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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.

mesel
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  • I started fixing the notation; can someone continue this? – Michael Hardy Mar 13 '14 at 17:05
  • 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? – Ben Voigt Mar 13 '14 at 17:05
  • oh,sorry I will fix it – mesel Mar 13 '14 at 17:07
  • 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 – imranfat Mar 13 '14 at 17:13
  • @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. – mesel Mar 13 '14 at 17:16
  • You like to brag... – evil999man Mar 13 '14 at 17:18
  • @Awesome:yes I am bragging things with the level of high school of math,I can suggest it to you ,it makes me more cool !!! – mesel Mar 13 '14 at 17:21
  • @mesel Do you know if there is any particular application of this identitiy in the realm of calculus? – imranfat Mar 13 '14 at 17:29
  • @imranfat:if you want to find derivative of $sin^2(x)$ by using defination of derivative,it would be helpful. – mesel Mar 13 '14 at 17:34

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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:

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

Ben Voigt
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