Challenge: The nastiest trig. identity that you have ever met!

• Jun 14th 2008, 07:33 PM
mathwizard
Challenge: The nastiest trig. identity that you have ever met!
Prove the identities

$\displaystyle \frac{cos(n\theta) sin[\frac{(n+1)\theta}{2}]}{sin(\theta/2)}$

$\displaystyle = \frac{cos[(n+1)\theta] cos\theta - cos\theta - cos[(n+1)\theta] + 1 + sin\theta sin[(n+1)\theta]}{2-2cos\theta}$

and

$\displaystyle \frac{sin(n\theta) - sin[(n+1)\theta] + sin\theta}{2-2cos\theta} = \frac{sin(n\theta/2)sin[(n+1)\theta/2]}{sin(\theta/2)}$

Good luck!
• Jun 15th 2008, 12:28 AM
red_dog
1.
$\displaystyle \displaystyle\frac{\cos(n+1)\theta\cos\theta-\cos\theta-\cos(n+1)\theta+1+\sin\theta\sin(n+1)\theta}{2-2\cos\theta}=$
$\displaystyle \displaystyle=\frac{\cos n\theta-\cos(n+1)\theta+1-\cos\theta}{2(1-\cos\theta)}=\frac{2\sin\frac{\theta}{2}\sin\frac{ (2n+1)\theta}{2}+2\sin^2\frac{\theta}{2}}{4\sin^2\ frac{\theta}{2}}=$
$\displaystyle \displaystyle=\frac{\sin\frac{(2n+1)\theta}{2}+\si n\frac{\theta}{2}}{2\sin\frac{\theta}{2}}=\frac{\s in\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{\sin \frac{\theta}{2}}$

2.
$\displaystyle \displaystyle\frac{\sin n\theta-\sin(n+1)\theta+\sin\theta}{2-2\cos\theta}=\frac{-2\sin\frac{\theta}{2}\cos\frac{(2n+1)\theta}{2}+2\ sin\frac{\theta}{2}\cos\frac{\theta}{2}}{4\sin^2\f rac{\theta}{2}}=$
$\displaystyle \displaystyle=\frac{\cos\frac{\theta}{2}-\cos\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} =\frac{\sin\frac{n\theta}{2}\sin\frac{(n+1)\theta} {2}}{\sin\frac{\theta}{2}}$