Question
Evaluate the following:
$(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)$
$(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)$
✓
Answer
We have to find: $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)$$$Since $\sin(90^\circ-\theta)=\cos\theta$
So, $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)=(\sin72^\circ)^2-(\cos18^\circ)^2$ $=[\sin(90^\circ-18^\circ)]^2-(\cos18^\circ)^2$ $=(\cos18^\circ)^2-(\cos18^\circ)^2$ $=\cos^218^\circ-\cos^218^\circ$ $=0$So value of $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)\text{ is }0$
So, $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)=(\sin72^\circ)^2-(\cos18^\circ)^2$ $=[\sin(90^\circ-18^\circ)]^2-(\cos18^\circ)^2$ $=(\cos18^\circ)^2-(\cos18^\circ)^2$ $=\cos^218^\circ-\cos^218^\circ$ $=0$So value of $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)\text{ is }0$
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