# deduce

• Aug 24th 2009, 08:47 PM
thereddevils
deduce
Prove that , for all values of a , sin a (cos 2a+cos 4a+cos 6a)=cos 4a sin 3a

i can prove this ..

Deduce that $\displaystyle \sin \frac{\pi}{12}=\frac{1}{\sqrt{6}+\sqrt{2}}$
i am not sure with this . i don see how i can use the prove above to solve this . Thanks for your help .
• Aug 24th 2009, 09:03 PM
Chris L T521
Quote:

Originally Posted by thereddevils
Prove that , for all values of a , sin a (cos 2a+cos 4a+cos 6a)=cos 4a sin 3a

i can prove this ..

Deduce that $\displaystyle \sin \frac{\pi}{12}=\frac{1}{\sqrt{6}+\sqrt{2}}$
i am not sure with this . i don see how i can use the prove above to solve this . Thanks for your help .

$\displaystyle \sin a=\frac{\cos\!\left(4a\right)\sin\!\left(3a\right) }{\cos\!\left(2a\right)+\cos\!\left(4a\right)+\cos \!\left(6a\right)}$

So $\displaystyle \sin\tfrac{\pi}{12}=\frac{\cos\tfrac{\pi}{3}\sin\t frac{\pi}{4}}{\cos\tfrac{\pi}{6}+\cos\tfrac{\pi}{3 }+\cos\tfrac{\pi}{2}}=\frac{\tfrac{1}{2\sqrt{2}}}{ \tfrac{\sqrt{3}}{2}+\tfrac{1}{2}}=\frac{1}{\sqrt{3 }\sqrt{2}+\sqrt{2}}=\frac{1}{\sqrt{6}+\sqrt{2}}$