# Proving trig identities

Callumnc1
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!

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Eh?

Callumnc1
Callumnc1
Eh?

Sorry what do you mean? I can explain.

Many thanks!

Callumnc1
Eh?
I have edited the post @PeroK . Dose that help?

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Sorry what do you mean? I can explain.

Many thanks!
What do you mean? What ##r##?

Callumnc1
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Include ##r## if you want. It will make no difference to the final identity.

Callumnc1
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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 ## ?

Callumnc1
Callumnc1
What do you mean? What ##r##?

##r## is the modulus

Many thanks!

Callumnc1
Include ##r## if you want. It will make no difference to the final identity.

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

Many thanks!

Mentor
##\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 = ... ?

Callumnc1
Callumnc1
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 ## ?

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

Many thanks!

Callumnc1
##\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 = ... ?

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!

Callumnc1
##\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 = ... ?

r = 1

Many thanks!

Mentor
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)##.

Callumnc1
Callumnc1
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 ## ?

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

Many thanks!

Callumnc1
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)##.

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!

Callumnc1
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.

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!

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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.

Last edited:
scottdave and Callumnc1
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##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.

PeroK and Callumnc1
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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)##.

malawi_glenn and Callumnc1
Callumnc1
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.

I will try to prove it.

Many thanks!

Callumnc1
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.

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

Many thanks!

Homework Helper
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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.

scottdave and Callumnc1
Callumnc1
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)##.

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

Many thanks!

Callumnc1
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.

Sorry what do you mean?

Many thanks!

Callumnc1
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.

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!

FactChecker
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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,

Callumnc1
Callumnc1
Not quite right.

Details:

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

is wrong in a couple of ways,

Sorry how?

Many thanks!

Staff Emeritus
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Sorry how?

Many thanks!
You tell me.

Callumnc1
Callumnc1
You tell me.

I let ##x = \cos\theta## and ##y = i\sin\theta## from the complex circle.

Many thanks!

Staff Emeritus
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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.

Callumnc1
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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.

Callumnc1
Callumnc1
No.

##y## does not include the ##i##.

Take some time before blasting back with a thoughtless answer.

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!

Callumnc1
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.