pythagorean identity

• Jan 12th 2014, 03:36 PM
Jonroberts74
pythagorean identity
I have an example that is about proving

$\frac{sin \theta}{cot \theta cos^2 \theta - cot \theta} - sec \theta} = 0$

it factors the denominator of the left term into $\frac{sin \theta}{cot \theta ( cos^2 \theta - 1)}$

then says replace $(cos^2 \theta - 1)$ with $sin^2 \theta$

Is this correct?

wouldn't $cos^2 \theta - 1 = -sin^2 \theta$ ??
• Jan 12th 2014, 05:04 PM
romsek
Re: pythagorean identity
Quote:

Originally Posted by Jonroberts74
I have an example that is about proving

$\frac{sin \theta}{cot \theta cos^2 \theta - cot \theta} - sec \theta} = 0$

it factors the denominator of the left term into $\frac{sin \theta}{cot \theta ( cos^2 \theta - 1)}$

then says replace $(cos^2 \theta - 1)$ with $sin^2 \theta$

Is this correct?

wouldn't $cos^2 \theta - 1 = -sin^2 \theta$ ??

you are correct.
• Jan 12th 2014, 05:43 PM
Soroban
Re: pythagorean identity
Hello, Jonroberts74!

The given equation is not true.

Quote:

$\text{Prove: }\:\frac{\sin\theta}{\cot\theta\cos^2\!\theta -\cot\theta} + \sec\theta} \:=\: 0$
. . . . . . . . . . . . . . . . . . . . . ${\color{red}\uparrow}$

We have: . $\frac{\sin\theta}{\cot\theta(\cos^2\theta-1)} + \sec\theta \;=\;\frac{\sin\theta}{\cot\theta(\text{-}\sin^2\theta)} + \sec\theta \;=\;\dfrac{\sin\theta}{\dfrac{\cos\theta}{\sin \theta}(\text{-}\sin^2\theta)}+ \sec\theta$

. . . . . $=\; \dfrac{\sin\theta}{\text{-}\cos\theta\sin\theta}+\sec\theta \;=\;\frac{\text{-}1}{\cos\theta} + \sec\theta \;=\; \text{-}\sec\theta + \sec\theta \;=\;0$
• Jan 13th 2014, 10:46 PM
Jonroberts74
Re: pythagorean identity
so many books with errors.