Answer
(a) $sin~2x = 2sin~x~cos~x$
(b) $cos(x+a) = cos~x~cos~a-sin~x~sin~a$
Work Step by Step
(a) We can differentiate both sides of the equation:
$cos~2x = cos^2~x-sin^2~x$
$-2sin~2x = 2cos~x(-sin~x)-2sin~x(cos~x)$
$-2sin~2x = -4sin~x~cos~x$
$sin~2x = 2sin~x~cos~x$
(b) We can differentiate both sides of the equation with respect to $x$:
$sin(x+a) = sin~x~cos~a+cos~x~sin~a$
$cos(x+a) = cos~x~cos~a-sin~x~sin~a$
