Question

How do you find the exact value of the six trigonometric functions of 330 degrees?

Answer 1

In this question, we are given the value of degree i.e., 330 degrees and we are asked to find the all other trigonometric ratio. First we will know the basic trigonometric functions, i.e., sin, cos and Now after we have the value of $\sin \theta$and $\cos \theta$, by these values $\tan \theta$ can be calculated, and other trigonometric ratios can be calculated by finding the reciprocal of these trigonometric ratios.

Complete step by step solution:
The angles which lie between ${0^o}$ and ${90^o}$ are said to lie in the first quadrant. The angles between ${90^o}$ and ${180^o}$ are in the second quadrant, angles between ${180^o}$ and ${270^o}$ are in the third quadrant and angles between ${270^o}$ and ${360^o}$ are in the fourth quadrant.
In the first quadrant, the values for sin, cos and tan are positive.
In the second quadrant, the values for sin are positive only.
In the third quadrant, the values for tan are positive only.
In the fourth quadrant, the values for cos are positive only.
Now given degree is 330,
We have to find the sin value of the given angle, first write the angle as the difference of two angles, i.e., $360 - 30 = 330$,
We know that ${360^o}$ lies in fourth quadrant and sine is negative in fourth quadrant,
So now we have,
$\Rightarrow \sin {330^o} = \sin \left( {{{360}^o} - {{30}^o}} \right)$,
Using the identity $\sin \left( {360 - \theta } \right) = - \sin \theta$ we get,
$\Rightarrow \sin {330^o} = - \sin \left( {{{30}^o}} \right)$,
And now using the trigonometric table we know that $\sin {30^o} = \dfrac{1}{2}$, using the identity we get,
$\Rightarrow \sin {330^o} = - \dfrac{1}{2}$,
Now we have to find the cosine of the given angle,
We know that ${360^o}$ lies in fourth quadrant and cosine is positive in fourth quadrant,
So now we have,
$\Rightarrow \cos {330^o} = \cos \left( {{{360}^o} - {{30}^o}} \right)$,
Using the identity $\cos \left( {360 - \theta } \right) = \cos \theta$ we get,
$\Rightarrow \cos {330^o} = \cos \left( {{{30}^o}} \right)$,
And now using the trigonometric table we know that $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$, using the identity we get,
$\Rightarrow \cos {330^o} = \dfrac{{\sqrt 3 }}{2}$,
Now using the trigonometric identities we get,
$\Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$,
Now substituting the value of $\sin \theta$ and $\cos \theta$ we get,
Here $\theta = {330^o}$, and the angle is in fourth quadrant, and tan is negative in fourth quadrant,
$\Rightarrow \tan {330^o} = - \dfrac{{\sin {{330}^o}}}{{\cos {{330}^o}}}$,
Now substituting the value, we get,
$\Rightarrow \tan {330^o} = - \dfrac{{\dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}$,
Now simplifying we get,
$\Rightarrow \tan {330^o} = - \dfrac{1}{{\sqrt 3 }}$,
Now as $\cot \theta$ is the inverse of $\tan \theta$, we get,
$\Rightarrow \cot \theta = \dfrac{1}{{\tan \theta }}$,
Now substituting the value of $\tan {330^o} = - \dfrac{1}{{\sqrt 3 }}$, we get,
$\Rightarrow \cot {330^o} = \dfrac{1}{{ - \dfrac{1}{{\sqrt 3 }}}}$,
Now simplifying we get,
$\Rightarrow \cot {330^0} = - \sqrt 3$, as the angle in the fourth quadrant and cot is negative in the fourth quadrant.
Now we know that $\csc \theta$ is the inverse of $\sin \theta$, we get,
$\Rightarrow \csc \theta = \dfrac{1}{{\sin \theta }}$,
Now substituting the value of$\sin {330^o} = - \dfrac{1}{2}$, we get,
$\Rightarrow \csc {330^o} = \dfrac{1}{{ - \dfrac{1}{2}}}$,
Now simplifying we get,
$\Rightarrow \csc {330^o} = - 2$,
And we know that $\sec \theta$ is the inverse of $\cos \theta$ we get,
$\Rightarrow \sec \theta = \dfrac{1}{{\cos \theta }}$,
Now substituting the value of $\cos {330^o} = \dfrac{{\sqrt 3 }}{2}$, we get,
$\Rightarrow \sec {0^o} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$,
Now simplifying we get,
$\Rightarrow \sec {330^o} = \dfrac{2}{{\sqrt 3 }}$.
The value of trigonometric ratios are,$\cos {330^o} = \dfrac{{\sqrt 3 }}{2}$, $\sin {330^o} = - \dfrac{1}{2}$, $\tan {330^o} = \dfrac{{ - 1}}{{\sqrt 3 }}$, $\cot {330^o} = - \sqrt 3$, $\sec {330^o} = \dfrac{2}{{\sqrt 3 }}$, $\csc {330^o} = - 2$.

$\therefore$ The value of trigonometric ratios are $\cos {330^o} = \dfrac{{\sqrt 3 }}{2}$, $\sin {330^o} = - \dfrac{1}{2}$, $\tan {330^o} = \dfrac{{ - 1}}{{\sqrt 3 }}$, $\cot {330^o} = - \sqrt 3$, $\sec {330^o} = \dfrac{2}{{\sqrt 3 }}$, $\csc {330^o} = - 2$.

Note: Most of the trigonometry calculations are done by using trigonometric ratios. There are 6 trigonometric ratios present in trigonometry. Every other important trigonometry formula is derived with the help of these ratios.