Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

How do you convert \[(\theta )=\dfrac{\pi }{3}\] into cartesian form?

Answer
VerifiedVerified
558.3k+ views
Hint: cartesian coordinates describe the point how far it is from the x-axis and y-axis whereas polar coordinates describe how far the point from origin and how much angle it makes with the x-axis in an anticlockwise direction. To convert polar coordinates into cartesian we use a right-angle triangle and convert it into cartesian coordinates.

Complete step by step answer:
As per the given question, we need to convert \[\theta \] into cartesian form. To convert a point in polar form \[\left( r,\theta \right)\] to \[\left( x,y \right)\].
To convert polar coordinates \[\left( r,\theta \right)\] to cartesian coordinates \[\left( x,y \right)\], we use the relation
\[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
Here we have no r to consider, so from the relation \[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]. On substituting the \[\theta \] value in the equation. we get
\[\Rightarrow \dfrac{\pi }{3}={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
On applying tan on both sides, the equation becomes
\[\Rightarrow \tan \dfrac{\pi }{3}=\tan \left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)\]
We know that the value of \[\tan \dfrac{\pi }{3}=\sqrt{3}\]. On substituting this value we get the equation as \[\Rightarrow \sqrt{3}=\tan \left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)\]
We also know that \[\tan \left( {{\tan }^{-1}}a \right)=a\]. On substituting this the equation becomes
\[\Rightarrow \sqrt{3}=\dfrac{y}{x}\]
On shifting x to the other side. We can rewrite the equation as
\[\Rightarrow y=\sqrt{3}x\]
Here, we got a straight line since \[(\theta )=\dfrac{\pi }{3}\] takes all values of r.
Therefore, the cartesian form of \[(\theta )=\dfrac{\pi }{3}\] is \[y=\sqrt{3}x\].

Note:
In order to solve such types of questions, we need to have enough knowledge over how to convert from polar to cartesian coordinates and also inverse trigonometric functions and their properties. We also need to know the algebraic formulae to simplify the expressions. We also need to know the trigonometric values of basic angles. We must avoid calculation mistakes to get the correct solution.