# Establish the identity of $\cos(\pi - \theta) = - \cos(\theta)$ [closed]

I need to establish the identity but am not sure how:

$$\cos(\pi - \theta) = - \cos(\theta).$$

• What facts about the cosine function do you already know? Is $\theta$ limited to $0\le\theta\le\pi$ or can it be any real number? What kind of answer do you want: geometric, using previous identities, using infinite series, using differential equations, or other? – Rory Daulton Mar 15 '16 at 18:05

In general it holds that: $$\cos(a – b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
For $a=\pi, b=θ$ we have: $$\cos(\pi – θ) = \cos(\pi)\cos(θ) + \sin(\pi)\sin(θ)$$ Since $\sin(\pi)=0$ and $\cos(\pi)=-1$ , you have the desired result.

• You can obtain $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, $\cot x$ by typing \sin x, \cos x, \tan x, \csc x, \sec x, and \cot x, respectively, when you are in math mode. – N. F. Taussig Mar 15 '16 at 17:42
• @N.F.Taussig Thank you for the tip! I am not very familiar with the commands yet to be honest! – MathematicianByMistake Mar 15 '16 at 17:43
• In that case, you might benefit from reading this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 15 '16 at 17:46
• @N.F.Taussig Yet again, thank you! It seems very comprehensive and exhaustive of the subject. – MathematicianByMistake Mar 15 '16 at 17:50

It follows from the difference formula for the cosine that

\begin{align*}\cos{(\pi - \theta)} & ~=~ \cos{\pi}\cos{\theta} + \sin{\pi}\sin{\theta} \\ & ~=~ -\cos{\theta} \end{align*}

We have $$\cos(\pi-\theta)=\Re e^{i(\pi-\theta)}=\Re (e^{i\pi}e^{-i\theta})=\Re (-(\cos(\theta)-i\sin(\theta)))=-\cos\theta$$

• That's a more sleek way to do it. +1.. – MathematicianByMistake Mar 15 '16 at 17:45