1. ## Trig Identity

Where do I even start?

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

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

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

3. Originally Posted by Jhevon
Hint: $\displaystyle \sin 2x = 2 \sin x \cos x$ ....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 = $\displaystyle cos^2x - sin^2x$? so it becomes $\displaystyle cos^4x - sin^4x$?

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 = $\displaystyle cos^2x - sin^2x$? so it becomes $\displaystyle cos^4x - sin^4x$?

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

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

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

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

Oh I see...

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

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

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

Oh I see...

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

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

Thus, the identity is proved

7. How does

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

8. Originally Posted by millerst
How does

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

Let u = 2x for easier visualisation