Question

-cotx sin(-x) = cosx

Answer 1

Well, this uses the fact that sin(x) is an odd function. sin, tan, cot, and csc are all odd functions, meaning f(-x) = -f(x). This is the case with sin(x), too. SO basically, sin(-x) = -sinx. Factoring the negative out of the angle cancels the negative in front of cotx, leaving you with simply
cot(x)sin(x). Now just turn cot into terms of sines and cosines.

Answer 2

Thank you cause I wasn't sure how to get rid of the negative. What next?

Answer 3

-cotx(-sinx) = cosx
LHS

-cosx/sinx(-sinx) = cosx [cotx = cosx/sinx]

Answer 4

minus sign & sinx gets cancelled

Answer 5

Right now now I have cosx/sinx sinx = cosx. What next?

Answer 6

it's like this\[\frac{ cosx }{ sinx } \]

Answer 7

\[\frac{ cosx }{sinx } \times sinx\]

Answer 8

sinx cancels

Answer 9

Oh I got ya. The two sines multiply together leaving cosx = cosx

Answer 10

yep , u got it !!!

Answer 11

You sir are the man! I have a few more problems will you check em out and see what you can do?

Answer 12

Alright , I've got a few minutes

Answer 13

1-2cos^2x + cos^4x = sin^4x

Answer 14

is this ur question ?\[1 - 2\cos ^{2}x + \cos ^{4}x = \sin ^{4}x\]

Answer 15

Yea that's it

Answer 16

\[1-2\cos ^{2} + \cos ^{^{4}}x\]
\[1-\cos ^{2}x - \cos ^{2}x +\cos ^{4}x\]
\[1 - \cos ^{2} - \cos ^{2}x (1 - \cos ^{2}x)\]
\[1 - \cos ^{2}x - \cos ^{2}x(\sin ^{2}x)\] [1 - cos^2x = sin^2x]
\[\sin ^{2} + \cos ^{2}x - \cos ^{2}x - \cos ^{2}xsin ^{2}x\] [1 = sin^2x + cos^2x]
\[\sin ^{2}x - \cos ^{2}xsin ^{2}x\]
\[\sin ^{2}x(1- \cos ^{2}x)\]
sin^2x * sin^2x = sin^4x

Answer 17

WOw that was more than I expected. So cos^4x is the same as saying (1-cos^2x)?