Question
If $A = B = 45^\circ ,$show that:$\cos (A + B) = \cos A \cos B - \sin A \sin B$
✓
Answer
Given that $A = B = 45^\circ $
$\text{LHS} = \cos (A + B)$
$= \cos ( 45^\circ + 45^\circ )$
$= \cos 90^\circ $
$= 0$
$\text{RHS} = \cos A \cos B – \sin A \sin B$
$= \cos 45^\circ \cos 45^\circ – \sin 45^\circ \sin 45^\circ $
$=\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$
$=0$
$\text { LHS }=\text { RHS }$
$\text{LHS} = \cos (A + B)$
$= \cos ( 45^\circ + 45^\circ )$
$= \cos 90^\circ $
$= 0$
$\text{RHS} = \cos A \cos B – \sin A \sin B$
$= \cos 45^\circ \cos 45^\circ – \sin 45^\circ \sin 45^\circ $
$=\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$
$=0$
$\text { LHS }=\text { RHS }$
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