Question

What is the reference angle for $\theta ={{210}^{\circ }}$ ? Using that reference angle, determine the values for the six trigonometric functions of ${{210}^{\circ }}$ .

Answer 1

Hint: The reference angle for any given angle can be found by adding or subtracting ${{180}^{\circ }}$ to the given angle. We can express the given angle using the reference angle and use the trigonometric identities to find the value of all 6 trigonometric functions.

Complete step-by-step solution:
The angle given to us in the question is ${{210}^{\circ }}$. The reference angle can be found by either adding ${{180}^{\circ }}$, or by subtracting ${{180}^{\circ }}$. Therefore, for the given angle, the reference angle is,
$\begin{align}
& \theta ={{210}^{\circ }}-{{180}^{\circ }} \\
& \Rightarrow \theta ={{30}^{\circ }} \\
\end{align}$
Next we can find the six trigonometric functions by using the trigonometric identities.
${{210}^{\circ }}$ belongs to the third quadrant and sine function is negative in the third quadrant, therefore,
$\begin{align}
  &\sin \left( {{180}^{\circ }}+\theta \right)=-\sin \left( \theta \right) \\
 & \Rightarrow \sin \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=-\sin \left( {{30}^{\circ }} \right) \\
 &\Rightarrow \sin \left( {{210}^{\circ }} \right)=-\dfrac{1}{2} \\
\end{align}$
Cosine function is negative in the third quadrant, therefore,
$\begin{align}
  & \cos \left( {{180}^{\circ }}+\theta \right)=-\cos \left( \theta \right) \\
 & \Rightarrow \cos \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=-\cos \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \cos \left( {{210}^{\circ }} \right)=-\dfrac{\sqrt{3}}{2} \\
\end{align}$
Tangent function is positive in the third quadrant, therefore,
$\begin{align}
  & \tan \left( {{180}^{\circ }}+\theta \right)=\tan \left( \theta \right) \\
 & \Rightarrow \tan \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=\tan \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \tan \left( {{210}^{\circ }} \right)=\dfrac{1}{\sqrt{3}} \\
\end{align}$
Cosecant function is negative in the third quadrant, therefore,
$\begin{align}
  &\csc \left( {{180}^{\circ }}+\theta \right)=-\csc \left( \theta \right) \\
 & \Rightarrow \csc \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=-\csc \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \csc \left( {{210}^{\circ }} \right)=-2 \\
\end{align}$
Secant function is negative in the third quadrant, therefore,
$\begin{align}
  & \sec \left( {{180}^{\circ }}+\theta \right)=-\sec \left( \theta \right) \\
 & \Rightarrow \sec \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=-\sec \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \sec \left( {{210}^{\circ }} \right)=-\dfrac{2}{\sqrt{3}} \\
\end{align}$
Cotangent function is positive in the third quadrant, therefore,
$\begin{align}
  & \cot \left( {{180}^{\circ }}+\theta \right)=\cot \left( \theta \right) \\
 & \Rightarrow \cot \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=\cot \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \cot \left( {{210}^{\circ }} \right)=\sqrt{3} \\
\end{align}$

Note: While using the trigonometric identities one has to be careful in taking positive and negative signs as they vary widely in all quadrants for each of the trigonometric functions. To avoid this, it is better to know in which quadrants, which of the functions are positive and which are negative, this helps to make the simplification easy and error free.