Proof Help

[kaebun]

can someone show me how to get from sin3x to 3sinx(cosx)^2-3x-sinx
my teacher was like "you have done this enough that you can skip all the steps in between" and im preety sure ive never seen this before or at least understood how to do this. I know you can make sin3x= sin(2x+x)= sin2xcosx+cos2xsinx
but im not sure how that gets you to 3sinx(cosx)^2-3x-sinx or if thats even the right way to go :? [/tex]

 

[pka]

There is absolutely no way to go from \(\displaystyle \L
\sin (3x)\) to a function that contains \(\displaystyle \L
- 3x\) as a term.

Please review your post. As written it is not true!

 

[kaebun]

well thats whats was writen on the board :O ahhhh!

maybe you could show me how to solve sin3x=sinx

 

[pka]

One can show that \(\displaystyle \L
\sin (3x) = 4\sin (x)\cos ^2 (x) - \sin (x).\)

\(\displaystyle \L
\begin{array}{l}
\\
4\sin (x)\cos ^2 (x) - \sin (x) = \sin (x)\quad \Rightarrow \quad \sin (x) = 0\quad \& \quad \cos ^2 (x) = \frac{1}{2}. \\
\end{array}\)

 

[kaebun]

how do you get the first step?

One can show that \(\displaystyle \L
\sin (3x) = 4\sin (x)\cos ^2 (x) - \sin (x).\)

 

[pka]

\(\displaystyle \L
\sin (3x) = \sin (2x + x) = \sin (2x)\cos (x) + \cos (2x)\sin (x)\).

 

[kaebun]

i got that far but then what?

 

[TchrQbic]

i got that far but then what?

Now redefine sin(2x) and cos(2x) using the same sum of two angles identities and replace them, then simplify.