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We already know that

$\sin (\frac{3 \pi}{2}-A)=-\cos (A)$ but is there any method to prove it geometrically?

Could someone suggest something?

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  • $\begingroup$ Have you drawn a picture when $0 < A \le \frac\pi 2$? It's pretty suggestive. $\endgroup$ – pjs36 Apr 17 '16 at 3:56
  • $\begingroup$ @pjs36 Yes. But how would that take care of $\frac{3 \pi}{2}-A$ ? $\endgroup$ – Ananya Apr 17 '16 at 4:02
  • $\begingroup$ See the question "How to remember a particular class of trig identities". $\endgroup$ – Blue Apr 17 '16 at 4:36
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Assume $|A|<\pi$, and let ${\bf a}=(u,v)\in S^1$ represent the angle $A$. Then $\bar {\bf a}=(u,-v)$ represents the angle $-A$. In order to obtain the point ${\bf b}$ representing angle $B:=-A+{3\pi\over2}$ we have to rotate $\bar {\bf a}$ by ${3\pi\over2}$ counterclockwise, or what is the same thing: by ${\pi\over2}$ clockwise. Such a rotation has the effect $(x,y)\mapsto(y,-x)$ on points $(x,y)$. It follows that ${\bf b}=(-v,-u)$, so that we obtain $$\sin\left({3\pi\over2}-A\right)=\sin B=-u=-\cos A\ ,$$ as claimed.

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Please refer to your text book how the radius vector (always >0) rotates, and how trig functions change sign as ratios of lengths. Or in the following net available source,

RadiusVectorRotn

Sin is $>0$ in quadrants 1,2 and $<0$ in quadrants 3,4. To remember sign of trig function in each quadrant keep in mind: All Silver Tea Cups. Accordingly,

$$\sin (\frac{\pi}{2}-A)=+\cos (A)$$ $$\sin (\frac{2 \pi}{2}-A)=+\cos (A)$$ $$\sin (\frac{3 \pi}{2}-A)=-\cos (A)$$ $$\sin (\frac{4 \pi}{2}-A)=-\cos (A).$$

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