# 6.7.3 Trigonometric Functions Short Questions (Question 7 & 8)

Question 7:
It is given that   $\mathrm{sin}A=\frac{5}{13}\text{and}\mathrm{cos}B=\frac{4}{5}$ , where A is an obtuse angle and B is an acute angle.
Find
(a) tan A
(b) sin (A + B)
(c) cos (A B

Solution:
(a)
$\mathrm{tan}A=-\frac{5}{12}$

(b)
$\begin{array}{l}\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\\ \mathrm{sin}\left(A+B\right)=\left(\frac{5}{13}\right)\left(\frac{4}{5}\right)+\left(-\frac{12}{13}\right)\left(\frac{3}{5}\right)←\overline{)\begin{array}{l}\mathrm{cos}A=-\frac{12}{13}\\ \mathrm{sin}B=\frac{3}{5}\end{array}}\\ \mathrm{sin}\left(A+B\right)=\frac{4}{13}-\frac{36}{65}\\ \mathrm{sin}\left(A+B\right)=-\frac{16}{65}\end{array}$

(c)
$\begin{array}{l}\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B\\ \mathrm{cos}\left(A-B\right)=\left(-\frac{12}{13}\right)\left(\frac{4}{5}\right)+\left(\frac{5}{13}\right)\left(\frac{3}{5}\right)\\ \mathrm{cos}\left(A-B\right)=-\frac{33}{65}\end{array}$

Question 8:
If sin A = p, and 90° < A < 180°, express in terms of p
(a) tan A
(b) cos A
(c) sin 2A

Solution:
$\begin{array}{l}\text{Using Pythagoras Theorem,}\\ \text{Adjacent side}\\ =\sqrt{{1}^{2}-{p}^{2}}\\ =\sqrt{1-{p}^{2}}\end{array}$

(a)

$\mathrm{tan}A=-\frac{p}{\sqrt{1-{p}^{2}}}←\overline{)\begin{array}{l}\text{tan is negative at}\\ \text{second quadrant}\end{array}}$

(b)
$\mathrm{cos}A=-\sqrt{1-{p}^{2}}←\overline{)\begin{array}{l}\mathrm{cos}\text{is negative at}\\ \text{second quadrant}\end{array}}$

(c)
$\begin{array}{l}\mathrm{sin}A=2\mathrm{sin}A\mathrm{cos}A\\ \mathrm{sin}A=2\left(p\right)\left(-\sqrt{1-{p}^{2}}\right)\\ \mathrm{sin}A=-2p\sqrt{1-{p}^{2}}\end{array}$