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saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{3A}{sinA} + \frac{\cos3A}{\cos A}≡ 4 \cos 2A\]
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
cos(3A) = cos(2A+A) =4 cos³ A  3 cos A
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
taking the LCM and then using 2 sinA * cos A= sin 2A
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
and cos 2A= cos^2 2A 1
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure about your working but i completed till here: \[\frac{\cos A \sin 3A + \cos 3A \sin A}{\sin A \cos A}\]
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Then: \[\frac{\sin(A+3A)}{\sin A \cos A}\]idk what to do after this..
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
@jasonxx ?
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
well, i guess expand everything
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{\sin(4A)}{\frac{1}{2}\sin(2A)}\] This the next step? :) hmm fun problem.
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{4\sin(2A) \cos(2A)}{\sin(2A)}\] \[=4\cos(2A)\] Something like that... :D Yayyy team!
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
HOW? HOW? HOW?
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
sin(4A)=sin(2* 2A)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Oh like say what each step that's being performed? :D
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Last 3 steps, yes sir. @zepdrix ;D
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
and sin (2*2A) is like sin (2* X)= 2*2 sin (2A)cos(2A)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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 ^^
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
:)) thanks @zepdrix
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
I have this general formula with me: sin2A = 2sinAcosA But how did you guys converted sin4A into this?
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
Let's say A=X then its gona be sin (2X), now X=2A therefore its gona be the conversion mentioned above
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
I think i got it now. Thanks @jasonxx and @zepdrix . :)
 2 years ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
:) keep rockin
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Oh sorry. I wasn't going into much detail because I thought you were testing us :) lol
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Lol. No. this was the last problem of 26. i had the answers but wasn't getting the last step!
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix Where did the 1/2 in your second last step go? D:
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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.\[\frac{2\sin(2\cdot 2A)}{\sin(2A)} \quad = \quad \frac{2\cdot 2 \sin(2A)\cos(2A)}{\sin(2A)}\]
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Oh! Now it makes perfect sense to me!
 2 years ago
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