# Thread: Can someone help me solve this identity? :)

1. ## Can someone help me solve this identity? :)

I've tried many things, can't seem to work it out. Thank you!

2. Did you tried simplifyin' the LHS?

Factorise.

3. Yes, I've been working with the left hand side.

I've tried a lot of things.

4. $\displaystyle \frac{{\tan \theta - \sin \theta }} {{\sin ^3 \theta }} = \frac{{\sin \theta \left( {\dfrac{1} {{\cos \theta }} - 1} \right)}} {{\sin ^3 \theta }} = \frac{{1 - \cos \theta }} {{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}.$

Does that make sense?

5. How does sin^3theta = (1+costheta)(1-costheta)?

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

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

7. Hello, Jeavus

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

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 $\displaystyle {\color{blue}\sin\theta}$

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

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

8. Thank you all so much.