Proving Trig Identities with Complex Numbers

[Oblio]

Consider the complex number z=e_{i\theta} = cos\theta+isin\theta. By evaluating z^{2} two different ways, prove the trig identities cos2\theta = cos^{2}\theta - sin^{2}\theta and sin2\theta = 2sin\thetacos\theta.

A question about the approach to this question:
How do you guys approach the task of 'evaluating' something, when told to do so, like here.
I find myself doing random manipulations without knowledge of whether the road I'm on is even close to the correct path or not.
Evaluate seems like such a general instruction...

Anyways,
If I square z I get;
z^{2} = cos\theta^{2} + i^{2}sin\theta^{2}

z^{2} = cos\theta^{2} - sin\theta^{2}

If I sub z^{2} back in I get 0 so that's wrong.

Is the use of the 'e definition' necessary for this? I kind of want to see how I ought to be approaching this 'evaluation'.

Thanks, as always

 

[neutrino]

If I square z I get;
z^{2} = cos\theta^{2} + i^{2}sin\theta^{2}

How??

If z = \cos{\theta}+\imath\sin{\theta}, then z^2 = \left(\cos{\theta}+\imath\sin{\theta}\right)^2 \neq \cos^2{\theta}+\imath^2\sin^2{\theta}.

Square that. Then use the other way of evaluating a complex number raised to some power, i.e., de Moiver's formula. Equate the two.

 

[Oblio]

Am I allowed to do this?

isin\theta^{2}

=-sin\theta^{2} ?

 

[neutrino]

isin\theta^{2}

The way you have written is quite ambiguous. But if you mean \left(\imath\sin{\theta}\right)^2, then it is equal to -\sin^2{\theta}.

P.S. You need not enclose each term of LaTeX code within 'tex' tags. It is enough if you do it for each line of code. Example: Enclosing this - \left(\imath\sin{\theta}\right)^2 - within tex tags will produce \left(\imath\sin{\theta}\right)^2.

 

[Oblio]

The way you have written is quite ambiguous. But if you mean \left(\imath\sin{\theta}\right)^2, then it is equal to -\sin^2{\theta}.

P.S. You need not enclose each term of LaTeX code within 'tex' tags. It is enough if you do it for each line of code. Example: Enclosing this - \left(\imath\sin{\theta}\right)^2 - within tex tags will produce \left(\imath\sin{\theta}\right)^2.

lol thanks!
It just does that when I click the latex icon each time.. your saying you write it all out manually?

 

[Oblio]

I'm still stuck with:

z^{2} = cos^{2}\theta + 2isin^{2}\theta - sin^{2}\theta

You'll see I still did it the long way if you quote:P

 

[neutrino]

lol thanks!
It just does that when I click the latex icon each time.. your saying you write it all out manually?

While I mostly do it manually(click n' latex is only a recent feature), I'm not asking you to do it manually.

When I click on any icon, what I get is that piece of code (say, \frac{}{} for fraction) within tex codes. The cursor is still within the those tags, so when I click on another icon, only its code is printed within the same set of tags. When I want to do a new line, I move the cursor outside the first set of tex tags.

 

[neutrino]

I'm still stuck with:
z^{2} = cos^{2}\theta + 2isin^{2}\theta - sin^{2}\theta

Shouldn't it be \cos^2{\theta} + 2\imath\cos{\theta}\sin{\theta} - \sin^2{\theta}?

 

[Oblio]

Shouldn't it be \cos^2{\theta} + 2\imath\cos{\theta}\sin{\theta} - \sin^2{\theta}?

yeah, sorry. Typo

 

[neutrino]

Did you also find the other expression using de Moivre's theorem/formula?

 

[Oblio]

oh i equate the 2 just for one of the two I am looking for? I thought that was for the other equation

 

[neutrino]

What do you get when you equate the two?

 

[Oblio]

Shouldn't it be \cos^2{\theta} + 2\imath\cos{\theta}\sin{\theta} - \sin^2{\theta}?

I've gotten to

cos2\theta=cos2\theta-isin2\theta+2isin\thetacos\theta

 

[Oblio]

im realizing now that's wrong isn't it..

 

[neutrino]

I've gotten to

cos2\theta=cos2\theta-isin2\theta+2isin\thetacos\theta

When is a + ib = c + id? (a,b,c,d are real)

 

[Oblio]

whennn...
a=c, b=d?

 

[neutrino]

Right.

So when you equate cos(2t) + isin(2t) and cos^2(t) + 2isin(t)cos(t) - sin^2(t)...?

(using 't' instead of theta.)

 

[Oblio]

well if i use what I am trying to prove,
cos(2t) = cos^2(t)-sin^2(t)
and sin(2t) = 2sintcost

 

[Oblio]

but doesn't this method require that which I am proving, to prove it?

 

[neutrino]

well if i use what I am trying to prove,
cos(2t) = cos^2(t)-sin^2(t)
and sin(2t) = 2sintcost

but doesn't this method require that which I am proving, to prove it?

No. You're just equating the real and imaginary parts of the complex numbers. All you're using is the definition of equality of two complex numbers. (a = c, b = d)

 

[Oblio]

No. You're just equating the real and imaginary parts of the complex numbers. All you're using is the definition of equality of two complex numbers. (a = c, b = d)

So is it a valid proof to say that

cos2t +isin2t = cos^2(t) + 2isintcost - sin^2(t)

cos2t must = cos^2(t)-sin^2(t)
sin2t=2sintcost ?

 

[neutrino]

They must be equal since the corresponding real and imaginary components are equal.

I think writing the RHS as (cos^2(t) - sin^2(t)) + i(2sin(t)cos(t)) will make things familiar to you.

 

[Oblio]

Ok.

If the i's cancel leaving:
cos2t + sin2t = cos^2(t) - sin^2(t) + 2sin(t)cos(t)

There are all real parts. You don't think I'll need to prove the two RHS pieces?