6.7.4 Trigonometric Functions Short Questions (Question 9 & 10)

Question 9:

Given that sin θ =

\frac{3}{5}

, where θ is an acute angle, without using tables or a calculator, find the values of

(a) sin (180º + θ),

(b) cos (180º – θ),

(c) tan (360º + θ).

Solution:
(a)

\sin θ = \frac{3}{5} \cos θ = \frac{4}{5} \tan θ = \frac{3}{4}

sin (180º + θ)

= sin 180º cos θ + cos 180º sin θ

= (0) cos θ + (– 1) sin θ

= – sin θ

- \frac{3}{5}

(b)

cos (180º – θ)

= cos 180º cos θ + sin 180º sin θ

= (– 1) cos θ + (0) sin θ

= – cos θ

- \frac{4}{5}

(c)

\begin{array}{l} \tan (360^{\circ} + θ) \\ = \frac{\tan 360^{\circ} + \tan θ}{1 - \tan 360^{\circ} \tan θ} \\ = \frac{0 + \tan θ}{1 - (0) (\tan θ)} \\ = \tan θ \\ = \frac{3}{4} \end{array}

Question 10:

Prove each of the following trigonometric identities.

(a) cot²x – cot² x cos²x = cos² x

(b) \frac{\sec x}{\sec x - \cos x} = \cos e c^{2} x

Solution:

(a)

\begin{array}{l} LHS: \cot^{2} x - \cot^{2} x \cos^{2} x \\ = \cot^{2} x (1 - \cos^{2} x) \\ = \cot^{2} x (s i n^{2} x) \\ = \frac{\cos^{2} x}{s i n^{2} x} (s i n^{2} x) \\ = \cos^{2} x (RHS) \end{array}

(b)

\begin{array}{l} LHS: \frac{\sec x}{\sec x - \cos x} \\ = \frac{\frac{1}{\cos x}}{\frac{1}{\cos x} - \cos x} = \frac{\frac{1}{\cos x}}{\frac{1}{\cos x} - \frac{\cos^{2} x}{\cos x}} \\ = \frac{\frac{1}{\cos x}}{\frac{1 - \cos^{2} x}{\cos x}} = \frac{1}{\cos x} \times \frac{\cos x}{1 - \cos^{2} x} \\ = \frac{1}{1 - \cos^{2} x} = \frac{1}{s i n^{2} x} \\ = \cos e c^{2} x (RHS) \end{array}