sin n cos to the power of three

Joel · #1 $Old$ November 2nd, 2009, 06:57 AM

Hi all,

2cos³ѳ – cosѳ / sinѳcos²ѳ - sin³ѳ = cotѳ

How would I go about proving this?

I was thinking to turn cot into cos / sin so now:

2cos³ѳ – cosѳ / sinѳcos²ѳ - sin³ѳ = cosѳ / sinѳ

but a bit lost from here...

any help would be appreciated

Prove It · #2 $Old$ November 2nd, 2009, 07:38 AM

Quote:

Originally Posted by Joel $View Post$

Hi all,

2cos³ѳ – cosѳ / sinѳcos²ѳ - sin³ѳ = cotѳ

How would I go about proving this?

I was thinking to turn cot into cos / sin so now:

2cos³ѳ – cosѳ / sinѳcos²ѳ - sin³ѳ = cosѳ / sinѳ

but a bit lost from here...

any help would be appreciated

Is it $\frac{2\cos^3{\theta} - \cos{\theta}}{\sin{\theta}\cos^2{\theta} - \sin^3{\theta}} = \cot{\theta}$ ?

Joel · #3 $Old$ November 2nd, 2009, 07:48 AM

yes thats the question, sorry still not confident writing in LA

Prove It · #4 $Old$ November 2nd, 2009, 08:43 AM

$\frac{2\cos^3{\theta} - \cos{\theta}}{\sin{\theta}\cos^2{\theta} - \sin^3{\theta}} = \frac{\cos{\theta}(2\cos^2{\theta} - 1)}{\sin{\theta}(\cos^2{\theta} - \sin^2{\theta})}$

$= \frac{\cos{\theta}}{\sin{\theta}}\cdot\frac{2\cos^2{\theta} -1}{\cos^2{\theta} - \sin^2{\theta}}$

$= \cot{\theta}\cdot\frac{\cos^2{\theta} - (1 - \cos^2{\theta})}{\cos^2{\theta} - \sin^2{\theta}}$

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

$= \cot{\theta}$ .