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!

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

1 Answer 1


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*}

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    $\begingroup$ and also, should be "+" instead of "-" because $\sin(y-x)=-\sin(x-y)$ $\endgroup$
    – Arnaldo
    Commented Aug 16, 2017 at 19:34
  • $\begingroup$ Why $\cos(x)=\cos(x+(y-x))=\cos y$ ? $\endgroup$
    – g.kov
    Commented Aug 16, 2017 at 19:41
  • $\begingroup$ @g.kov My mistake. I will edit my answer, As a result, the formula that you read was even worse than what it seemed to me at first sight. $\endgroup$ Commented Aug 16, 2017 at 20:47
  • $\begingroup$ 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 $cos (y) cos (x - y) - sin (y) sin (x - y)$. $\endgroup$ Commented Aug 16, 2017 at 20:56
  • $\begingroup$ I see. I should have considered the angle (x - y) as a unit rather than something to be expanded / simplified by using the difference identities upon. If (x - y) had been (z) or (theta), it would have been evident. Thanks to everyone who looked at this problem, for all its mistakes! $\endgroup$ Commented Aug 17, 2017 at 3:40

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