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?
$\endgroup$ 55 Answers
$\begingroup$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 $\sin \alpha = \cos (\frac{\pi}{2} - \alpha)$
$\endgroup$ 1 $\begingroup$The word "cosine" was earlier written "co. sinus", short for "complementi sinus", which (in Latin) means "sine of the complement", i.e., $\sin (\frac{\pi}{2}-\theta)$.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 1 $\begingroup$An explanation without words:
For visual people like me, this animation makes the relationship intuitively understandable:
