Converting from rectangular to polar form

[ausmathgenius420]

I'm trying to convert the following to polar form:
[math]1+cot(\theta)i[/math]The modulus is then [math]\sqrt[2]{cosec^2}[/math]I would think the polar form is then just:
[math]cosec(\theta)cis(\theta)[/math]However the answers show:
[math]cosec(\theta)cis(\frac{\pi}{2}-\theta)[/math]
Can someone pls explain why this is?

 

[topsquark]

I'm trying to convert the following to polar form:
[math]1+cot(\theta)i[/math]The modulus is then [math]\sqrt[2]{cosec^2}[/math]I would think the polar form is then just:
[math]cosec(\theta)cis(\theta)[/math]However the answers show:
[math]cosec(\theta)cis(\frac{\pi}{2}-\theta)[/math]
Can someone pls explain why this is?

You are taking the inverse tangent of the cotangent function, not the tangent function so [imath]atn(cot( \theta )) \neq \theta[/imath].

-Dan

 

[Dr.Peterson]

I'm trying to convert the following to polar form: [imath]1+cot(\theta)i[/imath]
The modulus is then [imath]\sqrt[2]{cosec^2}[/imath]
I would think the polar form is then just: [imath]cosec(\theta)cis(\theta)[/imath]
However the answers show: [imath]cosec(\theta)cis(\frac{\pi}{2}-\theta)[/imath]

Can someone pls explain why this is?

Take an example. Suppose [imath]\theta=\frac{\pi}{6}[/imath], so that [imath]\cot(\theta)=\sqrt{3}[/imath].

Then [imath]1+cot(\theta)i=1+i\sqrt{3}[/imath]. Its modulus is indeed [imath]\sqrt{1^2+\sqrt{3}^2}=2=\cosec(\frac{\pi}{6})[/imath].

But its argument is [imath]\arctan(\frac{\sqrt{3}}{1})=\frac{\pi}{3}[/imath], not [imath]\frac{\pi}{6}[/imath].

Can you explain why you think it should be [imath]\theta[/imath]? Explaining your thinking can help you correct it!

 

[ausmathgenius420]

Take an example. Suppose [imath]\theta=\frac{\pi}{6}[/imath], so that [imath]\cot(\theta)=\sqrt{3}[/imath].

Then [imath]1+cot(\theta)i=1+i\sqrt{3}[/imath]. Its modulus is indeed [imath]\sqrt{1^2+\sqrt{3}^2}=2=\cosec(\frac{\pi}{6})[/imath].

But its argument is [imath]\arctan(\frac{\sqrt{3}}{1})=\frac{\pi}{3}[/imath], not [imath]\frac{\pi}{6}[/imath].

Can you explain why you think it should be [imath]\theta[/imath]? Explaining your thinking can help you correct it!

I'm unsure how to find the argument. The other textbook problems haven't involved trig components.

If I make an argand diagram, I get:
[imath]\theta=tan^-(cot(\theta))[/imath]

 

[Dr.Peterson]

I'm unsure how to find the argument. The other textbook problems haven't involved trig components.

If I make an argand diagram, I get:
[imath]\theta'=\tan^{-1}(\cot(\theta))[/imath]

I corrected what I think you meant there. You meant the inverse tangent, and you didn't mean to say both thetas are the same.

Now, what angle has a tangent equal to [imath]\cot(\theta)[/imath]? How are tangent and cotangent related?

But I'd still like you to answer my question: Why did you think the answer should be [imath]\theta[/imath]? Thinking about your own thinking is an important learning skill.

 

[ausmathgenius420]

I corrected what I think you meant there. You meant the inverse tangent, and you didn't mean to say both thetas are the same.

Now, what angle has a tangent equal to [imath]\cot(\theta)[/imath]? How are tangent and cotangent related?

But I'd still like you to answer my question: Why did you think the answer should be [imath]\theta[/imath]? Thinking about your own thinking is an important learning skill.

I'm not sure why I had assumed the argument was theta sorry.
And yes the corrections are right.

I know: [math]cot=\frac{1}{tan}[/math]
I graphed [imath]tan(x)[/imath] and [imath]cot(x)[/imath] in Desmos. [imath]tan(x)[/imath] has asymptotes of [imath]+-\frac{\pi}{2}[/imath] and an x intercept of 0. [imath]cot(x)[/imath] is translated [imath]\frac{\pi}{2}[/imath] units left/right and the shape is 'flipped'.

If generally, [imath]cot(x)=tan(\frac{\pi}{2}-x)[/imath] then [imath]\theta'=\frac{\pi}{2}-\theta[/imath]

But that general rule wouldn't account for the flipped shape would it?

 

[Dr.Peterson]

I'm not sure why I had assumed the argument was theta sorry.
And yes the corrections are right.

I know: [math]cot=\frac{1}{tan}[/math]
I graphed [imath]tan(x)[/imath] and [imath]cot(x)[/imath] in Desmos. [imath]tan(x)[/imath] has asymptotes of [imath]+-\frac{\pi}{2}[/imath] and an x intercept of 0. [imath]cot(x)[/imath] is translated [imath]\frac{\pi}{2}[/imath] units left/right and the shape is 'flipped'.

If generally, [imath]cot(x)=tan(\frac{\pi}{2}-x)[/imath] then [imath]\theta'=\frac{\pi}{2}-\theta[/imath]

But that general rule wouldn't account for the flipped shape would it?

It's the last relationship you mentioned between tan and cot that matters here, which I hinted at by bolding the "co". They are "co-functions", which means the cotangent of an angle is the tangent of its complement. So the angle whose tangent is cot(x) is the complement of x, namely pi/2 - x. And that's the answer you're looking for.

And, yes, that is the explanation for the "flipping". Replacing x with -x reflects a graph in the y-axis; and adding pi/2 shifts the graph horizontally (by half the period, in this case). Together, those do what you observed.

I think you explained it all nicely, so I'm not sure why you asked the last question.

 

[ausmathgenius420]

It's the last relationship you mentioned between tan and cot that matters here, which I hinted at by bolding the "co". They are "co-functions", which means the cotangent of an angle is the tangent of its complement. So the angle whose tangent is cot(x) is the complement of x, namely pi/2 - x. And that's the answer you're looking for.

And, yes, that is the explanation for the "flipping". Replacing x with -x reflects a graph in the y-axis; and adding pi/2 shifts the graph horizontally (by half the period, in this case). Together, those do what you observed.

I think you explained it all nicely, so I'm not sure why you asked the last question.

Sorry for the late reply.

I think you explained it all nicely, so I'm not sure why you asked the last question.

Yes I see the negative sign now . I will usually write trig functions with the x first inside the brackets. Something about tan(-x +pi/2) is clearer to me than tan(pi/2-x). Thanks