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saifoo.khan
 3 years ago
Easy trig problem below:
saifoo.khan
 3 years ago
Easy trig problem below:

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

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

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

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

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

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

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

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

zepdrix
 3 years ago
Best 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!

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

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

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Hmm 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 ^^

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

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

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

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

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

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

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

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