Question

Prove the identity :
sin (x-y)sin(x+y) = sin^2 x - sin^2 y

Answer 1

@ybarrap

Answer 2

\[\sin(A+B)=\sin A\cos B+\cos A\sin B\\\sin(A-B)=\sin A\cos B-\cos A\sin B\]

Answer 3

use
sin(a-b)=sin(a) cos(b)-cos(a) sin(b)
and
sin(a+b)=sin(a) cos(b)+cos(a) sin(b)
$$
\left(\sin(a) \cos(b)-\cos(a) \sin(b)\right)\times \left (\sin(a) \cos(b)+\cos(a) \sin(b)\right )
$$

Answer 4

\[\alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta)\]

Answer 5

perfect!

Answer 6

take
$$
\alpha=\sin a\cos b\\
\beta=\cos a \sin b
$$

Answer 7

okayy sorry I haven't gotten much to those symbols yet here

Answer 8

what do they represent in this problem?

Answer 9

They denote that for any \(\alpha,\beta,A,B\), the formulae still hold.

Answer 10

For example, \(\sin^2\theta+\cos^2\theta=1\) means that for any \(\theta\) the formula holds

Answer 11

I know it's not proper grammar but yeah

Answer 12

okayyy

Answer 13

I thought I was using x and y

Answer 14

thnx for the help though

Answer 15

okay sorry for the wrong hints, here is the answer:

Answer 16

\[\begin{array}{cl}
&\sin^2x-\sin^2y\\
=&(\sin x+\sin y)(\sin x-\sin y)\\
=&4\sin\frac{x+y}2\cos\frac{x-y}2\cos\frac{x+y}2\sin\frac{x-y}2\\
=&\left(2\sin\frac{x+y}2\cos\frac{x+y}2\right)\left(2\sin\frac{x-y}2\cos\frac{x-y}2\right)\\
=&\sin(x+y)\sin(x-y)
\end{array}\]

Answer 17

awesome thank you so much! @ybarrap I don't know if you're still typing or my computer froze but thank you too :)

Answer 18

I have two other questions which I'd like you guys to check out but I will write them in a different post

Answer 19

And if you want to try a longer version:
$$
\alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta)
$$
Is the same as
$$
\left(\sin(a) \cos(b)-\cos(a) \sin(b)\right)\times \left (\sin(a) \cos(b)+\cos(a) \sin(b)\right ) \star
$$
Where,
$$
\alpha=\sin a\cos b\\
\beta=\cos a \sin b
$$
This means that when you multiply out \(\star\)

$$
\sin(a+b)\sin(a-b)\\
\left(\sin(a) \cos(b)-\cos(a) \sin(b)\right)\times \left (\sin(a) \cos(b)+\cos(a) \sin(b)\right )\\
=(\sin a\cos b)^2-(\cos a \sin b)^2\\
\sin^2a\cos^2b-\cos^2a\sin^2b\\
(1/2 (1-\cos(2 a)))1/2 (cos(2 b)+1)-1/2 (cos(2 a)+1)(1/2 (1-\cos(2 b)))\\
=(1/4)((\cos 2b+1-\cos(2a)\cos(2b)-\\
\cos(2a)-\cos(2a)+\cos(2a)\cos(2b)-1+\cos(2b))\\
=(1/4)(2\cos 2b-2\cos2a)\\
=(1/2)(\cos 2b-\cos2a)\\
=(1/2)(cos^2(b)-sin^2(b) -cos^2(a)+sin^2(a))\\
=(1/2)( 1-sin^2(b)-sin^2b -1+sin^2a+sin^2a)\\
=(1/2)(-2\sin^2b+2\sin^2a)\\
=-\sin^2b+\sin^2a\\
$$

Answer 20

wow thank you both sooo much!! I think I'm understanding this better :)