What is the answer for question 13) ?

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Let `DeltaABC` be a right triangle with right angle C.

Without loss of generality, let `theta=m/_A` .

We want to show that `sintheta=cos(pi/2-theta)` .

We know that the acute angles of a right triangle are complementary. So `m/_B=90-m/_A=pi/2-theta` .

`sintheta=sinA=a/c=cosB=cos(pi/2-theta)` QED

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Does the relationship hold for `theta>pi/2` ?

The answer is yes. For any angle `theta` we can draw a right triangle in the correct quadrant and the ratios will be the same -- we must keep track of the sign however.

Another possible problem with the analogy is angles that are multiples of `pi/2` . We cannot use a right triangle to show the ratio -- however the relationship still holds as we know that `sin ((pi)/2)=cos0,sin pi=cos ((pi)/2), ` etc... from the unit circle.

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