# Why is cosine a sine function with offset pi/2?

I was looking at this question and stumbled across this answer stating that the picture proves "Why cosine is simply sine but offset by pi/2 radians" I realized I don't know why this is true. So! Why is cosine sine with an offset pi/2?

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Do you know what sine and cosine represent in the trigonometric circle? – user88595 Apr 1 '14 at 20:34
Yes I understand that they are positions on the unit circle. – Dudemanword Apr 1 '14 at 20:38
Do you know the addition and subtraction formulas for sine and cosine? Alternatively, following the suggestion of @user88595, can you use reflection in the line $y=x$ to relate the sine of $\theta$ to the cosine of $\pi/2-\theta$? – Lubin Apr 1 '14 at 20:43
@Dudemanword : So now look closely at how they vary when you make the point move anti-clockwise. You should realise that the value for sine is always a bit late because it is the same as the value for cosine but $\frac{\pi}{2}$ late. – user88595 Apr 1 '14 at 20:46
This really depends on how you've defined $\cos$ and $\sin$. Spivak would define $\sin$ and $\cos$ in terms of an antiderivative. Then he proves that $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$. So, $\sin(a + \pi/2) = \cos(a)$, since $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$. – Chris K Apr 1 '14 at 22:10

Think of it as right triangles, which is probably how you were first introduced to trigonometry.

We have a right triangle with right angle $\theta$ and acute angles $\beta$ and $\alpha$. We have $$\sin \alpha = \frac{\text{side opposite}}{\text{hypotenuse}}$$ and $$\cos \beta = \frac{\text{side adjacent}}{\text{hypotenuse}}.$$Notice that the side adjacent to $\beta$ is the same as the side opposite of $\alpha$, thus $\sin \alpha = \cos \beta$.

In addition, since this is a right triangle, $\alpha + \beta = \frac{\pi}{2}$. So we can make a substitution for $\beta = \frac \pi 2 - \alpha$ to say that $\sin \alpha = \cos (\frac{\pi}{2} - \alpha)$

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Wondering, why the -1? – Michael Tong Apr 1 '14 at 23:34

The word "cosine" was earlier written "co. sinus", short for "complementi sinus", which (in Latin) means "sine of the complement", i.e., $\sin (\frac{\pi}{2}-\theta)$.

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While true, I don't think this very adequately answers OP's question. – Michael Tong Apr 1 '14 at 22:44
@MichaelT: Eh? I would have thought it was obvious that I am referring to the fact that $\cos\theta = \sin (\frac{\pi}{2}\pm\theta)$ – MPW Apr 2 '14 at 3:42
Yes, it is obvious, but you do not provide a reason why they are separated by $\frac \pi 2$, rather you kind of just say "hey, their names agree!" – Michael Tong Apr 2 '14 at 5:31
@MichaelT: I was thinking more along the lines of "hey, that's the definition of cosine!". I suppose I wasn't very explicit. – MPW Apr 2 '14 at 11:31

This is easily verified in the trigonometric definition of the sine and cosine, i.e. in a right triangle.

The definition of the sine of an angle in a right triangle is the ratio of the side opposite the angle and the hypotenuse:

$$\sin\left(\angle A\right) = \frac a c$$

The definition of the cosine is the ratio of the side adjacent to the angle and the hypotenuse:

$$\cos\left(\angle A\right) = \frac b c$$

Observe that the side opposite the other angle, $\angle B$, is $b$, so we get from the definition:

$$\sin\left(\angle B\right) = \frac b c$$

But, since this is a right triangle:

$$\angle B = \frac \pi 2 - \angle A$$

giving what we wanted.

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This is basically the same as Michael T's answer, who submitted it first :) – Yoni Rozenshein Apr 1 '14 at 21:35