Proof of a trigonometric identity

Question

I need to prove that: $$1+\cos(a+b)-\cos(a-b)=\cos^2(a)+\cos^2(b)$$

I tried to start from both ways but it always took me back to the same one again I tried also to prove that their - equal zero but I didn't work also

Answer 1

I think the right expression should be $$1+\cos(a+b)\cos(a-b)=\cos^2(a)+\cos^2(b)$$

And here is the solution.

Let $$F=\sin^2(a)-\cos^2(b)$$ and $$G=\sin^2(b)-\cos^2(a)$$

Notice that $F=G$ since $\sin^2(a)-\cos^2(b)=1-\cos^2(a)-\cos^2(b)=\sin^2(b)-\cos^2(a)$

Thus, $$F+G=2F=(\sin^2(a)-\cos^2(a))+(\sin^2(b)-\cos^2(b))$$

$$2F=-(\cos(2a)+\cos(2b))=-2\cos(a+b)\cos(a-b)$$ $$F=-\cos(a+b)\cos(a-b)$$

$$\sin^2(a)-\cos^2(b)=-\cos(a+b)\cos(a-b)$$ $$1+\cos(a+b)\cos(a-b)=\cos^2(a)+\cos^2(b)$$

Answer 2

Your posted "identity" is not an identity. We do have that $$\begin{align} 1 + \cos(a+b) - \cos (a-b) & = 1 + (\cos a \cos b - \sin a \sin b) - (\cos a \cos b + \sin a \sin b)\\ \\ & = 1-2\sin a \sin b\end{align}$$

But in general (in terms of arbitrary $a, b$: $$1 - 2\sin a \sin b \neq \cos^2a +\cos^2 b$$

Answer 3

A simpler approach to show this is not an identity:

$$1+\cos(a+b)-\cos(a-b)=\cos^2(a)+\cos^2(b)$$

Let $a=b=0$:

$$1+\cos(0)-\cos(0)=\cos^2(0)+\cos^2(0)$$

$$1+1-1=1+1$$

$$1=2$$

So that's not an identity.