# Trig proof - urg

• May 13th 2009, 04:40 AM
AshleyT
Trig proof - urg
It is given that:
$
Sin7\theta = 7Sin{\theta} - 56Sin^3\theta + 112Sin^5\theta - 64Sin^7\theta
$

Hence, prove that the only real solution of the equation
$Sin 7 \theta = 7Sin\theta$
are given by
$
\theta = n\pi
$

Where n is an integer.
I am really stuck with this :(

Any help would be greatly appreciated.
Thankyou.
• May 13th 2009, 04:49 AM
bebrave
Math Forum - Ask Dr. Math
• May 13th 2009, 05:12 AM
AshleyT
Quote:

Originally Posted by bebrave
Math Forum - Ask Dr. Math

Thankyou very much for your reply however im find with putting Sin 7x into the equation by demoivres :), that was the first part of the question which i managed :)...its the proof part im struggling with :(.
• May 13th 2009, 06:06 AM
Plato
Use what you just proved. Substitute , substract $7\sin(\theta)$.
You then have $64\sin^7(\theta)-112\sin^5(\theta)+56\sin^3(\theta)=0$.
Show that has no real roots other than 0.
• May 13th 2009, 06:57 AM
AshleyT
Quote:

Originally Posted by Plato
Use what you just proved. Substitute , substract $7\sin(\theta)$.
You then have $64\sin^7(\theta)-112\sin^5(\theta)+56\sin^3(\theta)=0$.
Show that has no real roots other than 0.

Plato, you're awsome! Thanks again for another reply ;) :)