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

[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.

 

[ChiralSuperfields]

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## ?

 

[ChiralSuperfields]

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)

 

[ChiralSuperfields]

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.

 

[ChiralSuperfields]

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.

 

[ChiralSuperfields]

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.

 

[ChiralSuperfields]

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

 

[ChiralSuperfields]

##|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##

 

[ChiralSuperfields]

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.