Question

How do you find the six trigonometric functions of 390 degrees?

Answer 1

Hint: Assume the given angle as \[\theta \] and convert it from degrees to radian using the relation: - 180 degrees = \[\pi \] radian. Now, use the relation: - \[\cos \left( 2\pi +\theta \right)=\cos \theta \] to find the value of \[\cos \left( {{390}^{\circ }} \right)\]. Similarly, use the formula: - \[\sin \left( 2\pi +\theta \right)=\sin \theta \] to find the value of \[\sin \left( {{390}^{\circ }} \right)\]. Further, to find the value of \[\tan \theta \], use the relation: - \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Finally, find the remaining three trigonometric ratios given as: - \[\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\] and \[\cot \theta =\dfrac{1}{\tan \theta }\].

Complete step by step answer:
Here, we have been provided with angle 390 degrees and we are asked to find all the six trigonometric functions of this angle. But first let us convert the given angle in degrees into angle in radian.
Now, we know that 180 degrees = \[\pi \] radian, so using the unitary method, we have,
\[\Rightarrow {{1}^{\circ }}=\dfrac{\pi }{180}\] radian
\[\Rightarrow {{390}^{\circ }}=\dfrac{390\pi }{180}\] radian
\[\Rightarrow {{390}^{\circ }}=\dfrac{13\pi }{6}\] radian
This can be written as: -
\[\Rightarrow \dfrac{13\pi }{6}=\left( 2\pi +\dfrac{\pi }{6} \right)\]
Now, let us find the value of cosine of the given angle first, so we have,
\[\Rightarrow \cos \left( {{390}^{\circ }} \right)=\cos \left( 2\pi +\dfrac{\pi }{6} \right)\]
Using the formula: - \[\cos \left( 2\pi +\theta \right)=\cos \theta \], where \[\theta =\dfrac{\pi }{6}\], we get,
\[\begin{align}
  & \Rightarrow \cos \left( {{390}^{\circ }} \right)=\cos \left( 2\pi +\dfrac{\pi }{6} \right)=\cos \left( \dfrac{\pi }{6} \right) \\
 & \Rightarrow \cos \left( {{390}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Now, let us find the value of sine of the provided angle, so we have,
\[\Rightarrow \sin \left( {{390}^{\circ }} \right)=\sin \left( 2\pi +\dfrac{\pi }{6} \right)\]
Using the formula: - \[\sin \left( 2\pi +\theta \right)=\sin \theta \], we get,
\[\begin{align}
  & \Rightarrow \sin \left( {{390}^{\circ }} \right)=\sin \left( \dfrac{\pi }{6} \right) \\
 & \Rightarrow \sin \left( {{390}^{\circ }} \right)=\dfrac{1}{2} \\
\end{align}\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\], so we have,
\[\begin{align}
  & \Rightarrow \tan \left( {{390}^{\circ }} \right)=\dfrac{\sin \left( {{390}^{\circ }} \right)}{\cos \left( {{390}^{\circ }} \right)} \\
 & \Rightarrow \tan \left( {{390}^{\circ }} \right)=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \\
 & \Rightarrow \tan \left( {{390}^{\circ }} \right)=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Now, we know that \[\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\] and \[\cot \theta =\dfrac{1}{\tan \theta }\], so using these relations we get,
\[\begin{align}
  & \Rightarrow \sec \left( {{390}^{\circ }} \right)=\dfrac{1}{\cos \left( {{390}^{\circ }} \right)} \\
 & \Rightarrow \sec \left( {{390}^{\circ }} \right)=\dfrac{1}{\dfrac{\sqrt{3}}{2}} \\
 & \Rightarrow \sec \left( {{390}^{\circ }} \right)=\dfrac{2}{\sqrt{3}} \\
\end{align}\]
Moving ahead we have,
\[\begin{align}
  & \Rightarrow \csc \left( {{390}^{\circ }} \right)=\dfrac{1}{\sin \left( {{390}^{\circ }} \right)} \\
 & \Rightarrow \csc \left( {{390}^{\circ }} \right)=\dfrac{1}{\dfrac{1}{2}} \\
 & \Rightarrow \csc \left( {{390}^{\circ }} \right)=2 \\
\end{align}\]
At last we have,
\[\begin{align}
  & \Rightarrow \cot \theta =\dfrac{1}{\tan \theta } \\
 & \Rightarrow \cot \left( {{390}^{\circ }} \right)=\dfrac{1}{\tan \left( {{390}^{\circ }} \right)} \\
 & \Rightarrow \cot \left( {{390}^{\circ }} \right)=\dfrac{1}{\dfrac{1}{\sqrt{3}}} \\
 & \Rightarrow \cot \left( {{390}^{\circ }} \right)=\sqrt{3} \\
\end{align}\]
Hence, the value of all the six trigonometric functions can be given as: -
\[\begin{align}
  & \Rightarrow \cos \left( {{390}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \sin \left( {{390}^{\circ }} \right)=\dfrac{1}{2} \\
 & \Rightarrow \tan \left( {{390}^{\circ }} \right)=\dfrac{2}{\sqrt{3}} \\
 & \Rightarrow \sec \left( {{390}^{\circ }} \right)=\dfrac{2}{\sqrt{3}} \\
 & \Rightarrow \csc \left( {{390}^{\circ }} \right)=2 \\
 & \Rightarrow \cot \left( {{390}^{\circ }} \right)=\sqrt{3} \\
\end{align}\]

Note:
One may note that the angle \[\left( 2\pi +\dfrac{\pi }{6} \right)\], i.e., 390 degrees lies in the first quadrant and this is the reason that we have obtained positive values of all the six trigonometric ratios. You must remember the signs of all the trigonometric functions in all the four quadrants to solve the question otherwise you will get confused. Remember the basic formulas of expressions like: - \[\sin \left( a\pi +\theta \right)\], \[\cos \left( a\pi +\theta \right)\], \[\tan \left( a\pi +\theta \right)\] for certain integer values of ‘a’.