Question
Evaluate the following:
If A = 60° and B = 30°, verify that:
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
$\cos(\text{A}-\text{B})=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}\\=\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ$
$=\Big(\frac{1}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times\frac12\Big)=\Big(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\Big)$
$=2\times\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
$\therefore\ \cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$

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