Short Question 7 & 8

Question 7:
Solution:
(a)
$\tan A=-\frac{5}{12}$


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


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

Question 8:

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


(a)
$\tan A=-\frac{p}{\sqrt{1-p^2}}\leftarrow \boxed{\begin{array}{l}\text{tan is negative at}\\ \text{second quadrant}\end{array}}$


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


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