Question
Given $A = 60^\circ $ and $B = 30^\circ ,$prove that :$\cos (A + B) = \cos A \cos B - \sin A \sin B$
✓
Answer
Given$ A = 60^\circ $ and $B = 30^\circ $
$\text{LHS} = \cos(A+B)$
$= \cos(60^\circ + 30^\circ )$
$= cos90^\circ $
$=0$
$\text{RHS} = \cos A \cos B – \sin A \sin B$
$= \cos 60^\circ \cos 30^\circ – \sin 60^\circ \sin 30^\circ $
$=\frac{1}{2} \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \frac{1}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$\text { LHS }=\text { RHS }$
$\text{LHS} = \cos(A+B)$
$= \cos(60^\circ + 30^\circ )$
$= cos90^\circ $
$=0$
$\text{RHS} = \cos A \cos B – \sin A \sin B$
$= \cos 60^\circ \cos 30^\circ – \sin 60^\circ \sin 30^\circ $
$=\frac{1}{2} \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \frac{1}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$\text { LHS }=\text { RHS }$
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