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\[\frac{3A}{sinA} + \frac{\cos3A}{\cos A}≡ 4 \cos 2A\]

cos(3A) = cos(2A+A) =4 cos³ A - 3 cos A

taking the LCM and then using 2 sinA * cos A= sin 2A

and cos 2A= cos^2 2A -1

Then:
\[\frac{\sin(A+3A)}{\sin A \cos A}\]idk what to do after this..

well, i guess expand everything

\[\frac{\sin(4A)}{\frac{1}{2}\sin(2A)}\] This the next step? :) hmm fun problem.

\[\frac{4\sin(2A) \cos(2A)}{\sin(2A)}\]
\[=4\cos(2A)\]
Something like that... :D Yayyy team!

HOW? HOW? HOW?

sin(4A)=sin(2* 2A)

Oh like say what each step that's being performed? :D

Last 3 steps, yes sir. @zepdrix ;D

and sin (2*2A) is like sin (2* X)= 2*2 sin (2A)cos(2A)

:) keep rockin

Oh sorry. I wasn't going into much detail because I thought you were testing us :) lol

Lol. No. this was the last problem of 26. i had the answers but wasn't getting the last step!

@zepdrix Where did the 1/2 in your second last step go? D:

Oh! Now it makes perfect sense to me!