# How to go about proving that $\cos(\frac{\pi}{2}-x) = \sin(x)$?

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?

• 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. Commented Aug 22, 2015 at 23:19
• 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?
– Ian
Commented Aug 22, 2015 at 23:19
• Given the sin, or cos, you can take this as the definition of the other :P Commented Aug 22, 2015 at 23:24
• (This is for a calculus 1 class, in a trig review section) Commented Aug 22, 2015 at 23:29
• See this answer for intuition. Commented Aug 22, 2015 at 23:34

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*}

• This is the method I would recommend, too. Commented Aug 23, 2015 at 3:10

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.