Question

I GIVE MEDALS AND I RLYY NEED HELP?!? PLZ JUST TRY IT EVEN IF U CANT FINISH IT WILL STILL HELP ME OUT :) THANKS
sin6x= 2sinxcosx(3-4sin^2 x)(1-4sin^2 x)

Answer 1

please use the Euler formula, namely:
\[{e^{ix}} = \cos x + i\sin x\]
so we can write:
\[{e^i}^{6x} = \left( {\cos \left( {6x} \right) + i\sin \left( {6x} \right)} \right) = {\left( {\cos x + i\sin x} \right)^6}\]

Answer 2

please note that you have to develop the 6-th power

Answer 3

we havent learned that formula - i think its supposed to use double angle formulas

Answer 4

ok!

Answer 5

then you have to evaluate
\[\cos \left( {3x} \right) = \cos \left( {2x + x} \right) = ...?\]
first, and then you have to evaluate
\[\cos \left( {6x} \right) = \cos \left( {2 \cdot 3x} \right) = 2{\left( {\cos \left( {3x} \right)} \right)^2} - 1\]

Answer 6

im confused on why you are using cos (6x) because its for sin (6x)

Answer 7

my teacher started it by breaking the sin(6x) into sin(3x+3x) but they never finished it. could u try walking through the steps that way please?

Answer 8

oops.. you are right!

Answer 9

Please I re-write your identities

Answer 10

what do u mean rewrite them?

Answer 11

\[\begin{gathered}
\cos \left( {3x} \right) = \cos \left( {2x + x} \right) = ...? \hfill \\
\sin \left( {3x} \right) = \sin (2x + x) = ...? \hfill \\
\sin \left( {6x} \right) = \sin \left( {2 \cdot 3x} \right) = 2\sin \left( {3x} \right)\cos \left( {3x} \right) = ...? \hfill \\
\end{gathered} \]

Answer 12

please apply the above identities, in order to find your anser

Answer 13

okay so then would u break up 2sin(3x)cos(3x) into 2sin(2x+x)cos(2x+x) right?

Answer 14

yes!

Answer 15

2(sin2xcosx+cos2xsinx)(cos2xcosx-sin2xsinx) ?

Answer 16

what would u do after that?

Answer 17

please you have to apply the subsequent identities:
\[\begin{gathered}
\sin \left( {2x} \right) = 2\sin x\cos x \hfill \\
\cos \left( {2x} \right) = {\left( {\cos x} \right)^2} - {\left( {\sin x} \right)^2} \hfill \\
\end{gathered} \]

Answer 18

i know but i don't understand how it's going to get to the (3-4sin^2x)(1-4sin^2x) part..

Answer 19

@directrix could u help me out?

Answer 20

@michele_laino im rly confused could u just write out what the next line would look like... i feel like maybe its supposed to use the cos2x=1-sin^2x and 2cos^2x-1 instead of cos^2x-sin^2x because how else would u get the second half of the identity...?

Answer 21

please wait, I try

Answer 22

okay thank you

Answer 23

\[\begin{gathered}
2\left[ {2\sin x\cos x\cos x + \left( {1 - 2{{\left( {\sin x} \right)}^2}} \right)\sin x} \right] \cdot \left[ {\left( {1 - 2{{\left( {\sin x} \right)}^2}} \right)\cos x - 2\sin x\cos x\cos x} \right] = \hfill \\
= 2\left[ {2\sin x\cos x} \right] \hfill \\
\end{gathered} \]

Answer 24

thank you so much! but is there a missing cos in the first term and a missing sin in the last term? if so would u just multiply those in

Answer 25

nevermind!!

Answer 26

\[\begin{gathered}
= 2\left[ {2\sin x{{\left( {\cos x} \right)}^2} + \sin x - 2{{\left( {\sin x} \right)}^3}} \right] \cdot \hfill \\
\cdot \left[ {\cos x - 2{{\left( {\sin x} \right)}^2}\cos x - 2{{\left( {\sin x} \right)}^2}\cos x} \right] = \hfill \\
\end{gathered} \]

Answer 27

\[\begin{gathered}
= 2\left[ {2\sin x\left( {1 - {{\left( {\sin x} \right)}^2}} \right) + \sin x - 2{{\left( {\sin x} \right)}^3}} \right] \cdot \hfill \\
\cdot \left[ {\cos x - 4{{\left( {\sin x} \right)}^2}\cos x} \right] \hfill \\
\end{gathered} \]

Answer 28

and now it is easy to get your answer, please complete the computation

Answer 29

thank you a lot you really helped! ;)

Answer 30

thank you! :)