Question

Trigonometric Identities
Proof for \[\tan(x+y)\tan(x-y)=\frac{ \sin^2x-\sin^2y }{ \cos^2x-\sin^2y }\]

Answer 1

I tried and I used the left hand side one. I got this so far...
\[\frac{ (\sin^2x)(\cos^2y)-(\sin^2y)(\cos^2x) }{ (\cos^2x)(\cos^2y)-(\sin^2x)(\sin^2y) }\]

Answer 2

how did you get that for the left hand side
what identities did you use to get that?

Answer 3

Well, I first use tan(x+y) and tan(x-y) and expand them

Answer 4

Then I change all the tan into sin/cos

Answer 5

oh say you wrote in terms of sin and cos first then use sum and difference identity for cos and sin?

Answer 6

\[\frac{\sin(x+y)}{\cos(x+y)} \cdot \frac{\sin(x-y)}{\cos(x-y)} \\ \frac{\sin(x)\cos(y)+\sin(y)\cos(x)}{\cos(x)\cos(y)-\sin(x)\sin(y)} \cdot \frac{\sin(x)\cos(y)-\sin(y)\cos(x)}{\cos(x)\cos(y)+\sin(x)\sin(y)} \]
so you are multiplying conjugates so what you have should be correct so far unless I also made mistake :p

Answer 7

ok so we need to go a little further from what you have above ...
\[\frac{ (\sin^2x)(\cos^2y)-(\sin^2y)(\cos^2x) }{ (\cos^2x)(\cos^2y)-(\sin^2x)(\sin^2y) } \]
well we want the top just in terms of sin
so let's try to use pythagorean theorem on top for cos^2(y) and cos^2(x)

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

Answer 8

so we see we have the correct top not using Pythagorean identities

Answer 9

we can probably do something similar to the bottom

Answer 10

oh I see! :)

Answer 11

yep just tried it
I chose on bottom to rewrite the cos^2(y) and the sin^2(x) since our bottom didn't want those