I need help proving this identity!

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battlestar

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I do not even know where to begin...please help!

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]
 
battlestar said:
I do not even know where to begin...please help!

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]
Click to expand...

Hints:

sin(A)sin(B) = cos(A−B) − cos(A+B)2\displaystyle sin(A)sin(B) \ = \ \frac{cos(A-B) \ - \ cos(A+B)}{2}

and

sin2(A) = 1 − cos(2A)2\displaystyle sin^2(A) \ = \ \frac{1 \ - \ cos(2A)}{2}
 
Thank you so much, but I am still stuck...now I have

[cos(2) - cos(2x+2)-1+cos(2*x+1*2)] / 2

Is that correct? And if so what next?

Notice the correction above - and simplify
 
Hello, battlestar!

I did this one head-on . . . It takes a while, though.

Prove:   sin⁡xsin⁡(x+2)−sin⁡2(x+1) = cos⁡(2)−12\displaystyle \text{Prove: }\;\sin x \sin(x+2) -\sin^2(x+1) \:=\: \frac{\cos(2)-1}{2}
Click to expand...

\(\displaystyle \text{We have: }\:\sin x\son(x+2) - \sin^2(x+1)\)

. . =  sin⁡x(sin⁡xcos⁡2+sin⁡2cos⁡x)−(sin⁡xcos⁡1+sin⁡1cos⁡x)2\displaystyle =\;\sin x(\sin x\cos 2 + \sin 2\cos x) - (\sin x\cos 1 + \sin 1\cos x)^2

. . =  sin⁡2 ⁣xcos⁡2  +  sin⁡xcos⁡xsin⁡2  −  sin⁡2 ⁣xcos⁡21  −  2sin⁡xcos⁡xsin⁡1cos⁡1  −  sin⁡2 ⁣1cos⁡2 ⁣x\displaystyle =\;\sin^2\!x\cos 2 \;+\; \sin x\cos x\sin 2 \;-\; \sin^2\!x\cos^21 \;-\; 2\sin x\cos x\sin 1\cos 1 \;-\; \sin^2\!1\cos^2\!x

. . \(\displaystyle =\;\left(\sin^2\!x\cos2 - \sin^2\!x\cos^2\!1) + (\sin x\cos x\sin 2 - 2\sin x\cos x\sin1\cos1) - \sin^2\!1\cos^2\!x\)

. . \(\displaystyle =\;\sin^2\!x\left(\cos 2 - \cos^2\!1\right) + \sin x \cos x\,(\sin 2 - \underbrace{2\sin 1\cos 1}_{\text{This is }\sin 2}) - \sin^2\@1\cos^2\!x\)

. . =  sin⁡2 ⁣x(cos⁡2−1+cos⁡22)+sin⁡xcos⁡x(sin⁡2−sin⁡2)⏟This is 0−sin⁡2 ⁣1cos⁡2 ⁣x\displaystyle =\;\sin^2\!x\left(\cos 2 - \frac{1+\cos 2}{2}\right) + \sin x\cos x\underbrace{(\sin 2 - \sin 2)}_{\text{This is 0}} - \sin^2\!1\cos^2\!x

. . =  sin⁡2 ⁣x(cos⁡2−12)−sin⁡2 ⁣1cos⁡2 ⁣x\displaystyle =\;\sin^2\!x\left(\frac{\cos 2-1}{2}\right) - \sin^2\!1\cos^2\!x

. . =  −sin⁡2 ⁣x(1−cos⁡22)−sin⁡2 ⁣1cos⁡2 ⁣x\displaystyle =\;-\sin^2\!x\left(\frac{1-\cos 2}{2}\right) - \sin^2\!1\cos^2\!x

. . =  −sin⁡2 ⁣xsin⁡2 ⁣1−sin⁡2 ⁣1cos⁡2 ⁣x\displaystyle =\;-\sin^2\!x \sin^2\!1 - \sin^2\!1\cos^2\!x

. . =  −sin⁡2 ⁣1(sin⁡2 ⁣x+cos⁡2 ⁣x)⏟This is 1\displaystyle =\;-\sin^2\!1 \underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}

. . =  −sin⁡2 ⁣1\displaystyle =\;-\sin^2\!1

. . =  −(1−cos⁡22)\displaystyle =\;-\left(\frac{1 - \cos2}{2}\right)

. . =  cos⁡2−12\displaystyle =\;\frac{\cos 2 - 1}{2}

 
battlestar said:
I do not even know where to begin...please help!

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]
Click to expand...

sin(x)∗sin(x+2) = 12[cos(x+2−x)−cos(x+2+x)]\displaystyle sin(x)*sin(x+2) \ = \ \frac{1}{2}[cos(x+2-x) - cos(x+2+x)] .....................................(1)

sin2(x+1) = 12[1−cos(2x+2)]\displaystyle sin^2(x+1) \ = \ \frac{1}{2}[1-cos(2x+2)].........................................................(2)

Subtract (2) from (1)
 
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