help need on trigonomtery function

**rlaydab** · July 13th 2012, 11:18 AM

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

**Prove It** · July 13th 2012, 09:39 PM

Originally Posted by rlaydab

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

I can't tell whether your equation is $\displaystyle \begin{align*} \sin^{2n}{\theta} - \sin^{2(n-1)}{\theta} = \sin^2{\theta} \end{align*}$ or $\displaystyle \begin{align*} \sin^2{(n\theta)} - \sin^2{[(n-1)\theta]} = \sin^2{\theta} \end{align*}$ ...

**rlaydab** · July 14th 2012, 07:43 AM

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

**skeeter** · July 14th 2012, 07:48 AM

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.

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

it's true for n = 1 ... are you to prove that this is an identity? looks like a candidate for induction.

**rlaydab** · July 14th 2012, 08:08 AM

Can you help to solve it. Can LHS be proven to be = RHS

**skeeter** · July 14th 2012, 10:15 AM

the equation is not an identity, however (after playing with the calculator) ...

for $n = 2$

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

for $n = 3$

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

for $n = 4$

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

hmmm ...

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

... ???

**rlaydab** · July 14th 2012, 11:47 AM

Thanks but I thought we have to prove the LHS = RHS?
May be I am wrong.

**BobP** · July 14th 2012, 02:19 PM

It needs the use of three trig identities.

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

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

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

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 $\sin \theta$ as a common factor and then use (2) again on the expression within the brackets.
That leads to

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

Thread: help need on trigonomtery function

LinkBack

Thread Tools

Display

help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Re: help need on trigonomtery function

Similar Math Help Forum Discussions

Word problem: Cost function, Revenue Function, Profit function.

[SOLVED] Expectancy (mean) of Probability density function composed of delta/rect function

rational function, polynomial function or some other function

Trigonomtery Angles: Compound Angle Formulas

Need sum and difference on trigonomtery

Search Tags