For the first problem, note that
$\displaystyle sin(A - B) = sin(A)~cos(B) - sin(B)~cos(A)$
so we have:
$\displaystyle sin \left ( \frac{\pi}{3} - \alpha \right ) = sin \left ( \frac{\pi}{3} \right ) ~cos( \alpha ) - sin( \alpha) ~cos \left ( \frac{\pi}{3} \right )$
See if that helps.
-Dan
For the second problem, note that $\displaystyle sin(2x) = 2~sin(x)~cos(x)$, so we have
$\displaystyle 1 - sin^4(x) - sin^2(x) = \frac{1}{4}sin^2(x)$
$\displaystyle 1 - sin^4(x) - sin^2(x) = \frac{1}{4} \left ( 2~sin(x)~cos(x) \right ) ^2$
Multiply out the RHS and use $\displaystyle cos^2(x) = 1 - sin^2(x)$, then solve for sin(x).
You will get $\displaystyle sin^2(x) = \frac{1}{2}$
Note that there will be four solutions to this.
-Dan
Wow. Lots of posting going on in this thread!
The second problem is:
$\displaystyle 1 - \sin^4(x) - \sin^2(x) = \frac{1}{4}\sin^2(2x)$
This is not an identity to prove, but an equation to solve.
Using the double-angle identity for sine provided by topsquark, we have:
$\displaystyle 1 - \sin^4(x) - \sin^2(x) = \frac{1}{4}(2\sin(x)\cos(x))^2$
$\displaystyle 1 -\sin^4(x) - \sin^2(x) = \sin^2(x)\cos^2(x)$
$\displaystyle 1 - \sin^4(x) - \sin^2(x) = \sin^2(x)(1-\sin^2(x))$
$\displaystyle 1 - \sin^4(x) - \sin^2(x) = \sin^2(x)-\sin^4(x)$
$\displaystyle 1 - \sin^2(x) = \sin^2(x)$
$\displaystyle 1 = 2\sin^2(x)$
$\displaystyle \sin^2(x)=\frac{1}{2}$
$\displaystyle \sin(x)=\pm\frac{1}{\sqrt{2}}$
You will find 4 solutions on the interval $\displaystyle 0\le x<2\pi$. If you are not restricted to an interval, then you will have an infinite number of solutions, which is easily expressed as the integral multiple of an angle.