cos x sin y = 1/2 sin(sin(x + y) – sin (x - y))?

[baselramjet]

cos x sin y = 1/2 sin(sin(x + y) – sin (x - y))?

Can someone explain to me how I can prove that cos x sin y = 1/2 sin(sin(x + y) – sin (x - y))?

 

[galactus]

Use the addition and subtraction formulas and subtract.

\(\displaystyle \L\\sin(x+y)=sin(x)cos(y)+cos(x)sin(y)\)
\(\displaystyle \L\\sin(x-y)=sin(x)cos(y)-cos(x)sin(y)\)

Now, subtract.

 

[baselramjet]

Thats the part I am a bit fuzzy on... after applying both addition and subtraction identities

Ashley

 

[soroban]

Re: cos x sin y = 1/2 sin(sin(x + y) – sin (x - y))?

Hello, baselramjet!

Exactly where is your difficulty?

Prove that: \(\displaystyle \:\cos x\cdot\sin y \:= \:\frac{1}{2}\,\left[\sin(x\,+\,y)\,-\,\sin(x\,-\,y)\right]\)

We have the compound-angle identity: \(\displaystyle \:\sin(A\,\pm\,B)\;=\;\sin A\cdot\cos B\,\pm\,\cos A\cdot\sin A\)

We have: \(\displaystyle \:\begin{array}{cc}[1]\\ [2]\end{array}\;\begin{array}{ccc}\sin(x\,+\,y) & \,=\, & \sin x\cdot\cos y \,+\,\cos x\cdot\sin x \\ \sin(x\,-\,y) & = & \sin x\cdot\cos y \,-\, \cos x\cdot\sin y\end{aray}\)

Subtract \(\displaystyle [2]\) from [1]: \(\displaystyle \:\sin(x\,+\,y)\,-\,\sin(x\,-\,y) \:=\:2\cdot\cos x\cdot\sin y\)
. . You're fuzzy on a subtraction?

Therefore: \(\displaystyle \:\cos x\cdot\sin y \:=\:\frac{1}{2}\,\left[\sin(x\,+\,y)\,-\,\sin(x\,-\,y)\right]\)

 

[baselramjet]

Soroban

That makes sense now! I am a little slow when it comes to this stuff..its nice when people can draw me "stick figures" as explanations! LOL

Thank you!!!

Ashley