Wtf, trigonometry.

hai guise. :3
This problem makes absolutely no sense to me;

Rewrite the product as a sum or difference:
sin(8t) sin(4t)

use sin(2a) = 2sin(a)cos(a)

here,
sin(8t) = 2sin(4t)cos(4t)

2sin(4t)cos(4t)sin(4t) = 2((sin(4t))^2)cos(4t)
=2(1-cos^2(4t)cos(4t)
=(2-2cos^2(4t))cos(4t)
=2cos(4t) - 2cos^3(4t)

>>1
Can't be done. You're black.

>>2
Or use Euler's formula (or just look it up) to get:
\sin (a) \sin (b)=-\frac{1}{4} \left(e^{i
   a}-e^{-i a}\right) \left(e^{i b}-e^{-i
   b}\right)=\frac{1}{4} \left(-e^{-i
   (a-b)}+e^{i (a-b)}+e^{-i (a+b)}-e^{i
   (a+b)}\right)=\frac{1}{2} (\cos
   (a-b)-\cos (a+b))
So \sin (8t) \sin (4t) becomes \frac{1}{2} (\cos
   (4t)-\cos (12t))

>>4
Err...
\sin (a) \sin (b)=-\frac{1}{4} \left(e^{i
   a}-e^{-i a}\right) \left(e^{i b}-e^{-i
   b}\right)=\frac{1}{4} \left(-e^{-i
   (a-b)}+e^{i (a-b)}+e^{-i (a+b)}-e^{i
   (a+b)}\right)=\frac{1}{2} (\cos
   (a-b)-\cos (a+b))
\sin (8t) \sin (4t) = \frac{1}{2} (\cos
   (4t)-\cos (12t))

Sry, meant to say:
\sin (a) \sin (b)
=-\frac{1}{4} \left(e^{i
   a}-e^{-i a}\right) \left(e^{i b}-e^{-i
   b}\right)
=\frac{1}{4} \left(-e^{-i
   (a-b)}+e^{i (a-b)}+e^{-i (a+b)}-e^{i
   (a+b)}\right)
=\frac{1}{2} (\cos
   (a-b)-\cos (a+b))
\sin (8t) \sin (4t)=\frac{1}{2} (\cos
   (4t)-\cos (12t))

Ok, fuck formatting -_-