Question

Easy trig problem below:

Answer 1

?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

I'm not sure about your working but i completed till here:
\[\frac{\cos A \sin 3A + \cos 3A \sin A}{\sin A \cos A}\]

Answer 7

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

Answer 8

@jasonxx ?

Answer 9

well, i guess expand everything

Answer 10

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

Answer 11

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

Answer 12

HOW? HOW? HOW?

Answer 13

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

Answer 14

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

Answer 15

Last 3 steps, yes sir. @zepdrix ;D

Answer 16

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

Answer 17

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 ^^

Answer 18

:)) thanks @zepdrix

Answer 19

I have this general formula with me:

sin2A = 2sinAcosA
But how did you guys converted sin4A into this?

Answer 20

Let's say A=X then its gona be sin (2X), now X=2A therefore its gona be the conversion mentioned above

Answer 21

I think i got it now. Thanks @jasonxx and @zepdrix . :)

Answer 22

:) keep rockin

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.\[\frac{2\sin(2\cdot 2A)}{\sin(2A)} \quad = \quad \frac{2\cdot 2 \sin(2A)\cos(2A)}{\sin(2A)}\]

Answer 27

Oh! Now it makes perfect sense to me!