SOLUTION: cos2x+cos4x ----------- -- = sin2x-sin4x a. cot x b.-cot x c. sin x d.- tan x

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Question 1170448: cos2x+cos4x
-------------- =
sin2x-sin4x
a. cot x
b.-cot x
c. sin x
d.- tan x
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%28cos%282x%29%2Bcos%284x%29%29%2F+%28sin%282x%29-sin%284x%29%29

use identities:
sin%282x%29=+2sin%28x%29cos%28x%29
cos%282x%29=cos%5E2%28x%29+-+sin%5E2%28x%29
sin%284x%29=4sin%28x%29cos%5E3%28x%29+-+4sin%5E3%28x%29cos%28x%29
cos%284x%29=sin%5E4%28x%29+%2B+cos%5E4%28x%29+-+6sin%5E2%28x%29+cos%5E2%28x%29

%28cos%282x%29%2Bcos%284x%29%29%2F+%28sin%282x%29-sin%284x%29%29
=.............use sin%5E4%28x%29+%2B+cos%5E4%28x%29=2+cos%5E4%28x%29+-+2+cos%5E2%28x%29+%2B+1

=
= ......use cos%5E2%28x%29+-+sin%5E2%28x%29=cos%282x%29

= ....simplify

=

=...simplify

=cos%28x%29+%2F%28-sin%28x%29+%29

= -cot%28x%29




Answer by ikleyn(52864) About Me  (Show Source):
You can put this solution on YOUR website!
.

For much simpler and much more straightforward way to solve the problem see the solution under this link

https://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.1170383.html

https://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.1170383.html


The referred solution is the CANONICAL way to solve this problem.