# help need on trigonomtery function

• Jul 13th 2012, 11:18 AM
rlaydab
help need on trigonomtery function
Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ
• Jul 13th 2012, 09:39 PM
Prove It
Re: help need on trigonomtery function
Quote:

Originally Posted by rlaydab
Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ

I can't tell whether your equation is \displaystyle \displaystyle \begin{align*} \sin^{2n}{\theta} - \sin^{2(n-1)}{\theta} = \sin^2{\theta} \end{align*} or \displaystyle \displaystyle \begin{align*} \sin^2{(n\theta)} - \sin^2{[(n-1)\theta]} = \sin^2{\theta} \end{align*}...
• Jul 14th 2012, 07:43 AM
rlaydab
Re: help need on trigonomtery function
The second bottom one is correct. N or n-1 or theta is not in the exponent/power. Squared power is for sine only.
• Jul 14th 2012, 07:48 AM
skeeter
Re: help need on trigonomtery function
Quote:

Originally Posted by rlaydab
The second bottom one is correct. N or n-1 or theta is not in the exponent/power. Squared power is for sine only.

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

it's true for n = 1 ... are you to prove that this is an identity? looks like a candidate for induction.
• Jul 14th 2012, 08:08 AM
rlaydab
Re: help need on trigonomtery function
Can you help to solve it. Can LHS be proven to be = RHS
• Jul 14th 2012, 10:15 AM
skeeter
Re: help need on trigonomtery function
the equation is not an identity, however (after playing with the calculator) ...

for $\displaystyle n = 2$

$\displaystyle \sin^2(2\theta) - \sin^2{\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{4}$

for $\displaystyle n = 3$

$\displaystyle \sin^2(3\theta) - \sin^2{2\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{6}$

for $\displaystyle n = 4$

$\displaystyle \sin^2(4\theta) - \sin^2{3\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{8}$

hmmm ...

conjecture that $\displaystyle \sin^2(n\theta) - \sin^2[(n-1)\theta] = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{2n}$

... ???
• Jul 14th 2012, 11:47 AM
rlaydab
Re: help need on trigonomtery function
Thanks but I thought we have to prove the LHS = RHS?
May be I am wrong.
• Jul 14th 2012, 02:19 PM
BobP
Re: help need on trigonomtery function
It needs the use of three trig identities.

$\displaystyle \sin A + \sin B=2\sin \left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)..................(1)$

$\displaystyle \sin A - \sin B=2\cos \left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)..................(2)$

$\displaystyle \sin 2A = 2\sin A \cos A................................................. .......(3)$

Briefly, take a difference of two squares on the LHS, simplify the two brackets using (1) and (2) and then use (3) (twice) to simplify the resulting expression. Move the resulting expression to the RHS, remove $\displaystyle \sin \theta$ as a common factor and then use (2) again on the expression within the brackets.
$\displaystyle 2\sin \theta \cos n\theta \sin(n-1)\theta =0$ from which $\displaystyle \sin\theta=0,$ or $\displaystyle \cos n\theta=0$ or $\displaystyle \sin (n-1)\theta = 0.$