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

• member 731016
In summary, proving trigonometric identities involves using algebraic manipulations and identities to show that two expressions are equal. This process is not directly related to the unit circle, as it is more focused on manipulating equations, but the unit circle can be used to help visualize and understand the relationships between trigonometric functions.
SammyS said:
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
SammyS said:
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.

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!

Callumnc1 said:

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
SammyS said:
It does not give the correct answer with the ##i## included, even if you square the ##\sin \theta##.

What is ##\displaystyle i^2## ?

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

Thank you!

FactChecker said:
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 and FactChecker
malawi_glenn said:
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 !

Callumnc1 said:
Oh, so ## r = \sqrt {\cos^2\theta + (i^2\sin\theta)} = \sqrt { \cos^2\theta + \sin^2\theta} = 1##
That's just so wrong!

member 731016
Callumnc1 said:
Thank you for your help @malawi_glenn !
You need to stop cluttering up your threads with all these unnecessary thanks.

chwala, Mark44 and member 731016
PeroK said:
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.

PeroK said:
That's just so wrong!
Agree!

PeroK said:
You need to stop cluttering up your threads with all these unnecessary thanks.
Callumnc1 said:
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.

chwala, malawi_glenn, member 731016 and 3 others
Mark44 said:
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.

Callumnc1 said:
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

Last edited:
chwala
##|z|^2= z \bar z = r(\cos \theta + i\sin \theta) r(\cos \theta - i \sin \theta) = ... ## please continue

malawi_glenn said:
##|z|^2= z \bar z = r(\cos \theta + i\sin \theta) r(\cos \theta - i \sin \theta) = ... ## please continue
malawi_glenn said:
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!

malawi_glenn
Post #47:
Callumnc1 said:
Ok I will say thank you less in precalculus forums.
Post #50:
Callumnc1 said:
Thank you for your replies @malawi_glenn !
<snip>
Many thanks!
Less ##\ne## more -- just saying.

chwala and member 731016
ChiralSuperfields said:
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 and malawi_glenn
kuruman said:
The equation should begin with ##|z|^2=\dots##
True, thanks for pointing that out @kuruman!

chwala
SammyS said:
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.

member 731016
ChiralSuperfields said:
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.

member 731016

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