Trig proof

**AshleyT** · May 13th 2009, 04:40 AM

It is given that:
$<br /> Sin7\theta = 7Sin{\theta} - 56Sin^3\theta + 112Sin^5\theta - 64Sin^7\theta<br />$

Hence, prove that the only real solution of the equation
$Sin 7 \theta = 7Sin\theta$
are given by
$<br /> \theta = n\pi<br />$
Where n is an integer.
I am really stuck with this

Any help would be greatly appreciated.
Thankyou.

**bebrave** · May 13th 2009, 04:49 AM

Math Forum - Ask Dr. Math
this link can help you

**AshleyT** · May 13th 2009, 05:12 AM

Originally Posted by bebrave

Math Forum - Ask Dr. Math
this link can help you

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

.

**Plato** · May 13th 2009, 06:06 AM

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.

**AshleyT** · May 13th 2009, 06:57 AM

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

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