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During a question involving proving a trigonometric identity, I was given help in which one of the lines showed that $\sin x \sin(\pi/2-x)$ equals $\sin x \cos x$? Could anyone please explain to me/show me why this is the case?

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2 Answers

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The sine and cosine function are very closely related. In fact, the one is just a horizontal shift of the other.

You have then that $\cos(x) = \sin(x+\frac{\pi}{2})$, which is seen by a horizontal shift to the right of the cosine function by $\frac{\pi}{2}$.

Furthermore, you have that $\cos(x)=\cos(-x)$ due to the fact that the cosine function is even. Thus, it follows that $\cos(x)=\cos(-x)=\sin(-x+\frac{\pi}{2})$.

The identity then follows that $\sin(x)\sin(\frac{\pi}{2}-x)=\sin(x)\cos(x)$

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it is true because $\cos(\pi/2-x) = \sin x$. If you ask why this is true? to make thing easy for you to follow, assume $0 < x < \pi/2$, then you let $A = x, B = \pi/2 - x $, then $A, B$ are complimentary angles in a right $\triangle$, and using SOHCAHTOA you conclude that $\sin A = \cos B$.

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