1. ## Trigometric identity problem.

prove
$sin^4t=3/8-1/2*cos2t+1/8*cos4t$
get rid of the fraction so $3-4cos2t+8cos4t$
$3-4(1-2sin^2t)+8(1-2sin^2t)^2$
$7+8sin^2t-32sin^2t+32sin^4t$
$7-24sin^2t+32sin^4t$
this looks nothing like $sin^4$...

2. Originally Posted by dorkymichelle
prove
$sin^4t=3/8-1/2*cos2t+1/8*cos4t$
get rid of the fraction so $3-4cos2t+8cos4t$
$3-4(1-2sin^2t)+8(1-2sin^2t)^2$
$7+8sin^2t-32sin^2t+32sin^4t$
$7-24sin^2t+32sin^4t$
this looks nothing like $sin^4$...
Remember that

$\sin^4{t} = (\sin^2{t})^2 = \bigg(\frac{1-\cos{2t}}{2}\bigg)^2$

which expands to

$\frac{1}{4}\bigg(1-2\cos{2t}+\cos^2{2t}\bigg)$

Now expand

$\cos^2{2t} = \frac{\cos{4t} +1}{2}$

plug that back in and you get

$\frac{1}{4}\bigg(1-2\cos{2t}+\frac{\cos{4t} +1}{2}\bigg)$

simplify that and you get your identity

3. isnt $sin^4 = (sin^2)(sin^2)$?
so wouldn't it be $(1-cos^2)^2$?
Where did that formula come from?

4. Originally Posted by dorkymichelle
isnt $sin^4 = (sin^2)(sin^2)$?
so wouldn't it be $(1-cos^2)^2$?
Where did that formula come from?

$\cos{2t} = \cos^2{t} - \sin^2{t}$

$\cos{2t} = (1-sin^2{t}) - \sin^2{t}$

$\cos{2t} = 1-2sin^2{t}$

$\cos{2t}-1 = -2sin^2{t}$

$\frac{\cos{2t}-1}{-2} = sin^2{t}$

$\frac{1-\cos{2t}}{2} = sin^2{t}$