# Thread: Trig. Identity Proof: Sin2X CotX- 2Sin^2X = 2 Cos2X ...?

1. ## Trig. Identity Proof: Sin2X CotX- 2Sin^2X = 2 Cos2X ...?

Hello everyone. I am new to this site.

Now, here is the situation. I am in 11th grade math III, and we started Trig. Identities recently. I am having some problems with Trig. Identity Proofs.

I watched a "Math Tutor" DVD about it, and they said: "You can treat Trig. Identity Proofs like an equation, because of the equal sign." And thus, they worked the problem on both sides of the equal sign.

However, my math teacher told me: "NO, Andrew. You can absolutely canNOT work on both sides of a Trig. Identity proof!" ...Thus, my homework was all wrong...

She told me to leave the simpler side alone and to work the "complicated side".

Could someone help me with this one problem?
I got it started and am stuck.

"Prove this statement is an Identity for all values of the angle for which the expressions are defined"

Sin2X CotX- 2Sin^2X = 2 Cos2X

Sin 2 X (1/Tan X)- 2 Sin^2X
-1 Sin X (1/Tan X)

Did I at least start it out right? Hopefully, I did!
Please try to help explain this to me...

2. The thing with trigonometric identities is that you can work both sides as long as they generally don't interfere with each other.

$\sin (2x) \cot (x) - 2\sin^{2} (x) = 2\cos (2x)$ ?

Looking at the sin(2x)cot(x), convert everything into terms of sinx and cosx (this trick generally puts you in the right direction for most of these identities that you will encounter):
$\text{Left Hand Side} = \underbrace{\left(2\sin (x) \cos (x) \right)}_{\sin (2x)} \cdot \left( \frac{\cos x}{\sin x}\right) - 2\sin^{2} (x)$

Notice what cancels? The sin x does. So:
$\text{LHS} = 2\cos^{2} x - 2\sin^{2} x = 2\left(\cos^{2} x - \sin^{2} x \right)$ (Look familiar ?)

3. Thank you very much for the help!

I appreciate it.

4. Hello, mroats!

Welcome aboard!

$\sin2x\cot x - 2\sin^2\!x \;=\;2\cos2x$
We need a few basic identities:
. . $\sin2\theta \:=\:2\sin\theta\cos\theta \qquad\quad \cot\theta \:=\:\frac{\cos\theta}{\sin\theta} \qquad\quad \cos2\theta \:=\:\cos^2\!\theta - \sin^2\!\theta$

$\text{The left side is: }\;\underbrace{\sin2x}_{\swarrow\;\;}\underbrace{\ cot x}_{\;\;\searrow} - 2\sin^2\!x$
. . . . . $= \;(2\sin x\cos x)\left(\frac{\cos x}{\sin x}\right) - 2\sin^2\!x$

. . . . . $= \;2\cos^2\!x - 2\sin^2\!x$

. . . . . $=\;2\underbrace{(\cos^2\!x - \sin^2\!x)}$

. . . . . $= \qquad\; 2\overbrace{\cos2x}$

5. Thank you for expanding soroban! =)