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By drawing a right triangle it is obvious that $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$. I'm trying to prove to myself that this is true for all values of $\theta$ by following the reasoning on this page where the sine graph is inverted vertically by plugging in $-\theta$ and then shifted to the right by $\frac{\pi}{2}$ to get the cosine graph. To shift a graph to the right by $x$, don't you have to subtract $x$ from the input value?

I'm getting $\cos \theta = \sin\left(-\theta - \frac{\pi}{2}\right)$ which is of course wrong and Wolfram shows me is the vertically-inverted cosine graph.

I know this is ridiculously simple and something to do with the $-\theta$ parameter to $\sin$ meaning that a right shift then needs the shift value added, but I can't see clearly why.

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If you compose $f$ with $x\rightarrow x+T, T>0$, you shift its graph to the left, because lower values of x are (almost?) universally put on the left. You may check that on a graphing software -like Wolfram Alpha which is what you probably meant. – plm May 15 '12 at 11:23
On the page you link to the author forgets to mention that putting the graph upside down is only replacing $t$ with $-t$ for the sine function because it is an odd function $sin(-t)=-sin(t)$. In general this does not hold, you invert the graph sending $f$ to $-f$ (not $t$), in particular $cos(-t)=cos(t)$, which allows to have the "symmetric" formulas $sin(t)=cos(\pi/2-t)$, $cos(t)=sin(\pi/2-t)$. – plm May 15 '12 at 11:37
I looked at your profile, being a CAD developer I now guess you rather meant Mathematica, rather than Wolfram Alpha, sorry. – plm May 15 '12 at 11:46
@plm: I did mean Wolfram Alpha, sorry for confusion. We don't have anything like Mathematica or MATLAB here :( – PeteUK May 15 '12 at 13:35
up vote 2 down vote accepted

Let $\gamma$ denote the graph of the function $f$, hence $\gamma=\{(x,f(x))\mid x\in\mathbb R\}$. Let $a$ denote a real number and $\alpha=(a,0)$. The set $\gamma+\alpha$ is $$ \gamma+\alpha=\{u+\alpha\mid u\in\gamma\}=\{(x+a,f(x))\mid x\in\mathbb R\}=\{(t,f(t-a))\mid t\in\mathbb R\}, $$ hence $\gamma+\alpha$ is the graph of the function $f_a:x\mapsto f(x-a)$.

In the present case, $f(x)=-\sin(x)$, $a=+\pi/2$ (remember the shift is to the right hence $a\gt0$), hence $f_a(x)=-\sin(x-\pi/2)=\sin(\pi/2-x)$, as desired.

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I'm trying to go through this step by step. Am I right in saying $\gamma$ is a set of ordered pairs? If so, what does it mean to say $u + a$ where $u\in\gamma$? Looks like this is adding a to the first element of $u$, right? – PeteUK May 15 '12 at 13:33
You can probably ignore my comment as I guess the answer is you've provided the definition of addition of a member of $\gamma$ and a real number (to be the first element of the ordered pair added with the real number). I've gone through your answer and it makes sense to me now. Thank you. – PeteUK May 15 '12 at 13:54
Excellent remark, I was sloppy. See revised version. – Did May 15 '12 at 14:29
Appreciate the response and edit. – PeteUK May 15 '12 at 14:44

Alternatively, one can use

$$\sin (a-b)=\sin a \cos b - \sin b \cos a$$

Then $a=\pi/2$ and $b=\theta$,

$$\sin (\pi/2-\theta)=\sin \pi/2 \cos \theta - \sin \theta \cos \pi/2$$

$$\sin (\pi/2-\theta)=1 \cdot \cos \theta - \sin \theta \cdot0$$

$$\sin (\pi/2-\theta)=\cos \theta $$

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I like this other way of looking at it, thanks. – PeteUK May 15 '12 at 16:04

Beware that you compose $sin$ first with $t\rightarrow -t$ and then the resulting function with the right-shift $t-\frac{\pi}{2}$, so you get the desired $cos(t)=sin(-(t-\frac{\pi}{2}))=sin(\frac{\pi}{2}-t)$.

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Proofs are already given so I'm not giving a mathematical proof. Indeed this is not a mathematical proof because from here you can only conclude that $\sin(\frac\pi2-\theta)=\cos(\theta)$ for $\theta<\frac\pi2$.

enter image description here

Consider the above figure

$\sin(\frac \pi2-\theta)=\frac ah$ and $\cos(\theta)=\frac ah$

$\therefore \sin(\frac\pi2-\theta)=\cos(\theta)$

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