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I have very little experience writing proofs so I don't know how to begin.

I recognize that the statement is always true, but I can't go about proving it without using circular reasoning.

How could I write a proof for this?

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    $\begingroup$ It depends where you start and level of rigour required. If you have this identity, it's straight forward to show formally: $\cos(a + b) = \cos a \cos b - \sin a \sin b$. But I think an intuitive approach going back to the unit circle is more instructive. $\endgroup$
    – Simon S
    Aug 22, 2015 at 23:19
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    $\begingroup$ If you have geometry: why is the cosine of one of the acute angles in a right triangle equal to the sine of the other? $\endgroup$
    – Ian
    Aug 22, 2015 at 23:19
  • $\begingroup$ Given the sin, or cos, you can take this as the definition of the other :P $\endgroup$ Aug 22, 2015 at 23:24
  • $\begingroup$ (This is for a calculus 1 class, in a trig review section) $\endgroup$
    – Bassinator
    Aug 22, 2015 at 23:29
  • $\begingroup$ See this answer for intuition. $\endgroup$
    – pjs36
    Aug 22, 2015 at 23:34

2 Answers 2

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We can prove this using the well known identity $$\cos (a-b) = \cos a\cos b + \sin a\sin b$$ Here $a = \dfrac{\pi}{2}, b = x$ so now we have \begin{align*} \cos\left(\frac{\pi}{2}-x\right) &= \cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x \\ &=0\cdot\cos x + 1\cdot \sin x \\ &= \sin x \end{align*}

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  • $\begingroup$ This is the method I would recommend, too. $\endgroup$ Aug 23, 2015 at 3:10
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Geometrically:

given a right triangle with angles $A$, $B$, and $C=\pi/2$ and sides $a$, $b$, and $c$ opposite the respective angles, you have $$\sin(A)=\frac{a}{c}$$ and $$\cos(B)=\frac{a}{c}$$ And we know that $$B=\pi-A-\pi/2$$ so the result is immediate.


Using Euler's formula:

We know that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, so that $$e^{i(\pi/2-x)}=\cos(\pi/2-x)+i\sin(\pi/2-x)$$ But, this is the same as $$e^{i(\pi/2-x)}=e^{i\pi/2}e^{-ix}=i(\cos(-x)+i\sin(-x))\\ =i\cos(x)+\sin(x)$$

Since two complex numbers are the same only when their real and imaginary parts are respectively equal, the result holds.

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