Converting from rectangular to polar form

[ausmathgenius420]

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

 

[topsquark]

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

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

-Dan

 

[Dr.Peterson]

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

Can someone pls explain why this is?

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

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

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

Can you explain why you think it should be $\theta$? Explaining your thinking can help you correct it!

 

[ausmathgenius420]

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

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

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

Can you explain why you think it should be $\theta$? 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:
$\theta=tan^-(cot(\theta))$

 

[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:
$\theta'=\tan^{-1}(\cot(\theta))$

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 $\cot(\theta)$? How are tangent and cotangent related?

But I'd still like you to answer my question: Why did you think the answer should be $\theta$? 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 $\cot(\theta)$? How are tangent and cotangent related?

But I'd still like you to answer my question: Why did you think the answer should be $\theta$? 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: $$cot=\frac{1}{tan}$$
I graphed $tan(x)$ and $cot(x)$ in Desmos. $tan(x)$ has asymptotes of $+-\frac{\pi}{2}$ and an x intercept of 0. $cot(x)$ is translated $\frac{\pi}{2}$ units left/right and the shape is 'flipped'.

If generally, $cot(x)=tan(\frac{\pi}{2}-x)$ then $\theta'=\frac{\pi}{2}-\theta$

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

If generally, $cot(x)=tan(\frac{\pi}{2}-x)$ then $\theta'=\frac{\pi}{2}-\theta$

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