Proving $\sin (x)=\cos (90^\circ-x)$ I'm interested in the different ways of proving this, any proof is welcome.
I understand one way is the cosine sum/difference formula, another is using a right angled triangle. Are there any others?
Thanks.
 A: This depends on which definitions of sine and cosine you prefer to work with.
If you use the unit-circle definition

$(\cos \theta,\sin \theta)$ are the coordinates of the point on the unit circle reached by going a distance of $\theta$ counterclockwise along the unit circle from its intersection with the positive $x$ axis.

then the property you want is almost immediate. Replacing the angle $\theta$ by $90^\circ-\theta$ amounts to reflecting the point in the line $x=y$, which is the same as swapping the axes, so the cosine becomes the sine and vice versa.

As an example of a proof with a quite different flavor, suppose we have defined the sine and cosine by power series. (Since you're working in degrees, that's very unlikely to be what you have done, through -- so if this part of the answer makes no sense to you, feel free to ignore it). Then it would be more natural to proceed along the lines of:


*

*The derivative of $\sin$ is $\cos$, and the derivative of $\cos$ is $-\sin$.
(This can be shown by term-by-term differentiation of the power series).

*The functions $x \mapsto \cos x$ and $x \mapsto \sin(\pi/2 - x)$ both satisfy the ordinary differential equation $y''=-y$.
(Easily verified by calculus given the point above).

*They both satisfy $f(0)=1$ and $f'(0)=0$.
(Here we need to know that the sine and cosine of $\pi/2$ is, which again depends on our definition of $\pi$. It is not uncommon to define $\pi$ by the fact that $\cos(\pi/2)=0$, and the "idiot formula" then yields $\sin(\pi/2)=\pm 1$, but it must be positive because $\sin'(0)>0$, and the sign of $\sin$ cannot have changed before the first zero of $\cos$, which by definition is $\pi/2$).

*Therefore, by the Picard-Lindelöf theorem, they are the same function.
A: Here's a cute picture: 


Thanks to $w|\alpha$
A: First and easiest:
open the bracket of $\cos (90-x)$ using $\cos (a+b)$ formula , you get $\sin x$;
Second Refer graphs!
Third (:P )
We can visualise the same by rotation of a complex number in complex plane.
Multiply any $Z$ with $e^{i\pi/2}$ and we get the same complex no. just in other quadrant . Implying $\cos (90-x)=\sin x$
A: Take a right angled triangle $ABC$ with angle $B=90,A=x$ This implies angle $C=90-x$.
$\sin x=BC/AC$
Similarly $\cos (C)=\cos (90-x)=BC/AC$
Using the fact that $\sin $ and $\cos $ are height/hypotenuse and base / hypotenuse respectively. 
A: Cosine satisfies the identity $\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$. In your case, we have
$$
\cos(90^\circ-x)=\cos(90^\circ)\cos(x)+\sin(90^\circ)\sin(x)=0\cdot\cos(x)+1\cdot\sin(x)=\sin(x).
$$
I suppose you'd have to verify that identity to convince yourself.
A: Plot the curves $y_1 = \sin x$ and the curve $y_2=\cos x$. You will realize that the $\cos x$ graph is somewhat "ahead" of the $\sin x$ graph. Actually, that "lead" is $90^o$ or $\frac \pi 2$. All you need to do is to translate the $y_2$ "backwards" by that difference. That is one very useful way of looking at it.
A: Some proof using the euler identity: $$e^{it} = \cos(t)+i\sin(t)$$
Note that $sin(t) = \Im (e^{it})$ and $\cos(\frac \pi 2 -x) = \Re (e^{i(\frac \pi 2 -t)})$. Let's take a look at $$sin(t)-i\cos(t)=-ie^{it} = e^{i\frac{-\pi}2}e^{it} = e^{i(t-\frac \pi 2)} = \cos(t-\frac \pi 2)+i \sin(t-\frac \pi 2)$$
Comparing real and imaginary parts we get $$\sin(t) = \cos(t-\frac \pi 2) = \cos(\frac \pi 2-t)$$ as $$\cos(x) = \cos(-x) ~~x\in \mathbb R$$
