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When I was in high school, I had proved that $$\sin^2(x)-\sin^2(y)=\sin(x-y)\sin(x+y) $$

I think it is beautiful since it resembles the identity $a^2-b^2=(a+b)(a-b)$.

But I can not find it in sources,I think it should be known.(of course,it is not something remarkable just beatuful)

If anyone finds it in somewhere,I would be thankful also.

Here is the proof:

$$\sin(x+y)\sin(x-y)=(\sin(x)\cos(y)+\cos(x)\sin(y))(\sin(x)\cos(y)-\cos(x)\sin(y))$$ $$=\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)$$ $$=\sin^2(x)\cos^2(y)-\sin^2(y)(1-\sin^2(x))$$ $$=\sin^2(x)\cos^2(y)-\sin^2(y)+\sin^2(x)sin^2(y)$$ $$=\sin^2(x)(\cos^2(y)+\sin^2(y))-\cos^2(y)) $$ $$=\sin^2(x)-\sin^2(y)$$ If it is already known, it will not be surprise for me. I hope you would like this.

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  • $\begingroup$ I started fixing the notation; can someone continue this? $\endgroup$
    – Michael Hardy
    Mar 13, 2014 at 17:05
  • $\begingroup$ Some major typos in your proof... the term (sin(x)cos(y)−sin(x)cos(y)) is just zero. Second to last line has mismatched parentheses. I presume you did this right on paper and made a mistake typing it in? $\endgroup$
    – Ben Voigt
    Mar 13, 2014 at 17:05
  • $\begingroup$ oh,sorry I will fix it $\endgroup$
    – mesel
    Mar 13, 2014 at 17:07
  • $\begingroup$ It is a nice identity, but what is its purpose? Is there any direct application to it? That may answer your question why it is not found frequently in books. BTW, there are LOTS of cool trig identities. +1 though $\endgroup$
    – imranfat
    Mar 13, 2014 at 17:13
  • $\begingroup$ @imranfat: you can find $sin(15)$ by setting $x=45$ and $y=15$ of course you can find it in may way but it is very easy to remember. $\endgroup$
    – mesel
    Mar 13, 2014 at 17:16

1 Answer 1

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This identity appears in my CRC Standard Mathematical Tables in the Trigonometry chapter under the heading Angle-sum and angle-difference relations. Actually several similar ones appear:

$$\sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta = \cos^2 \beta - \cos^2 \alpha$$ $$\cos(\alpha + \beta) \cos(\alpha - \beta) = \cos^2 \alpha - \sin^2 \beta = \cos^2 \beta - \sin^2 \alpha$$

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