Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I go about proving this? I know one method is:

$\eqalign{ \cos (90^\circ + \theta ) &\equiv \cos90^\circ \cos\theta - \sin90^\circ \sin\theta \cr & \equiv (0)(\cos\theta ) - (1)(\sin \theta ) \cr & \equiv - \sin \theta \cr} $

I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.

Thanks alot.

share|cite|improve this question
Most of the answers to your earlier question generalize straightforwardly to this case. – Henning Makholm Apr 2 '13 at 21:44
@HenningMakholm Could you explain how please, this is confusing me i cant get my head around it, thank you! – seeker Apr 2 '13 at 21:51
<pedant>$\cos 90 \neq 0$, but $\cos \frac\pi2 = \cos90^\circ = 0$</pedant> – kahen Apr 2 '13 at 23:04
Ah I missed the degrees symbol? I'll be careful next time. – seeker Apr 2 '13 at 23:06
up vote 5 down vote accepted

I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.

The picture represents the trigonometric unit circle. The angle $\theta=\angle AOP$ is represented in the first quadrant, but it is a generic orientate angle. $|AP|=\sin\theta$. $\angle AOQ=\theta+90^\circ$. The right triangle $[O,A,P]$ is similar to the right triangle $[O,B,Q]$, because $\angle BQO=\angle AOP=\theta$. Since $|OP|=|OQ|=1$, both triangles are congruent and $$\cos(\theta+90^\circ)=-|OB|=-|AP|=-\sin\theta.$$

enter image description here

share|cite|improve this answer
Thanks you very much! – seeker Apr 2 '13 at 23:06
@Assad: You are welcome! – Américo Tavares Apr 2 '13 at 23:10
how OB=Cos(90°+θ)? – Yogesh Tripathi Jan 23 at 16:12
@user3788135 $\cos (\theta +90^{\circ})<0 $ and $\cos (\theta+90^{\circ})=-|OB|/|OQ|=-|OB|/1=-|OB|$. – Américo Tavares Jan 23 at 22:27
I think OB has to be cos (90° - θ). – Yogesh Tripathi Jan 24 at 3:51

The point $P$ on the unit circle corresponding to $\theta$ is $\langle\cos\theta,\sin\theta\rangle$. Assume for a moment that $P$ is not on either coordinate axis; then it’s on the line $y=(\tan\theta)x=\frac{\sin\theta}{\cos\theta}x$.

The point $Q$ corresponding to $\theta+\frac{\pi}2$ is on the perpendicular line through the origin, whose slope is $-\frac{\cos\theta}{\sin\theta}$. But $Q$ is also the point $\left\langle\cos\left(\theta+\frac{\pi}2\right),\sin\left(\theta+\frac{\pi}2\right)\right\rangle$, so


The two sides of $(1)$ are evidently the two square roots of $\cot^2\theta$, and they have the additional property that the squares of numerator and denominator sum to $1$, so they must be identical except for sign. Thus, either




and consideration of quadrants rules out the former. The case in which $P$ is on a coordinate axis is easily handled separately.

share|cite|improve this answer
The point Q is in the second quadrant of the unit circle as the angle $\theta + {\pi \over 2}$ is obtuse (if im not mistaken?), shouldnt the corresponding cosine coordinate be negative? i.e. $ - cos(\theta + {\pi \over 2})$? – seeker Apr 2 '13 at 21:58
@Assad: $Q$ can be in any quadrant; it depends on $\theta$, which need not be in the first quadrant. If $\theta$ is in the first quadrant, $\sin\theta>0$, so $\cos\left(\theta+\frac{\pi}2\right)=-\sin\theta<0$, which is exactly what you want. – Brian M. Scott Apr 2 '13 at 22:00
+1 I strongly favor students learning trig functions in terms of the unit circle. – Calvin Lin Apr 2 '13 at 22:16
@Calvin: For a very first exposure the triangle does have the advantage of familiarity, but I certainly wish that there were a much greater emphasis on the unit circle approach early on. Sadly, of the three (triangle, circle, analytic) it was the one with which most of my students were least comfortable. – Brian M. Scott Apr 2 '13 at 22:33

Remember that sine and cosine are co-functions, meaning that they are connected through complementary angles (this is evident for $\newcommand{\dg}{^\circ} 0\dg \le \theta \le 90\dg$ from geometry of a right triangle). $$\begin{align} \cos \theta &= \sin(90\dg - \theta)\\ \sin \theta &= \cos(90\dg - \theta) \end{align} $$

Now, sine is an odd function, which means that $$\sin(-\theta) = -\sin\theta \qquad \text{for all } \theta.$$

Putting these together:

$$ \cos(\theta + 90\dg) = \cos(90\dg - (-\theta)) = \sin(-\theta) = -\sin\theta. $$

share|cite|improve this answer

Using Euler's formula we get: $$\cos(90 + \theta) = \cos (\pi/2 + \theta) = \frac{e^{i(\pi/2 + \theta)}+e^{-i(\pi/2 + \theta)}}{2} = \frac{e^{i\theta} e^{i\pi/2}+e^{-i\theta} e^{-i\pi/2}}{2} = \frac{ie^{i\theta} - ie^{-i\theta}}{2} = -\frac{e^{i\theta} - e^{-i\theta}}{2i} = -\sin(\theta) $$

share|cite|improve this answer
This is probably not elementary enough for someone taking precalculus and trigonometry for the first time :-) – oldrinb Apr 2 '13 at 21:56

$\sin$ is the vertical position. $\cos$ is the horizontal position. If you add 90 to the angle $\theta$, that's equivalent to rotating the coordinate system by 90°: the axes switch places.

share|cite|improve this answer

So you have cos(x+90)=-sin(x) You know that cos(90-x)=sin(x), and hopefully this make sense why (if not, imagine a right triangle with angles x and 90-x. Take the sin x and cos(90-x) and you will see they will be the same). You also know that cos (w)=cos(-w) (you can see why this is by imagining w and -w on the unit circle). So let's factor a -1 from cos(90-x). We are left with cos(x-90)=-sin(x). Now remember that cos(x)=-cos(x+180). So let's apply this to our problem. -cos(x-90+180)=sin(x), so cos(x+90)=-sin(x)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.