Proving Trigonometric Identities Using Addition Formula And Double Angle Formulae (Part 1)

Posted on April 22, 2020 by user

Example 2:

Prove each of the following trigonometric identities.

\begin{array}{l} (a) \frac{1 + \cos 2 x}{\sin 2 x} = \cot x \\ (b) \cot A \sec 2 A = \cot A + \tan 2 A \\ (c) \frac{s i n x}{1 - c o s x} = \cot \frac{x}{2} \end{array}

Solution:

(a)

\begin{array}{l} L H S \\ = \frac{1 + \cos 2 x}{\sin 2 x} \\ = \frac{1 + (2 \cos^{2} x - 1)}{2 \sin x \cos x} \\ = \frac{2 \cos^{2} x}{2 \sin x \cos x} \\ = \frac{\cos x}{\sin x} \\ = \cot x = R H S (proven) \end{array}

(b)

\begin{array}{l} R H S \\ = \cot A + \tan 2 A \\ = \frac{\cos A}{\sin A} + \frac{\sin 2 A}{\cos 2 A} \\ = \frac{\cos A \cos 2 A + \sin A \sin 2 A}{\sin A \cos 2 A} \\ = \frac{\cos A (\cos^{2} A - \sin^{2} A) + \sin A (2 \sin A \cos A)}{\sin A \cos 2 A} \\ = \frac{\cos^{3} A - \cos A \sin^{2} A + 2 \sin^{2} A \cos A}{\sin A \cos 2 A} \\ = \frac{\cos^{3} A + \cos A \sin^{2} A}{\sin A \cos 2 A} \\ = \frac{\cos A (\cos^{2} A + \sin^{2} A)}{\sin A \cos 2 A} \\ = \frac{\cos A}{\sin A \cos 2 A} \leftarrow \sin^{2} A + \cos^{2} A = 1 \\ = (\frac{\cos A}{\sin A}) (\frac{1}{\cos 2 A}) \\ = \cot A \sec 2 A \end{array}

(c)

\begin{array}{l} L H S \\ = \frac{s i n x}{1 - c o s x} \\ = \frac{2 s i n \frac{x}{2} \cos \frac{x}{2}}{1 - (1 - 2 s i n^{2} \frac{x}{2})} \leftarrow \begin{array}{l} \sin x = 2 s i n \frac{x}{2} \cos \frac{x}{2}, \\ \cos x = 1 - 2 \sin^{2} \frac{x}{2} \end{array} \\ = \frac{2 s i n \frac{x}{2} \cos \frac{x}{2}}{2 s i n^{2} \frac{x}{2}} = \frac{\cos \frac{x}{2}}{s i n \frac{x}{2}} \\ = \cot \frac{x}{2} = R H S (proven) \end{array}