1. ## Trig Identity

Where do I even start?

$4(sin^2x)(cos^2x) + cos^2(2x) = 1$

2. Originally Posted by millerst
Where do I even start?

$4(sin^2x)(cos^2x) + cos^2(2x) = 1$
Hint: $\sin 2x = 2 \sin x \cos x$ ....what happens if we square this equation?

3. Originally Posted by Jhevon
Hint: $\sin 2x = 2 \sin x \cos x$ ....what happens if we square this equation?
Then I can use an identity, but would it make more sense to factor out the cosx or something... I'm so lost...

or cos2X = $cos^2x - sin^2x$? so it becomes $cos^4x - sin^4x$?

4. Originally Posted by millerst
Then I can use an identity, but would it make more sense to factor out the cosx or something... I'm so lost...

or cos2X = $cos^2x - sin^2x$? so it becomes $cos^4x - sin^4x$?

by the way, you squared your identity incorrectly: $(\cos^2 x - \sin^2 x)^2 {\color{red}\ne} \cos^4 x - \sin^4 x$, since we (should) know that $(a + b)^2 = a^2 + 2ab + b^2$

5. Originally Posted by Jhevon
Hint: $\sin 2x = 2 \sin x \cos x$ ....what happens if we square this equation?

$\sin 2x = 2 \sin x \cos x$ so squared it becomes:

$4\sin^2x\cos^2x$?

Oh I see...

So $\sin 2x = 2 \sin x \cos x + cos^2/2x = 1$
$\sin 2x = 2 \sin x \cos x + (cos2x)^2 = 1$
So $2 \sin x \cos x + (cos^2 - sin^2)^2 = 1$

6. Originally Posted by millerst
$\sin 2x = 2 \sin x \cos x$ so squared it becomes:

$4\sin^2x\cos^2x$?

Oh I see...

So $\sin 2x = 2 \sin x \cos x + cos^2/2x = 1$
$\sin 2x = 2 \sin x \cos x + (cos2x)^2 = 1$
So $2 \sin x \cos x + (cos^2 - sin^2)^2 = 1$
huh? you're thinking way too hard, Miller

$\begin{matrix} LHS & = & 4 \sin^2 x \cos^2 x + \cos^2 2x \\ & = & \sin^2 2x + \cos^2 2x \\ & = & 1 \\ & = & RHS \end{matrix}$

Thus, the identity is proved

7. How does

$sin^2(2x) + cos^2(2x)$ = 1

8. Originally Posted by millerst
How does

$sin^2(2x) + cos^2(2x)$ = 1
The Pythagorean identity. Probably the most basic in trig.

Let u = 2x for easier visualisation