Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ

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- Jul 13th 2012, 11:18 AMrlaydabhelp need on trigonomtery function
Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ

- Jul 13th 2012, 09:39 PMProve ItRe: help need on trigonomtery function
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 AMrlaydabRe: 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 AMskeeterRe: help need on trigonomtery function
- Jul 14th 2012, 08:08 AMrlaydabRe: help need on trigonomtery function
Can you help to solve it. Can LHS be proven to be = RHS

- Jul 14th 2012, 10:15 AMskeeterRe: 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 AMrlaydabRe: 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 PMBobPRe: 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.

That leads to

$\displaystyle 2\sin \theta \cos n\theta \sin(n-1)\theta =0$ from which $\displaystyle \sin\theta=0,$ or $\displaystyle \cos n\theta=0$ or $\displaystyle \sin (n-1)\theta = 0.$