1. ## show identity

hey im trying to show that $\displaystyle sin(X+Y)sin(X-Y) = sin^2X-sin^2Y$ .

any idea how i can go with it?

basically firstly i have:

$\displaystyle sin(X+Y)sin(X-Y) = sinXcosY+cosXsinY \cdot sinXcosY-cosXsinY =$

then where can i go? cheers

2. Originally Posted by jvignacio
hey im trying to show that $\displaystyle sin(X+Y)sin(X-Y) = sin^2X-sin^2Y$ .

any idea how i can go with it?

basically firstly i have:

$\displaystyle sin(X+Y)sin(X-Y) = {\color{red}(}sinXcosY+cosXsinY{\color{red})} \cdot {\color{red}(}sinXcosY-cosXsinY{\color{red})} =$

then where can i go? cheers
First, I put needed parentheses into your setup.

This can be foiled out easily, since it can be recognized as the difference of two squares. You should have

\displaystyle \begin{aligned}\sin^2X\cos^2Y-\cos^2X\sin^2Y&=\sin^2X\left(1-\sin^2Y\right)-\left(1-\sin^2X\right)\sin^2Y\\&=\sin^2X-\sin^2X\sin^2Y-\sin^2Y+\sin^2X\sin^2Y\\&=\sin^2X-\sin^2Y\end{aligned}

Does this make sense?

3. Originally Posted by Chris L T521
First, I put needed parentheses into your setup.

This can be foiled out easily, since it can be recognized as the difference of two squares. You should have

\displaystyle \begin{aligned}\sin^2X\cos^2Y-\cos^2X\sin^2Y&=\sin^2X\left(1-\sin^2Y\right)-\left(1-\sin^2X\right)\sin^2Y\\&=\sin^2X-\sin^2X\sin^2Y-\sin^2Y+\sin^2X\sin^2Y\\&=\sin^2X-\sin^2Y\end{aligned}

Does this make sense?
ahh yes yes that does make sence but quick question. Does $\displaystyle cos^2X = (1-sin^2X)$ ?

4. Originally Posted by jvignacio
ahh yes yes that does make sence but quick question. Does $\displaystyle cos^2X = (1-sin^2X)$ ?
Yes, since $\displaystyle \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X$

5. Originally Posted by Chris L T521
Yes, since $\displaystyle \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X$
of course simple algebra. just gotta know your identities..

thanks Chris

6. Originally Posted by Chris L T521
Yes, since $\displaystyle \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X$
Hi chris quick question.

what is $\displaystyle (sinXcosY) \cdot (-cosXsinY)$ ?

cheers

7. Originally Posted by jvignacio
Hi chris quick question.

what is $\displaystyle (sinXcosY) \cdot (-cosXsinY)$ ?

cheers
Its just $\displaystyle -\sin X\cos X\sin Y\cos Y$.