Submitted by Grainne on Mon, 10/02/2017 - 00:29
Forums:
In question 2ii the hints and partial solutions say you should end up with 0=2s^3-2s^2-s+1 however I don't understand why it's +1 since 2cos2A=2(1-2sin^2A)=2-4sin^2A and so I'm confused about why it's +1 rather than +2. Thanks for the help.
2ii
We have:
\begin{align*}
\sin 3A-\sin A &=2 \cos 2A \\
3 \sin A - 4\sin^3 A - \sin A &= 2 (1 - 2 \sin^2 A) \\
2 \sin A - 4 \sin^3 A &= 2 - 4 \sin^2 A \\
\sin A - 2 \sin^3 A &= 1 - 2 \sin^2 A
\end{align*}
Basically, it is "+1" as you can divide throughout by $2$.