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Unit Circle, important angles

December 5, 2016 Mouctar Trigonometry 0

Complementary angles

It has been said that complementary angles have their sum $\frac{\pi}{2}\; or\; 90^\circ$

Looking at the following figure, we can see that $\triangle ABB' \cong \triangle D'FA$

$\angle BAB' =\angle AD'F=\frac{\pi}{2}-\angle FAD'$

$\sin \angle BAD'=FD'$

However, by CPCTC, $FD'=EA=\cos \angle BAB'$

This shows that:

$\sin (\frac{\pi}{2}-\alpha)=\cos \alpha$

$\cos (\frac{\pi}{2}-\alpha)=\sin \alpha$

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