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

[member 731016]

Homework Statement

Pls see the statement below

Relevant Equations

Binomial theorem

Why when proving trig identities,

Do we assume that r = 1 from $rcis\theta = r[\cos\theta + i\sin\theta]$? This makes me think that this is somehow it is related the unit circle.

Note: I am trying to prove the $cos3\theta$ identity and am curious why we assume that the modulus is 1.

Many thanks!

 

[PeroK]

Eh?

 

[member 731016]

Eh?

Thank you for your reply @PeroK!

Sorry what do you mean? I can explain.

Many thanks!

 

[member 731016]

Eh?

I have edited the post @PeroK . Dose that help?

 

[PeroK]

Thank you for your reply @PeroK!

Sorry what do you mean? I can explain.

Many thanks!

What do you mean? What $r$?

 

[PeroK]

Include $r$ if you want. It will make no difference to the final identity.

 

[SammyS]

Homework Statement:: Pls see the statement below
Relevant Equations:: Binomial theorem

Why when proving trig identities,
View attachment 323655
Do we assume that r = 1 from $rcis\theta = r[\cos\theta + i\sin\theta]$? This makes me think that this is somehow it is related the unit circle.

Note: I am trying to prove the $cos3\theta$ identity and am curious why we assume that the modulus is 1.

Many thanks!

Here we go again.

No. We don't need to assume that the modulus of $\displaystyle \cos(\theta)+i\sin(\theta)$ is $1$ . You should be able to show it.

What is the modulus of $\displaystyle x+iy$ ?

 

[member 731016]

What do you mean? What $r$?

Thank you for your reply @PeroK!

$r$ is the modulus

Many thanks!

 

[member 731016]

Include $r$ if you want. It will make no difference to the final identity.

Thank you for your reply @PeroK!

Sorry I forgot to mention that $r$ is just some scalar (modulus)

Many thanks!

 

[Mark44]

$\cos(\theta) + i\sin(\theta)$ is the polar form of a point P(x, y) on the unit circle. The ray from the origin to point P makes an angle of $\theta$ with the horizontal axis. This fact is something you would have learned in a precalc class...

Since P is a point on the unit circle, its modulus r = ... ?

 

[member 731016]

Here we go again.

No. We don't need to assume that the modulus of $\displaystyle \cos(\theta)+i\sin(\theta)$ is $1$ . You should be able to show it.

What is the modulus of $\displaystyle x+iy$ ?

Thank you for your reply @SammyS!

Sorry what do you mean show it? Do you mean by by trying to do the proof when r = 2 say?

Many thanks!

 

[member 731016]

$\cos(\theta) + i\sin(\theta)$ is the polar form of a point P(x, y) on the unit circle. The ray from the origin to point P makes an angle of $\theta$ with the horizontal axis. This fact is something you would have learned in a precalc class...

Since P is a point on the unit circle, its modulus r = ... ?

Thank you for your reply @Mark44!

I think I did learn. But I though the point on a unit circle was $(\cos(\theta), \sin(\theta))$ not with the imaginary component? Are you talking about a unit circle in the complex plane? I can see that would have a point $(\cos(\theta), i\sin(\theta))$.

Many thanks!

 

[member 731016]

$\cos(\theta) + i\sin(\theta)$ is the polar form of a point P(x, y) on the unit circle. The ray from the origin to point P makes an angle of $\theta$ with the horizontal axis. This fact is something you would have learned in a precalc class...

Since P is a point on the unit circle, its modulus r = ... ?

Thank you for your reply @Mark44!

r = 1

Many thanks!

 

[Mark44]

Do we assume that r = 1 from $rcis\theta = r[\cos\theta + i\sin\theta]$?

No.

But the problem you posted doesn't include r. It is a proof of the identity $(\cos(\theta) + i \sin(\theta))^3 = \cos(3\theta) + i\sin(3\theta)$.

 

[member 731016]

Here we go again.

No. We don't need to assume that the modulus of $\displaystyle \cos(\theta)+i\sin(\theta)$ is $1$ . You should be able to show it.

What is the modulus of $\displaystyle x+iy$ ?

Thank you for your reply @SammyS!

$r = \sqrt {x^2 + y^2}$

Many thanks!

 

[member 731016]

No.

But the problem you posted doesn't include r. It is a proof of the identity $(\cos(\theta) + i \sin(\theta))^3 = \cos(3\theta) + i\sin(3\theta)$.

Thank you for your reply @Mark44!

I am wondering whether the identity can be proved in general for any r. That is, if we let
$(r[\cos(\theta) + i\sin(\theta)])^3$.

EDIT: I believe it would work, but why?

Many thanks!

 

[Mark44]

I am wondering whether the identity can be proved in general for any r.

Not for just any old r, but it can be proved for integer values of r. This is what De Moivre's Theorem is concerned with. See https://math.libretexts.org/Bookshe...Moivres_Theorem_and_Powers_of_Complex_Numbers.

 

[member 731016]

Not for just any old r, but it can be proved for integer values of r. This is what De Moivre's Theorem is concerned with. See https://math.libretexts.org/Bookshelves/Precalculus/Book:_Trigonometry_(Sundstrom_and_Schlicker)/05:_Complex_Numbers_and_Polar_Coordinates/5.03:_DeMoivres_Theorem_and_Powers_of_Complex_Numbers.

Thank you for your reply @Mark44!

I will check that out. Sorry, I was meant to post that question I had (accidently in the intro physics forum) about complex identity in this forum.

Many thanks!

 

[malawi_glenn]

You basically need to understand where those identities come from $z^n + \dfrac{1}{z^n} = 2 \cos (n\theta)$ in your (now) closed thread https://www.physicsforums.com/threads/another-way-to-find-trig-identities.1050812/

Here is a guide you can follow to prove the identity above.
1) any complex number $z$ where $|z| = 1$ can by definition be written as $\cos \theta + \mathrm{i} \sin \theta$.
2) $z^n = \cos (n\theta) + \mathrm{i} \sin (n\theta)$ where $n$ is a positive integer. This one is pretty simple to prove, you have probably done so in class. Otherwise, it is pretty simple to show using a proof by induction, e.g. show that if $z^p = \cos (p\theta) + \mathrm{i} \sin (p\theta)$ where $p$ is a positive integer, then show the following also holds: $z^{p+1} = \cos ((p+1)\theta) + \mathrm{i} \sin ((p+1)\theta)$
3) Next things to understand, is that $\dfrac{1}{z^n} = \cos (-n\theta) + \mathrm{i} \sin (-n\theta)$ for $n$ any integer, positive or negative (and 0 of course). This is also pretty simple to show. Use that cos is even function and that sin is odd function. Then show that this indeed is equal to $\dfrac{1}{z^n}$.
4) Now you are almost there. Should be straightforward to show that $z^n + \dfrac{1}{z^n} = 2 \cos (n\theta)$ at this point.

 

[SammyS]

Thank you for your reply @SammyS!

$r = \sqrt {x^2 + y^2}$

Many thanks!

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.

 

[FactChecker]

You basically need to understand where those identities come from $z^n + \dfrac{1}{z^n} = 2 \cos (n\theta)$ in your (now) closed thread https://www.physicsforums.com/threads/another-way-to-find-trig-identities.1050812/

Here is a guide you can follow to prove the identity above.
1) any complex number $z$ can by definition be written as $\cos \theta + \mathrm{i} \sin \theta$.

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 $\theta = \arg(z)$.

 

[member 731016]

You basically need to understand where those identities come from $z^n + \dfrac{1}{z^n} = 2 \cos (n\theta)$ in your (now) closed thread https://www.physicsforums.com/threads/another-way-to-find-trig-identities.1050812/

Here is a guide you can follow to prove the identity above.
1) any complex number $z$ can by definition be written as $\cos \theta + \mathrm{i} \sin \theta$.
2) $z^n = \cos (n\theta) + \mathrm{i} \sin (n\theta)$ where $n$ is a positive integer. This one is pretty simple to prove, you have probably done so in class. Otherwise, it is pretty simple to show using a proof by induction, e.g. show that if $z^p = \cos (p\theta) + \mathrm{i} \sin (p\theta)$ where $p$ is a positive integer, then show the following also holds: $z^{p+1} = \cos ((p+1)\theta) + \mathrm{i} \sin ((p+1)\theta)$
3) Next things to understand, is that $\dfrac{1}{z^n} = \cos (-n\theta) + \mathrm{i} \sin (-n\theta)$ for $n$ any integer, positive or negative (and 0 of course). This is also pretty simple to show. Use that cos is even function and that sin is odd function. Then show that this indeed is equal to $\dfrac{1}{z^n}$.
4) Now you are almost there. Should be straightforward to show that $z^n + \dfrac{1}{z^n} = 2 \cos (n\theta)$ at this point.

Thank you for your reply @malawi_glenn!

That is very helpful!

I will try to prove it.

Many thanks!

 

[member 731016]

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.

Thank you for your reply @SammyS!

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

 

[FactChecker]

Homework Statement:: Pls see the statement below
Relevant Equations:: Binomial theorem

Why when proving trig identities,
View attachment 323655
Do we assume that r = 1 from $rcis\theta = r[\cos\theta + i\sin\theta]$? This makes me think that this is somehow it is related the unit circle.

Yes, it is. $[\cos(\theta) + i \sin(\theta)]^3 = [e^{i \theta}]^3$. Now use the properties of exponents and put it back into the cos(), i sin() form.

 

[member 731016]

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 $\theta = \arg(z)$.

Thank you for your reply @FactChecker !

True it is $z = |z|e^{i\theta}$

Many thanks!

 

[member 731016]

Yes, it is. $[\cos(\theta) + i \sin(\theta)]^3 = [e^{i \theta}]^3$. Now use the properties of exponents and put it back into the cos(), i sin() form.

Thank you for your reply @FactChecker!

Sorry what do you mean?

Many thanks!

 

[member 731016]

Yes, it is. $[\cos(\theta) + i \sin(\theta)]^3 = [e^{i \theta}]^3$. Now use the properties of exponents and put it back into the cos(), i sin() form.

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!

 

[SammyS]

Thank you for your reply @SammyS!

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

Not quite right.

Details:

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

is wrong in a couple of ways,

 

[member 731016]

Not quite right.

Details:

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

is wrong in a couple of ways,

Thank you for your reply @SammyS!

Sorry how?

Many thanks!

 

[SammyS]

Thank you for your reply @SammyS!

Sorry how?

Many thanks!

You tell me.

 

[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!