# 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)

$\mathrm{sin}\theta =\frac{3}{5}\text{}\mathrm{cos}\theta =\frac{4}{5}\text{}\mathrm{tan}\theta =\frac{3}{4}$

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)
$\begin{array}{l}\mathrm{tan}\left({360}^{\circ }+\theta \right)\\ \\ =\frac{\mathrm{tan}{360}^{\circ }+\mathrm{tan}\theta }{1-\mathrm{tan}{360}^{\circ }\mathrm{tan}\theta }\\ \\ =\frac{0+\mathrm{tan}\theta }{1-\left(0\right)\left(\mathrm{tan}\theta \right)}\\ =\mathrm{tan}\theta \\ =\frac{3}{4}\end{array}$

Question 10:
Prove each of the following trigonometric identities.
(a) cot2 x – cot2 x cos2x = cos2 x
$\text{(b)}\frac{\mathrm{sec}x}{\mathrm{sec}x-\mathrm{cos}x}=\mathrm{cos}e{c}^{2}x$

Solution:
(a)
$\begin{array}{l}\text{LHS:}{\mathrm{cot}}^{2}x-{\mathrm{cot}}^{2}x{\mathrm{cos}}^{2}x\\ ={\mathrm{cot}}^{2}x\left(1-{\mathrm{cos}}^{2}x\right)\\ ={\mathrm{cot}}^{2}x\left(si{n}^{2}x\right)\\ =\frac{{\mathrm{cos}}^{2}x}{si{n}^{2}x}\left(si{n}^{2}x\right)\\ ={\mathrm{cos}}^{2}x\text{(RHS)}\end{array}$

(b)
$\begin{array}{l}\text{LHS:}\frac{\mathrm{sec}x}{\mathrm{sec}x-\mathrm{cos}x}\\ =\frac{\frac{1}{\mathrm{cos}x}}{\frac{1}{\mathrm{cos}x}-\mathrm{cos}x}=\frac{\frac{1}{\mathrm{cos}x}}{\frac{1}{\mathrm{cos}x}-\frac{{\mathrm{cos}}^{2}x}{\mathrm{cos}x}}\\ =\frac{\frac{1}{\mathrm{cos}x}}{\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{cos}x}}=\frac{1}{\mathrm{cos}x}×\frac{\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}\\ =\frac{1}{1-{\mathrm{cos}}^{2}x}=\frac{1}{si{n}^{2}x}\\ =\mathrm{cos}e{c}^{2}x\text{(RHS)}\end{array}$