# Proving trigonometric cofunction identities for angles larger than $\pi / 2$

Background

For the sine and cosine functions, we have

$$\sin(\frac{\pi}{2} - x) = \cos(x)$$

$$\cos(\frac{\pi}{2} - x) = \sin(x)$$

For $$x\in[0, \pi/2]$$, this can be easily shown using a right triangle, but every textbook I have leaves it there and simply assures me that it can be shown for $$x>\pi/2$$.

My thoughts

I've tried looking at this from the unit circle perspective, but I have a hard time proving what seems apparent enough. If I first take the angle $$\pi/2$$, then subtract the initial angle $$x$$ (let's call the resulting angle $$y$$), the rectangle with side lengths $$\sin y, \cos y$$ seems to be a rotated form of the rectangle formed in the same way for $$x$$.

Question

How can this be proven to hold for all angles $$x$$?

• Related: When the trig functions moved from the right triangle to the unit circle?. In particular, see my answer. – Blue Apr 5 at 21:24
• Since $\sin x=\sin (x+2n\pi)$ and $\cos x=\cos (x+2n\pi)$ for any $n\in \Bbb Z$ and any $x,$ observe that it suffices to prove your formulas only for all $x\in [-\pi,\pi]$. You can show geometrically that for $|x|\le \pi$ we have $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$ and $\cos (\pi\pm x)=-\cos x$ and $\sin (x\pm \pi)=-\sin x,$ and then use these. For example, if $x=-2\pi/3$ then $\sin x=$ $-\sin (-2\pi/3+\pi)=$ $-\sin (\pi/3)=$ $-\cos (\pi/2-\pi/3)=$ $+\cos (\pi/2-\pi/3-\pi)=$ $\cos (-5\pi/6)=$ $\cos (+5\pi/6)=$ $\cos (\pi/2-x)$. – DanielWainfleet Apr 6 at 4:24

These identities can be shown by using the trigonometric addition formula, as the other answer shows. However, this may not help with the geometric insight. You can see this with either the triangle perspective, or the unit circle perspective. Here is how.

The angle in this triangle at ACB is $$\theta$$. The angle at BAC therefore, is $$\frac\pi2-\theta$$. The side BC is equal to $$\cos\theta$$, and is also equal to $$\sin\left(\frac\pi2-\theta\right)$$. Similarly, the side AB is equal to $$\sin\theta$$, and is also equal to $$\cos\left(\frac\pi2-\theta\right)$$. So we're done.

Now for the unit circle perspective. You have the unit circle, and a line at angle $$\theta$$. Draw this line and a downward vertical line where it touches the circle, forming a right angled triangle with angles $$\theta,\frac\pi2-\theta,\pi$$, and a known side, the hypotenuse, with length $$1$$. Now draw a line at angle $$\frac\pi2-\theta$$, meaning it makes an angle of $$\theta$$ with the y-axis. Where it touches the circle, draw a horizontal line to form a right triangle. This triangle is congruent to the first since it has all the same angles, and one length the same. So all the side lengths are the same. Compare the relevant side lengths and you'll find the identities as required.

For $$\frac{\pi}{2}\le x\le \pi$$, use the identities $$cos(x)=cos(x-\frac{\pi}{2}+\frac{\pi}{2})=cos(x-\frac{\pi}{2})cos(\frac{\pi}{2})-sin(x-\frac{\pi}{2})sin(\frac{\pi}{2})=-sin(x-\frac{\pi}{2})$$ and $$sin(x)=sin(x-\frac{\pi}{2}+\frac{\pi}{2})=cos(x-\frac{\pi}{2})sin(\frac{\pi}{2})+sin(x-\frac{\pi}{2})cos(\frac{\pi}{2})=cos(x-\frac{\pi}{2})$$

Continue this procedure for intervals $$\pi\le x\le \frac{3\pi}{2}$$ and $$\frac{3\pi}{2}\le x \le 2\pi$$. The identities you want immediately fall out.

To determine the value of sin or cosine, we draw perpendicular from the end of the radius (or the point on the circle) to x axis. and we calculate from the right angle formed in this way.

Does not matter whether the angle is acute or obtuse.

Suppose the location of the point (on the circle) is (x,y). The $$\sin\theta$$ is determined by the ratio of the y and radius of circle (r) which is $$\dfrac{y}{r}$$, always.

Here we need to take the magnitude as well as sign of y, r has only magnitude (distance, scalar). Similarly true for $$\cos\theta$$ which is $$\dfrac{x}{r}$$

Suppose the point is on second quadrant. We can express this as $$\sin(\pi/2+\theta)$$, which can be written as

$$\sin(\pi/2-(-\theta) )$$ and as per identity it should be equal to $$\cos(-\theta)$$, where the point falls in fourth quadrant. We an readily confirm this is true (both magnitude and sign matches (both are positive) ).

Suppose the point is on third quadrant. We can express this as $$\sin(\pi+\theta)$$, which can be written as

$$\sin(\pi/2-(-(\pi/2+\theta)) )$$ and as per identity it should be equal to $$\cos(-(\pi/2+\theta))$$, where the point falls in (again) third quadrant. We an readily confirm this is true (both magnitude and sign matches (both are negtive) ).

Suppose the point is on fourth quadrant. We can express this as $$\sin(3\dfrac{\pi}{2}+\theta)$$, which can be written as

$$\sin(\pi/2-(-(\pi+\theta)) )$$ and as per identity it should be equal to $$\cos(-(\pi+\theta))$$, where the point falls in second quadrant. We an readily confirm this is true (both magnitude and sign matches (both are negtive) ).

For $$\theta \gt 2\pi$$ it repeats.