Question

Help please! Will give medal and fan!

Answer 1

hint:
\[\begin{gathered}
\sin \left( {3\alpha } \right) = \sin \left( {\alpha + 2\alpha } \right) = \hfill \\
= \sin \alpha \cos \left( {2\alpha } \right) + \cos \alpha \sin \left( {2\alpha } \right) \hfill \\
\end{gathered} \]

Answer 2

What do I do after that?

Answer 3

we have to use these identities:
\[\begin{gathered}
\cos \left( {2\alpha } \right) = 1 - 2{\left( {\sin \alpha } \right)^2} \hfill \\
\sin \left( {2\alpha } \right) = 2\sin \alpha \cos \alpha \hfill \\
\end{gathered} \]

Answer 4

hint:
\[\begin{gathered}
\sin \left( {3\alpha } \right) = \sin \left( {\alpha + 2\alpha } \right) = \hfill \\
= \sin \alpha \cos \left( {2\alpha } \right) + \cos \alpha \sin \left( {2\alpha } \right) = \hfill \\
= \sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) + \cos \alpha \left( {2\sin \alpha \cos \alpha } \right) \hfill \\
\end{gathered} \]

Answer 5

please continue

Answer 6

So I cross out sin a on both sides, right?

Answer 7

no, we have to continue those equalities, namely:

Answer 8

\[\begin{gathered}
\sin \left( {3\alpha } \right) = \sin \left( {\alpha + 2\alpha } \right) = \hfill \\
= \sin \alpha \cos \left( {2\alpha } \right) + \cos \alpha \sin \left( {2\alpha } \right) = \hfill \\
= \sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) + \cos \alpha \left( {2\sin \alpha \cos \alpha } \right) = \hfill \\
= \sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) + 2\sin \alpha {\left( {\cos \alpha } \right)^2} = \hfill \\
= \sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) + 2\sin \alpha \left( {1 - {{\left( {\sin \alpha } \right)}^2}} \right) = ...? \hfill \\
\end{gathered} \]

Answer 9

I have substituted:
\[\cos \alpha = 1 - {\left( {\sin \alpha } \right)^2}\]

Answer 10

oops..
\[{\left( {\cos \alpha } \right)^2} = 1 - {\left( {\sin \alpha } \right)^2}\]

Answer 11

I am confused what to do at this point. Do I need to start simplifying or substituting the factors that you've given me

Answer 12

you have to compute the multiplications indicated by the parentheses, for example:
\[\sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) = \sin \alpha - 2{\left( {\sin \alpha } \right)^3}\]

Answer 13

Oh okay. So (sina-2(sina)^3)+sina^2?

Answer 14

no, the othr term is:
\[2\sin \alpha \left( {1 - {{\left( {\sin \alpha } \right)}^2}} \right) = 2\sin \alpha - 2{\left( {\sin \alpha } \right)^3}\]

Answer 15

so, what is:
\[\sin \alpha - 2{\left( {\sin \alpha } \right)^3} + 2\sin \alpha - 2{\left( {\sin \alpha } \right)^3} = ...?\]

Answer 16

hint:
we have to compute this sum:
\[\sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) + 2\sin \alpha \left( {1 - {{\left( {\sin \alpha } \right)}^2}} \right)\]

Answer 17

is it ok?

Answer 18

It's the same on both sides do I just put 2 in front of it since it is doubled?

Answer 19

the first term is :
\[\sin \alpha \left( {1 - 2{{\left( {\sin \alpha } \right)}^2}} \right) = \sin \alpha - 2{\left( {\sin \alpha } \right)^3}\]

Answer 20

and the second term is:
\[2\sin \alpha \left( {1 - {{\left( {\sin \alpha } \right)}^2}} \right) = 2\sin \alpha - 2{\left( {\sin \alpha } \right)^3}\]

Answer 21

so, we have to compute this sum:
\[{\text{first term + second term}} = \sin \alpha - 2{\left( {\sin \alpha } \right)^3} + 2\sin \alpha - 2{\left( {\sin \alpha } \right)^3} = ...?\]

Answer 22

@kkbrookly

Answer 23

2sina^2-4(sina)^3?

Answer 24

better is
3 sina^2-4(sina)^3

Answer 25

Oh okay. I was one off. Now I substitute the terms or do I continue to simplify it?

Answer 26

we have finished since the expression which we have got is equal to the right side of your identity

Answer 27

Oh okay! So it is an identity! Thank you!

Answer 28

thank you!