# How does $\cos x \cos(y − x) − \sin y \sin(x − y)$ reduce to $\cos x$ as a function of one variable?

I'm working from a review book, and I can't figure out how $\cos (x) \cos(y − x) − \sin (y) \sin(x − y)$ reduces to just $\cos (x)$. In the review book, the steps go from the initial problem to $\cos [y + (x - y)]$ to $\cos (x)$. How do the sines get eliminated? Are the angles of the cosine really just factored?

At this point in the review book, the identities that have been covered are the Pythagorean, reciprocal, cofunction, identities for negatives, and addition, so solutions that use those would be the most helpful ... but I'll take anything.

Thanks for any help you can give!

• Do you know of the formula used to compute the sine or cosine or a difference? Commented Aug 16, 2017 at 18:58

Since$$(\forall\alpha,\beta\in\mathbb{R}):\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta,$$you have\begin{align*}\cos(x)&=\cos\bigl(y+(x-y)\bigr)\\&=\cos(y)\cos(x-y)-\sin(y)\sin(x-y).\end{align*}This is quite different from what you wrote. I suspect that the author of the text that you read meant $\cos(y)$ and not $\cos(x)$. If I'm right, then it's just a matter of noticing that\begin{align*}\cos(y)&=\cos\bigl(x+(y-x)\bigr)\\&=\cos(y)\cos(x-y)-\sin(y)\sin(x-y).\end{align*}
• and also, should be "+" instead of "-" because $\sin(y-x)=-\sin(x-y)$ Commented Aug 16, 2017 at 19:34
• Why $\cos(x)=\cos(x+(y-x))=\cos y$ ? Commented Aug 16, 2017 at 19:41
• Yes, I made a couple of typos. The original question was asked of alpha and beta; I should have never have tried to change them to x and y. The problem should have read $cos (y) cos (x - y) - sin (y) sin (x - y)$. Commented Aug 16, 2017 at 20:56