Proving trig identities -- Is the method related to the unit circle?

[member 731016]

You tell me.

Thank you for your reply @SammyS!I let $x = \cos\theta$ and $y = i\sin\theta$ from the complex circle.

Many thanks!

 

[SammyS]

Thank you for your reply @SammyS!I let $x = \cos\theta$ and $y = i\sin\theta$ from the complex circle.

Many thanks!

No.

$y$ does not include the $i$.

Take some time before blasting back with a thoughtless answer.

 

[FactChecker]

Thank you for your reply @FactChecker!

Sorry never mind. I think I understand now.

$[e^{i\theta}]^3 = e^{3\theta i}$ which from the definition of Euler's formula gives
$\cos3\theta + i\sin3\theta$ interesting that we don't have to use de moiré's theorem.Many thanks!

I guess it depends on what class this is for, and if you are allowed to use Euler's formula. Many people consider Euler's formula to be the most important equation in mathematics.

 

[member 731016]

No.

$y$ does not include the $i$.

Take some time before blasting back with a thoughtless answer.

Thank you for your reply @SammyS!

I just realized that before I read you reply! For some reason I was thinking that it need an i for a point on a unit circle in complex plane.

Many thanks!

 

[member 731016]

I guess it depends on what class this is for, and if you are allowed to use Euler's formula. Many people consider Euler's formula to be the most important equation in mathematics.

Thank you for your reply @FactChecker !

This is for a linear algebra class.

Many thanks!

 

[SammyS]

Not quite right.

Details:

$\displaystyle \sqrt {\cos^2\theta + (i^2\sin\theta)}$

is wrong in a couple of ways,

You still haven't correctly stated what is wrong with this.

BY the way, the rest of what you had was correct, That is; $\displaystyle r=\sqrt {\cos^2\theta+\sin^2\theta}=1$
is correct.

 

[member 731016]

You still haven't correctly stated what is wrong with this.

BY the way, the rest of what you had was correct, That is; $\displaystyle r=\sqrt {\cos^2\theta+\sin^2\theta}=1$
is correct.

Thank you for your reply @SammyS!

What was wrong was the $i$, if I remove it, it solves the problem. It is kind of cool that it still gave the correct answer with the i, which makes me wonder.

Many thanks!

 

[SammyS]

Thank you for your reply @SammyS!

What was wrong was the $i$, if I remove it, it solves the problem. It is kind of cool that it still gave the correct answer with the i, which makes me wonder.

Many thanks!

It does not give the correct answer with the $i$ included, even if you square the $\sin \theta$.

What is $\displaystyle i^2$ ?

 

[member 731016]

It does not give the correct answer with the $i$ included, even if you square the $\sin \theta$.

What is $\displaystyle i^2$ ?

Thank you for your reply @SammyS!

Oh true! $i^2 = -1$ - I see now!

Thank you!

 

[malawi_glenn]

No, this is only true if $|z| = 1$. In general,
$z = |z|e^{i \arg(z)} = |z|(\cos(\arg(z))+i\sin(\arg(z))) = |z|(\cos(\theta)+i\sin(\theta))$, where $\theta = \arg(z)$.

Yeah true I forgot to write "where $|z| = 1$" it was very late for me (in little sweden)

 

[member 731016]

Yeah true I forgot to write "where $|z| = 1$" it was very late for me (in little sweden)

Thank you for your help @malawi_glenn !

 

[PeroK]

Oh, so $r = \sqrt {\cos^2\theta + (i^2\sin\theta)} = \sqrt { \cos^2\theta + \sin^2\theta} = 1$

That's just so wrong!

 

[PeroK]

Thank you for your help @malawi_glenn !

You need to stop cluttering up your threads with all these unnecessary thanks.

 

[member 731016]

You need to stop cluttering up your threads with all these unnecessary thanks.

Thank you for your replies @PeroK!

I like to be thankful to your guys spending you time helping me.

 

[member 731016]

That's just so wrong!

Agree!

 

[Mark44]

You need to stop cluttering up your threads with all these unnecessary thanks.

Thank you for your replies @PeroK!

I like to be thankful to your guys spending you time helping me.

That's a nice sentiment, but I agree with @PeroK that putting in a thank you reply in every post is way overdoing it.

 

[member 731016]

That's a nice sentiment, but I agree with @PeroK that putting in a thank you reply in every post is way overdoing it.

Ok I will say thank you less in precalculus forums.

 

[malawi_glenn]

Ok I will say thank you less in precalculus forums.

No need in making a new reply just to say thanks. A "like" is just as good as showing apprechiation.

We also get a notice that you have replied and thus we check if you have more questions. With likes we just assume it was a thanks

 

[malawi_glenn]

$|z|^2= z \bar z = r(\cos \theta + i\sin \theta) r(\cos \theta - i \sin \theta) = ...$ please continue

 

[member 731016]

$|z|^2= z \bar z = r(\cos \theta + i\sin \theta) r(\cos \theta - i \sin \theta) = ...$ please continue

No need in making a new reply just to say thanks. A "like" is just as good as showing apprechiation.

We also get a notice that you have replied and thus we check if you have more questions. With likes we just assume it was a thanks

Thank you for your replies @malawi_glenn !

$|z| = r^2[\cos^2\theta + \sin^2\theta] = r^2[1] = r^2$

Many thanks!

 

[Mark44]

Post #47:

Ok I will say thank you less in precalculus forums.

Post #50:

Thank you for your replies @malawi_glenn !
<snip>
Many thanks!

Less $\ne$ more -- just saying.

 

[kuruman]

Thank you for your replies @malawi_glenn !

$|z| = r^2[\cos^2\theta + \sin^2\theta] = r^2[1] = r^2$

Many thanks!

The equation should begin with $|z|^2=\dots$

 

[member 731016]

The equation should begin with $|z|^2=\dots$

True, thanks for pointing that out @kuruman!

 

[chwala]

What do you mean by "Thanks"?

I intended that you apply that to $\displaystyle \cos(\theta)+i\sin(\theta)$ and show that its Modulus is indeed $1$ .

Math is a not a spectator sport. You must practice and do, not just watch and cheer.

@SammyS you made my day! I'll pass this to my students.

' Math is a not a spectator sport. You must practice and do, not just watch and cheer".

Nice one mate.

 

[chwala]

True, thanks for pointing that out @kuruman!

@ChiralSuperfields i would suggest on a light note that you avoid the many thanks remarks in your posts and instead stick to the material/work at hand. That is how you will gain respect and momentum in the subject. Get down to work! Cheers mate.