Question

a) verify that sin 90 = sin 30 cos 60 + cos 60 sin 60

b) show that cos 38 cos 52 - sin 38 sin 52 = 0

Answer 1

$$\begin{gathered} \left(1\right). \\ \mathrm{LHS}=\sin 90°=1 \\ \mathrm{RHS}=\sin 30°\cos 60°+\cos 30°\sin 60° \\ =\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2} \\ =\frac{1}{4}+\frac{3}{4} \\ =\frac{1+3}{4}=\frac{4}{4}=1 \\ \mathrm{So},\mathrm{LHS}=\mathrm{RHS} \\ \left(2\right). \\ \mathrm{LHS}=\cos 38°\cos 52°-\sin 38°\sin 52° \\ =\cos 38°\cos \left(90°-38°\right)-\sin 38°\sin \left(90°-38°\right) \\ =\cos 38°\sin 38°-\sin 38°\cos 38° \\ =\sin 38°\cos 38°-\sin 38°\cos 38° \\ =0 \\ =\mathrm{RHS} \end{gathered}$$