Question

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x...
I can solve it pretty far and then i get stumpped

Answer 1

i got down to...
sinx(sinx)- cos^2x/sinx

Answer 2

@agent0smith

Answer 3

You're on the right track. Just turn that (sinx - [cos^2(x)/sinx] part into one fraction.

Answer 4

so you have \[\Large \frac{ \sin^2 x }{ 1 } - \frac{ \cos^2 x }{ \sin x } = \frac{ \sin^2 x *\sin x}{ 1*\sin x } - \frac{ \cos^2 x }{ \sin x }\]

Answer 5

no...

Answer 6

Yeah, you do... that's the left hand side.

Answer 7

ok...

Answer 8

sin^3x-cos^2x/(sinx)?

Answer 9

sinx[sinx - [cos^2(x)/sinx]] becomes sinx[(sin^2(x) - cos^2(x))/sinx]. The sinx's then cancel and you go from there.

Answer 10

Wait, no you didn't distribute the sinx. I continued from where you were, which is missing a sinx.

Answer 11

what...what sinx's cancel...why

Answer 12

where is it missing the sin x :/