Question

How do you find the value of $\cot \left( -{{150}^{\circ }} \right)$ ? \[\]

Answer 1

Hint: We recall the definition of sine, cosine and cotangent trigonometric ratios. We recall that we can convert from co-tangent to sine and cosine using $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ . We recall the negative angle $\cos \left( -\theta \right)=\cos \theta ,\sin \left( -\theta \right)=-\sin \theta $. We reduction formula by ${{180}^{\circ }}$ that is $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta ,\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ for $\theta ={{30}^{\circ }}$ to get the result. \[\]

Complete step by step answer:
We know that in the right angled triangle the side opposite to the right angled triangle is called hypotenuse denoted as $h$, the vertical side is called perpendicular denoted as $p$ and the horizontal side is called the base denoted as $b$.
We know from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of side opposite to the angle to the hypotenuse, cosine of an angle is the ratio of side adjacent to the angle (excluding hypotenuse) to the hypotenuse and co-tangent of the angle is the ratio of the adjacent side to opposite side. So we have for angle $\theta $
\[\sin \theta =\dfrac{p}{h},\cos \theta =\dfrac{b}{h},\cot \theta =\dfrac{b}{p}\]
So we have
\[\cot \theta =\dfrac{b}{p}=\dfrac{\dfrac{b}{h}}{\dfrac{p}{h}}=\dfrac{\cos \theta }{\sin \theta }\]
We are asked to find the value of $\cot \left( -{{150}^{\circ }} \right)$. We convert it sine and cosine to have for $\theta =-{{150}^{\circ }}$;
\[\cot \left( -{{150}^{\circ }} \right)=\dfrac{\cos \left( -{{150}^{\circ }} \right)}{\sin \left( -{{150}^{\circ }} \right)}\]
We know from reflection identities for negative angle orientation that $\cos \left( -\theta \right)=\cos \theta ,\sin \left( -\theta \right)=-\sin \theta $ for $\theta ={{150}^{\circ }}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\dfrac{\cos \left( {{150}^{\circ }} \right)}{-\sin \left( {{150}^{\circ }} \right)} \\
 & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\dfrac{\cos \left( {{180}^{\circ }}-{{30}^{\circ }} \right)}{-\sin \left( {{180}^{\circ }}-{{30}^{\circ }} \right)} \\
\end{align}\]
We use the reduction formula $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta ,\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ for $\theta ={{30}^{\circ }}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\dfrac{-\cos \left( {{30}^{\circ }} \right)}{-\sin \left( {{30}^{\circ }} \right)} \\
 & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\dfrac{\cos \left( {{30}^{\circ }} \right)}{\sin \left( {{30}^{\circ }} \right)} \\
\end{align}\]
We put the known values $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}$ in the above step to have the required value as ;
\[\Rightarrow \cot \left( -{{150}^{\circ }} \right)=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}=\sqrt{3}\]

Note:
We note that we can directly find using reflection identity $\cot \left( -\theta \right)=-\cot \theta $ to have
\[\begin{align}
  & \cot \left( -{{150}^{\circ }} \right)=-\cot {{150}^{\circ }} \\
 & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=-\cot \left( {{180}^{\circ }}-{{30}^{\circ }} \right) \\
\end{align}\]
We use reduction formula $\cot \left( {{180}^{\circ }}-\theta \right)=-\cot \theta $ for $\theta ={{30}^{\circ }}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=-\left( -\cot {{30}^{\circ }} \right) \\
 & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\cot {{30}^{\circ }} \\
 & \Rightarrow \cot \left( -{{150}^{\circ }} \right)=\sqrt{3} \\
\end{align}\]
We note that reduction formula $\left( {{180}^{\circ }}-\theta \right)$ is different from shift formula $\left( {{180}^{\circ }}+\theta \right)$. The shift formulae are given as $\cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta ,\sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta $. We should remember quadrant values to remember easily.