Short Question 19

Question 19 (4 marks):
It is given that cos α = t where t is a constant and 0^o ≤ α ≤ 90^o.
Express in terms of t
(a) sin (180^o + α),
(b) sec 2α.

Solution:
(a)




$\begin{array}{l}\text{sin}\left({180}^o+\alpha \right)\\ =\sin 180\cos \alpha +\cos 180\sin \alpha \\ =0-\sin \alpha \\ =-\sin \alpha \\ =-\sqrt{1-t^2}\end{array}$

(b)
$\begin{array}{l}\sec 2\alpha =\frac{1}{\cos 2\alpha }\\ =\frac{1}{2{\cos }^2\alpha -1}\\ =\frac{1}{2t^2-1}\end{array}$