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Easy trig problem below:

Mathematics
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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

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Other answers:

taking the LCM and then using 2 sinA * cos A= sin 2A
and cos 2A= cos^2 2A -1
I'm not sure about your working but i completed till here: \[\frac{\cos A \sin 3A + \cos 3A \sin A}{\sin A \cos A}\]
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)
Hmm so we're taking advantage of Sine's Double Angle Formula, On top, in one direction, on the bottom in the other direction. Yah looks like Jason has the right idea also ^^
:)) thanks @zepdrix
I have this general formula with me: sin2A = 2sinAcosA But how did you guys converted sin4A into this?
Let's say A=X then its gona be sin (2X), now X=2A therefore its gona be the conversion mentioned above
I think i got it now. Thanks @jasonxx and @zepdrix . :)
:) 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:
So we're dividing by a fraction, let's fix that a sec. \[\frac{\sin(4A)}{\frac{1}{2}\sin(2A)} \quad = \quad \frac{2\sin(4A)}{\sin(2A)}\] Then we apply the Sine Double Angle Formula to the top part.\[\frac{2\sin(2\cdot 2A)}{\sin(2A)} \quad = \quad \frac{2\cdot 2 \sin(2A)\cos(2A)}{\sin(2A)}\]
Oh! Now it makes perfect sense to me!

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