Trig Identity

**millerst** · April 13th 2009, 01:59 PM

Where do I even start?

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

**Jhevon** · April 13th 2009, 02:03 PM

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?

**millerst** · April 13th 2009, 02:07 PM

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 $cos^4x - sin^4x$ ?

**Jhevon** · April 13th 2009, 02:09 PM

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 $cos^4x - sin^4x$ ?

please reread my post (note that i did not mention cos(2x)) and answer the question i asked

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 $(a + b)^2 = a^2 + 2ab + b^2$

**millerst** · April 13th 2009, 02:12 PM

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

**Jhevon** · April 13th 2009, 02:24 PM

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

**millerst** · April 13th 2009, 02:40 PM

How does

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

**e^(i*pi)** · April 13th 2009, 02:50 PM

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

Thread: Trig Identity

LinkBack

Thread Tools

Display

Trig Identity

Similar Math Help Forum Discussions

trig identity help...

Another trig identity

Trig Identity

Trig identity

Odd Trig Identity

Search Tags