# Can someone help me solve this identity? :)

• Nov 6th 2007, 01:52 PM
Jeavus
Can someone help me solve this identity? :)
http://img443.imageshack.us/img443/701/identityuf5.png

I've tried many things, can't seem to work it out. Thank you!
• Nov 6th 2007, 01:58 PM
Krizalid
Did you tried simplifyin' the LHS?

Factorise.
• Nov 6th 2007, 01:59 PM
Jeavus
Yes, I've been working with the left hand side.

I've tried a lot of things.
• Nov 6th 2007, 02:01 PM
Krizalid
$\displaystyle \frac{{\tan \theta - \sin \theta }} {{\sin ^3 \theta }} = \frac{{\sin \theta \left( {\dfrac{1} {{\cos \theta }} - 1} \right)}} {{\sin ^3 \theta }} = \frac{{1 - \cos \theta }} {{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}.$

Does that make sense?
• Nov 6th 2007, 02:13 PM
Jeavus
How does sin^3theta = (1+costheta)(1-costheta)?
• Nov 6th 2007, 02:20 PM
angel.white
$\displaystyle \frac{tan\theta - sin\theta}{sin^{3}\theta}$

1. because $\displaystyle tan\theta = \frac{sin\theta}{cos\theta}$, substitute that into $\displaystyle tan\theta$
$\displaystyle \frac{\frac{sin\theta}{cos\theta} - sin\theta}{sin^{3}\theta}$

2. Common denominator of the fractions in the numerator
$\displaystyle \frac{\frac{sin\theta}{cos\theta} - \frac{sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$

3. combine fractions in the numerator
$\displaystyle \frac{\frac{sin\theta-sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$

4. simplify (dividing a fraction by something means multiplying by it's inverse):
$\displaystyle \frac{sin\theta-sin\theta cos\theta}{cos\theta sin^{3}\theta}$

5. factor out $\displaystyle sin\theta$ in the numerator
$\displaystyle \frac{sin\theta (1 - cos\theta)}{cos\theta sin^{3}\theta}$

$\displaystyle sin\theta$ in the numerator and denominator cancel eachother out
$\displaystyle \frac{1 - cos\theta}{cos\theta sin^{2}\theta}$

6. Multiply the numerator and the denominator by $\displaystyle 1+cos\theta$
$\displaystyle \frac{(1 - cos\theta)(1 + cos\theta)}{cos\theta sin^{2}\theta (1 + cos\theta)}$

7. FOIL the numerator:
$\displaystyle \frac{1 + cos^{2}\theta}{(1 + cos\theta)cos\theta sin^{2}\theta}$

8. Using the formula $\displaystyle sin^{2}\theta + cos^{2}\theta = 1$ see that $\displaystyle sin^{2}\theta = 1 - cos^{2}\theta$ and substitute that into the equation

$\displaystyle \frac{1 - cos^{2}\theta}{(1 + cos\theta)cos\theta (1- cos^{2}\theta)}$

9. $\displaystyle 1- cos^{2}\theta$ is in the numerator and the denominator, so cancel them out
$\displaystyle \frac{1}{(1 + cos\theta)cos\theta}$
• Nov 6th 2007, 04:00 PM
Soroban
Hello, Jeavus

Quote:

How does sin^3theta = (1+costheta)(1-costheta)?
It doesn't . . .

Quote:

Originally Posted by Krizalid

$\displaystyle \frac{{\tan \theta - \sin \theta }}{{\sin ^3 \theta }}$

. . $\displaystyle = \;\frac{{\sin \theta \left( {\dfrac{1}{{\cos \theta }} - 1} \right)}}{{\sin ^3 \theta }}$ . . . . cancel a $\displaystyle {\color{blue}\sin\theta}$

. . $\displaystyle {\color{blue}= \;\dfrac{\dfrac{1}{\cos\theta} - 1}{\sin^2\theta} \;=\;\dfrac{\dfrac{1-\cos\theta}{\cos\theta}}{\sin^2\theta} \;=\;\frac{1-\cos\theta}{\cos\theta\sin^2\theta} \;=\;\frac{1-\cos\theta}{\cos\theta(1-\cos^2\theta)}}$

. . $\displaystyle =\; \frac{{1 - \cos \theta }} {{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}$

• Nov 6th 2007, 04:24 PM
Jeavus
Thank you all so much. :)