6.5.1 Proving Trigonometric Identities Using Addition Formula and Double Angle Formulae (Examples)

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)

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

Example 3:

(a) Given that

$\sin P = \frac{3}{5} and \sin Q = \frac{5}{13},$ such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).

(b) Given that

$\sin A = - \frac{3}{5} and \sin B = \frac{12}{13},$ such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A – B).

Solution:

(a)

$\begin{array}{l} \sin P = \frac{3}{5}, \cos P = \frac{4}{5} \\ \sin Q = \frac{5}{13}, \cos Q = - \frac{12}{13} \\ \cos (P + Q) \\ = \cos A \cos B - \sin A \sin B \\ = (\frac{4}{5}) (- \frac{12}{13}) - (\frac{3}{5}) (\frac{5}{13}) \\ = - \frac{48}{65} - \frac{15}{65} \\ = - \frac{63}{65} \end{array}$

(b)

$\begin{array}{l} \sin A = - \frac{3}{5}, \cos A = - \frac{4}{5} \\ \sin B = - \frac{5}{13}, \cos B = \frac{12}{13} \\ s i n (A - B) \\ = s i n A \cos B - c o s A \sin B \\ = (- \frac{3}{5}) (\frac{12}{13}) - (- \frac{4}{5}) (- \frac{5}{13}) \\ = - \frac{36}{65} - \frac{20}{65} \\ = - \frac{56}{65} \end{array}$

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