# Trigonometry/Polar

Shown is θ in the first quadrant. If r is parallel to the y axis, the angle, θ is a right angle: θ=π/2, x=0, and y=r.

While SohCahToa is a popular mnemonic, it is better to define three essential trig functions using polar coordinates. The figure shows how the sine and cosine functions can be defined for all θ, i.e., angles outside the first quadrant (0<θ<π/2).

 ${\displaystyle x=r\cos \theta }$ ${\displaystyle y=r\sin \theta }$

This calculates x and y if r and θ are known.

Example: Suppose θ is in the second quadrant, e.g., θ=3π/4, we see that x<0, and hence cos(3π/4)<0. On the other hand, sin(3π/4)>0 in the second quadrant, since y>0 for this value of θ.

The following calculates r and θ if x and y are known:

 ${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$ ${\displaystyle \theta =\tan ^{-1}{\frac {y}{x}}}$

The tangent and its inverse function are defined by

 ${\displaystyle \tan \theta ={\frac {y}{x}}}$ ${\displaystyle \tan ^{-1}\left({\frac {y}{x}}\right)=\theta }$

This simple relation between the tangent and its inverse holds only in the first quadrant (0<θ<π/2) because the inverse function is not well defined for all angles. Also,it must be emphasized that the exponent ${\displaystyle -1}$ on the ${\displaystyle \tan }$ function does NOT represent the multiplicative inverse:

${\displaystyle \tan ^{-1}\theta \neq {\frac {1}{\tan \theta }}}$

For this reason, some authors avoid the superscript "-1" and instead write the arctangent as arctan:

${\displaystyle \theta =\tan ^{-1}{\frac {y}{x}}\equiv \arctan {\frac {y}{x}}}$

Just as lines of constant x and y are used to illustrate a Cartesian coordinate system contours of constant r and θ are used to depict polar coordinates as shown above and to the right.

In physics these equations are often used to describe the components of a vector. If Ax and Ay are the components of the vector A:

 ${\displaystyle A_{x}=A\cos \theta }$ ${\displaystyle A_{y}=A\sin \theta }$ ${\displaystyle A={\sqrt {A_{x}^{2}+A_{y}^{2}}}}$ ${\displaystyle \theta =\tan ^{-1}{\frac {A_{y}}{A_{x}}}\Leftrightarrow \tan \theta ={\frac {A_{y}}{A_{x}}}}$