Question

How do you find the six trigonometric functions of \[150\] degrees?

Answer 1

Hint: To solve this Question, first we need to find out the quadrant in which the given degree is present.
Then Identify the hypotenuse, adjacent side, and opposite side of an acute angle in a right triangle.
Determine the six trigonometric ratios for a given angle in a right triangle.
Use a calculator to find the value of the six trigonometric functions for any acute angle.
Use a calculator to find the measure of an angle given the value of a trigonometric function.

Complete step by step answer:
Consider a unit circle, in which the given angle ${150^ \circ }$  is in the second quadrant, where cosine is negative and sine is positive.
Unit circle coordinates are given by $\left( {\cos \theta \,,\,\sin \theta } \right)$
This means that the coordinates for ${150^ \circ }$ are \[\left( { - \dfrac{{\sqrt 3 }}{2}\,,\,\dfrac{1}{2}} \right)\]

Now we know that the six trigonometric function are$\sin \theta $,$\cos \theta $ ,$\tan \theta $ ,$\cos ec\theta $ ,$\sec \theta $ and $\cot \theta $
First we know,
$i)\,\sin \theta = \dfrac{{opposite}}{{hypotenuse}}$
Here we know that the opposite side is $\dfrac{1}{2}$ and hypotenuse is $1$ because it is a unit circle.
Then substitute the value in the formula given above, then we get
$\therefore \sin \left( {{{150}^ \circ }} \right) = \left( {\dfrac{{\dfrac{1}{2}}}{1}} \right)$
$ \Rightarrow \sin \left( {{{150}^ \circ }} \right) = \left( {\dfrac{1}{2}} \right)$
$ii)\,\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}$
Here we know that the opposite side is $\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$ and hypotenuse is $1$. Then substitute the value in the formula given above, then we get
$\therefore \cos \theta = \left( {\dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{1}} \right)$
$ \Rightarrow \cos \theta = \left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
$iii)\,\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
In this case, we already found $\sin \left( {{{150}^ \circ }} \right)$ and$\cos \left( {{{150}^ \circ }} \right)$. Substitute that value.
$\therefore \tan \left( {{{150}^ \circ }} \right) = \left( {\dfrac{{\dfrac{1}{2}}}{{ - \dfrac{{\sqrt 3 }}{2}}}} \right)$
$ \Rightarrow \tan \left( {{{150}^ \circ }} \right) = \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$
$iv)\,\cos ec\theta = \left( {\dfrac{1}{{\sin \theta }}} \right)$
$\therefore \cos ec\left( {{{150}^ \circ }} \right) = \left( {\dfrac{1}{{\dfrac{1}{2}}}} \right)$
$ \Rightarrow \cos ec\left( {{{150}^ \circ }} \right) = 2$
$v)\,\sec \theta = \left( {\dfrac{1}{{\cos \theta }}} \right)$
$\therefore \sec \left( {{{150}^ \circ }} \right) = \left( {\dfrac{1}{{ - \dfrac{{\sqrt 3 }}{2}}}} \right)$
$ \Rightarrow \sec \left( {{{150}^ \circ }} \right) = \left( { - \dfrac{2}{{\sqrt 3 }}} \right)$
$vi)\,\cot \theta = \left( {\dfrac{1}{{\tan \theta }}} \right)$
$\therefore \cot \left( {{{150}^ \circ }} \right) = \left( {\dfrac{1}{{ - \dfrac{1}{{\sqrt 3 }}}}} \right)$
$ \Rightarrow \cot \left( {{{150}^ \circ }} \right) = \left( { - \sqrt 3 } \right)$
Hence we found out the six trigonometric functions for degree $150$ .

Note: The figure will help you to find the quadrant, degrees and signs of the unit circle which will be very useful to solve different problems. To find the trigonometric functions, it is very important to correctly write the signs and to find the quadrant in which the degree is present. And the trigonometric formulas which are mentioned above for six trigonometric functions are very important to memorize.