Question

Trigonometry Help

Prove the Identity

sin 6x tan 3x = 2sin^2 3x

This just isn't making too much sense to me :\

Answer 1

I've tried to work the problem a bit but i just dont think its correct to me

Answer 2

half angle identity is useful here

Answer 3

see i don't think ive ever learned anything about half angle identity, people keep saying it but it just doent ring any bells in my head.

Answer 4

You need a double-angle identity.

sin(2x) = 2sin(x)cos(x)

Use the above identity for sin 6x on the left side.

Think of sin 6x as sin (2 * 3x)

Answer 5

kinda weird but it works out most beautifully. i'm sure someone with the patience to use latex will explain it better than i can

Answer 6

alright thanks, let me give that a try mathstudent

Answer 7

erm not gettin it :\

Answer 8

Double angle identity: \( \sin 2x = 2 \sin x \cos x\)

\( \sin 6x \tan 3x ~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

Think of \(\sin 6x\) as \(\sin (2 \times 3x) \)

\(\sin (2 \times 3x) \tan 3x ~~~~~~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

\(2 \sin 3x \cos 3x \tan 3x ~~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

\(2 \sin 3x \cos 3x \dfrac{\sin 3x}{\cos 3x} ~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

\(2 \sin 3x \cancel{\cos 3x} \dfrac{\sin 3x}{\cancel{\cos 3x}~~1} ~~~~~~~~= 2 \sin^2 3x\)

\(2 \sin 3x \sin 3x ~~~~~~~~~~~~~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

\(2 \sin^2 3x ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 2 \sin^2 3x\)

Answer 9

you left but yeah i guess that makes sense, thank you very much

Answer 10

You're welcome.