Question

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2x

Answer 1

Proving the identity, right?

Answer 2

Yeah

Answer 3

I would start with turning tangent and cotangent into sines and cosines.

Answer 4

the sin and cos are separated

Answer 5

Yeah, I know what you mean with how you typed it. Alright, so cosines cancel on the left side of your parenthesis and the right becomes cos^2(x)/sinx. Think you notice what to do after that, or not sure?

Answer 6

Do you think you could show me by like typing out because I think I know what to do but I just want to be certain ?

Answer 7

Sure

Answer 8

Sure\[sinx(sinx - \frac{ \cos ^{2}x }{ sinx }) = 1-2\cos ^{2}x\]
So thats what it should look like.

Answer 9

Ok then would you distribute the sin to the other sin on the left side

Answer 10

Yes, you would distribute sinx to both terms inside of the parenthesis.

Answer 11

ok so it then would be sin^2 x- cosx /sin^2x

Answer 12

Sorry. And when you distribute sinx, itll cancel out in the second term:

Answer 13

could you show me

Answer 14

\[sinx(sinx - \frac{ \cos ^{2x} }{ sinx }) = 1-2\cos ^{2}x\]
This becomes:
\[(\sin ^{2}x - \frac{ sinxcos ^{2}x }{ sinx }) = 1-2\cos ^{2}x\]

Can you see from here?

Answer 15

I think so

Answer 16

Awesome. So once sinx cancels on that right term, know that sin^2(x) + cos^2(x) = 1. Using that fact, you can do a substitution that will lead to the finality of your proof : ) I have to head out now unfortunately. Good luck ^_^