Question

2(cos((x)/(2))cos((y)/(2))-sin((x)/(2))sin((y)/(2)))(sin((x)/(2))cos((y)/(2))-cos((x)/(2))sin((y)/(2)))=sin(x)-sin(y)

verify

Answer 1

Could you draw that using the drawing tool provided just below the comment box. {down the left of your comment box}

Answer 2

\[2(\cos \frac{ x }{ 2 }\cos \frac{ y }{ 2 }-\sin \frac{ x }{ 2}\sin \frac{ y }{ 2})(\sin\frac{ x }{ 2 }\cos \frac{ y }{ 2 }-\cos \frac{ x }{ 2 }\sin \frac{ y }{ 2 })=\sin x-\sin y\]

Answer 3

Using your trig angle identities:
\[\large \cos (a+b)=\cos a\cos b -\sin a\sin b\]
\[LHS=\cos \frac{x}{2}\cos \frac{y}{2}-\sin \frac{x}{2}\sin \frac{y}{2}=\cos(\frac{x}{2}+\frac{y}
{2})\]
\[\large \sin (a-b)=\sin a\cos b -\cos a\sin b\]
\[\sin \frac{x}{2}\cos \frac{y}{2}-\cos \frac{x}{2}\sin \frac{y}{2}=\sin (\frac{x}{2}-\frac{y}{2})\]
Since:
\[\cos(a+b)\sin(a-b)=\sin (a)\cos (a)-\sin(b)\cos(b)\]
Therefore you have this:
\[\large 2\cos(\frac{x}{2}+\frac{y}{2})\sin (\frac{x}{2}-\frac{y}{2})=2(\sin \frac{x}{2}\cos \frac{x}{2}-\sin\frac{y}{2}\cos\frac{y}{2})\]
\[\large =2\sin \frac{x}{2}\cos \frac{x}{2}-2\sin\frac{y}{2}\cos\frac{y}{2}\]
Using double angle results for sine:
\[\large \sin2x =2\sin x\cos x\]
\[\large =\sin 2[\frac{x}{2}] -\sin 2[\frac{y}{2}]\]
\[\large =\sin x-\sin y\]
\[\large =RHS\]

Answer 4

Forgot a bracket in the 6th last line. There should be an extra bracket on the LHS.

Answer 5

Sorry about that.

Answer 6

Excellent!!!! Thanks so much...I was thinking that I would have to break each one down but the deeper into the rabbit hole I went the more convoluted it became!