# Thread: help need on trigonomtery function

1. ## help need on trigonomtery function

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

2. ## Re: help need on trigonomtery function

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*}...

3. ## 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.

4. ## Re: help need on trigonomtery function

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.

5. ## Re: help need on trigonomtery function

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

6. ## Re: help need on trigonomtery function

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 $\frac{\pi}{2n}$

... ???

7. ## Re: help need on trigonomtery function

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

8. ## Re: help need on trigonomtery function

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