Trigonometric Functions Short Questions (Question 7 & 8)

Question 7:
It is given that   , 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:

(a)

(b)

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