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

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

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

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
and cos 2A= cos^2 2A 1
 one year 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: \[\frac{\cos A \sin 3A + \cos 3A \sin A}{\sin A \cos A}\]
 one year 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..
 one year ago

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

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

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

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

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

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

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

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

jasonxx Group TitleBest ResponseYou've already chosen the best response.2
:) keep rockin
 one year 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
 one year 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!
 one year 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:
 one year 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.\[\frac{2\sin(2\cdot 2A)}{\sin(2A)} \quad = \quad \frac{2\cdot 2 \sin(2A)\cos(2A)}{\sin(2A)}\]
 one year ago

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