Question

Find the value of sin 2A, cos 2A when cos A=3/5

Answer 1

cos A=B/H
i.e P=?
B=3
H=5
by Pythagoras' theorem
P^2=H^2-B^2
P^2=5^2- 3^2
P^2=25-9
P^2=16
P=
$\sqrt{16}$
P=4
therefore, sinA =P/H
i.e.4/5
so value of cos ^2 A=(3/5)^2=9/25
and sin ^2 A=(4/5)^2 =16/25