Question

Prove the following identities:

Answer 1

sin(x-y)sin(x+y) = sin^2x - sin^2y

Answer 2

Here's what I have so far (not sure if it's right...)
sin(x-y)sin(x+y) = (sinxcosy - cosxsiny)(sinxcosy + cosxsiny)
(sinxcosy)^2 - (cosxsiny)^2
Then I'm thinking of using the product to sum identities... how am I doing so far? It seems a bit too complicated, but I dunno.

Answer 3

[sin(x)cos(y) - cos(x)sin(y)][sin(x)cos(y)] + cos(x)sin(y)] = sin^2 (x) - sin^2 (y)

you should know that (a-b)(a+b) = a^2 - b^2 so:

[sin(x)cos(y)]^2 - [cos(x)sin(y)]^2 = sin^2 (x) - sin^2 (y)

sin^2 (x) [ 1 - sin^2 (y)] - sin^2 (y) [1- sin^2 (x)] = sin^2 (x) - sin^2 (y)

sin^2 (x) - sin^2 (y)sin^2(x) - sin^2(y) + sin^2 (y)sin^2(x) = sin^2 (x) - sin^2 (y)

final result is:

sin^2(x) - sin^2 (y) = sin^2 (x) - sin^2 (y)

I hope you understand

Answer 4

Thank you so much! That is very helpful, I'll go through it. Do you mind looking over the steps I tried and seeing if I was doing it correctly or not? I posted what I had so far above.

Answer 5

From this point,\[\large\rm (\sin x \cos y)^2 - (\cos x \sin y)^2\]\[\large\rm \sin^2x \color{orangered}{\cos^2y}-\color{orangered}{\cos^2x}\sin^2y\]you need to get rid of all the cosines, ya?
So I would apply your `Pythagorean Identity` to these orange parts.

Answer 6

what you did it's OK.... if you have a doubt about my solution method, let me know please my friend

Answer 7

sin^2x(1 - sin^2y) - (1 - cos^2x)(sin^2y)
Like this zep?

Answer 8

Whoops, that should be (1 - sin^2x) in the second part ;)

Answer 9

ya ya ya, good stuff

Answer 10

and mooooore distributing D:
stuff should cancel out.

Answer 11

sin^2x(1 - sin^2y) - (1 - sin^2x)(sin^2y)
sin^2x - sin^2y(sin^2x) - (sin^2y - sin^2x(sin^2y))
It looks like a mess :P But do you know what to do from here?

Answer 12

yes

Answer 13

abbles, I am providing you of a word archive with the specific points, every single step well explained
do you agree?

Answer 14

hmmm.. what?

Answer 15

wait a second, I'm starting solving from your post

Answer 16

what you need to do from there is just apply distributive property, then you'll note there is a possitive\[\sin^2 (x)\sin^2 (y)\] and a negative one, so just cancel them out

Answer 17

So like this?
sin^2x - sin^2ysin^2x - sin^2y + sin^2xsin^2y
I see what you mean!
I would be left with sin^2x - sin^2y
Exactly like the original :) Thanks!

Answer 18

yay!

Answer 19

dare yourself! try to get the left expression starting from sin^2 (y) - sin^2(x) at the beginning... are you able to do it, abbles?