Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I need help to prove that:


I know that $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$, but I don't know how to get to this step without memorizing this, can you possibly draw some sort of triangle to get this information? I only know the proof that $\cos^2(x)+\sin^2(x)=1$ by using the Pythagoras theorem on a right-angled triangle.

share|cite|improve this question
"I don't know how to get to this step without memorizing this"... your use of pronouns makes it hard to tell where you need help. – The Chaz 2.0 Oct 1 '13 at 12:43
up vote 3 down vote accepted

First, you have these identities: $$\cos({\pi \over 2}+x)=-\sin(x),\ \sin({\pi \over 2}+x)=\cos(x)$$

Now consider two right angled triangles glued on their opposites, making a big triangle. With that, do some labeling on the triangles as shown in the picture attached, you can find the areas of the triangles. Triangle (The area of a triangle is ${1 \over 2}AB\sin(C)$, with $C$ being the included angle in between $A$ and $B$.)

Note that the area of the big triangle is the sum of the areas of the small triangles. $$\therefore {1 \over 2}ae\sin(x+y)={1 \over 2}ac\sin(x)+{1 \over 2}ce\sin(y)$$ $$\therefore \sin(x+y)={{ac}\over {ae}}\sin(x)+{{ce}\over {ae}}\sin(y)$$ $$\because {{ac}\over {ae}}={{c}\over {e}}=\cos(y),\ {{ce}\over {ae}}={{c}\over {a}}=\cos(x)$$ $$\therefore \sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$$ Now, you replace $x$ by ${\pi \over 2} +x$. $$\therefore \sin({\pi \over 2} +x+y)=\cos(y)\sin({\pi \over 2} +x)+\cos({\pi \over 2} +x)\sin(y)$$ With the identities on top, you will get $$\cos(x+y)=\cos(y)\cos(x)-\sin(x)\sin(y)$$ Using this formula and also ${\cos}^2(x)+{\sin}^2(x)=1$, you can get the required double angle for $cosine$. The identities in the post can be deduced: From observation of a single right angled triangle, $$\cos({\pi \over 2}-x)=\sin(x),\ \cos({\pi \over 2}-x)=\sin(x)$$ Replace $x$ by $-x$ and note that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, ($\because$ $sine$ and $cosine$ are odd and even functions respectively) you will have the identities. :)

share|cite|improve this answer
Hint: use \sin for a non-italicized $\sin$ – The Chaz 2.0 Oct 1 '13 at 12:41
This is exactly the answer I was looking for. Thanks for the quick response! – Ryan Oct 1 '13 at 19:14

In $cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$ put $\beta=\alpha$. Then $cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)=\cos^2(\alpha)-(1-\cos^2(\alpha))=2\cos^2(\alpha)-1.$.

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.