I need help proving this identity!

[battlestar]

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

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]

 

[Deleted member 4993]

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

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]

Hints:

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

and

\(\displaystyle sin^2(A) \ = \ \frac{1 \ - \ cos(2A)}{2}\)

 

[battlestar]

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

 

[soroban]

Hello, battlestar!

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

\(\displaystyle \text{Prove: }\;\sin x \sin(x+2) -\sin^2(x+1) \:=\: \frac{\cos(2)-1}{2}\)

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

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

. . \(\displaystyle =\;\sin^2\!x\cos 2 \;+\; \sin x\cos x\sin 2 \;-\; \sin^2\!x\cos^21 \;-\; 2\sin x\cos x\sin 1\cos 1 \;-\; \sin^2\!1\cos^2\!x\)

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

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

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

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

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

. . \(\displaystyle =\;-\sin^2\!x \sin^2\!1 - \sin^2\!1\cos^2\!x\)

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

. . \(\displaystyle =\;-\sin^2\!1\)

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

. . \(\displaystyle =\;\frac{\cos 2 - 1}{2}\)

 

[Deleted member 4993]

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

sin(X)sin(x+2)-sin^2(x+1) = [cos(2)-1]/[2]

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

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

Subtract (2) from (1)