Question

sin 3Q + sin Q = 2 sin 2Q cos Q

Answer 1

Are we ask to prove it ???

Answer 2

Yes

Answer 3

Any guess ????

Answer 4

What do u mean by any guess?

Answer 5

i know i can split sin 3Q= sin(2Q+Q)

Answer 6

okay

Answer 7

but i am stucked after that.

Answer 8

just a moment

Answer 9

Here at right hand side just use trigonometric formula
sin(a) + sin(b) = 2* sin((a+b)/2) * cos((a-b)/2
Hope this will help you

Answer 10

i dont see any relation?

Answer 11

We are having left hand side as sin3Q + sinQ compare it with sina + sinb
So a = --- ? b = --- ?

Answer 12

so a = 3Q & b = Q plug in that formula .,,& simplify it

Answer 13

can someone please show me the steps?
sin3QcosQ + cos3QsinQ ?

Answer 14

Q=x
sin(3x)+sin(x)=2sin(2x)cos(x)

sin(3x)+sin(x)=2sin((3x+x)/2)cos((3x-x)/2)= =2sin(2x)cos(x)!!
I use the trigonometric
(sin(a+b)=2sin((a+b)/2)cos((a-b)/2)

Answer 15

SIN(2Q+Q)+SIN_(2Q-Q)

Answer 16

phaniraj61: your answer is 2sinQcosQ ?

Answer 17

You're trying to show that \[\sin(3x) + \sin(x) = 2\cos(x)\sin(2x)\]

You can expand \(\sin(3x)\) via identity \(\sin(u+v) = \sin(u)\cos(v) + \cos(u)\sin(v)\) with \(u=2x,v=x\):
\[\sin(3x) = \sin(2x+x) = \sin(2x)\cos(x) + \cos(2x)\sin(x)\]our left side becomes
\[\sin(2x)\cos(x) + \cos(2x)\sin(x)+\sin(x) = 2\cos(x)\sin(2x)\]Next, use the identity \[\cos(2x) = 2\cos^2(x)-1\]

\[\sin(2x)\cos(x) + (2\cos^2(x)-1)\sin(x)+\sin(x) = 2\cos(x)\sin(2x)\]\[\sin(2x)\cos(x) + 2\cos^2(x)\sin(x)\cancel{ -\sin(x)+\sin(x) }= 2\cos(x)\sin(2x)\]Now just use \[\sin(2x) = 2\sin(x)\cos(x)\]to convert the middle term into the right form and you're all done.