Question

If A = 60$°$ B = 30$°$ then show that cos(A$-$B) = cosA.cosB + sin A. sin B

Answer 1

Given:
A = 60$°$
B = 30$°$

To show:
cos (A – B) = cos A cos B + sin A sin B

LHS = cos (A – B)
= cos (60$°$ – 30$°$)
= cos 30$°$
= $\frac{\sqrt{3}}{2}$ (From the table)

Now, RHS = cos A.cos B + sin A.sin B
= cos 60$°$ cos 30$°$ + sin 60$°$ sin 30$°$
$$\begin{gathered} =\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2} \\ =\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ =\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2} \\ \therefore \mathrm{LHS}\hspace{0.17em}=\mathrm{RHS} \end{gathered}$$

Hence, proved