Question

How do you find the 6 trigonometric functions of 120 degrees?

Answer 1

Hint: Now to find all trigonometric values we will first find the value of $\sin \left( {{120}^{\circ }} \right)$ by using the formula $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ . Now once we have the value of $\sin \left( {{120}^{\circ }} \right)$ we will use the formula ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to find the value of $\cos \left( {{120}^{\circ }} \right)$ . Now we know that the \[\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}\] hence we will find the value of \[\tan \left( {{120}^{\circ }} \right)\] . Now again know that cot, sec and cosec are reciprocal of tan, cos and sin respectively. Hence we will easily find the value of $\cot \left( {{120}^{\circ }} \right)$ , $\sec \left( {{120}^{\circ }} \right)$ and $\cos ec\left( {{120}^{\circ }} \right)$ . Hence we have all the required values.

Complete step by step solution:
Now let us fist consider the trigonometric function sin.
Let us find the value of $\sin \left( {{120}^{\circ }} \right)$
Now we can write the function as $\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)$ .
Now we know that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.$
Hence we will use the formula to expand $\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)$ .
$\Rightarrow \sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)=\sin \left( {{90}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\cos \left( {{90}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$
$\Rightarrow \sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)=1\times \dfrac{\sqrt{3}}{2}+0\times \dfrac{1}{2}=\dfrac{\sqrt{3}}{2}$
Hence the value of $\sin \left( {{120}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ .
Now let us find the value of $\cos \left( {{120}^{\circ }} \right)$ .
Now we know the relation between sin and cos which is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Hence we have
$\begin{align}
  & {{\sin }^{2}}\left( {{120}^{\circ }} \right)+{{\cos }^{2}}\left( {{120}^{\circ }} \right)=1 \\
 & \Rightarrow {{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{\cos }^{2}}\left( {{120}^{\circ }} \right)=1 \\
 & \Rightarrow \dfrac{3}{4}+{{\cos }^{2}}\left( {{120}^{\circ }} \right)=1 \\
 & \Rightarrow {{\cos }^{2}}\left( {{120}^{\circ }} \right)=1-\dfrac{3}{4} \\
 & \Rightarrow {{\cos }^{2}}\left( {{120}^{\circ }} \right)=\dfrac{1}{4} \\
 & \Rightarrow \cos \left( {{120}^{\circ }} \right)=\pm \dfrac{1}{2} \\
\end{align}$
Now we know that ${{120}^{\circ }}$ is in the second quadrant where cos is negative.
Hence we have $\cos \left( {{120}^{\circ }} \right)=-\dfrac{1}{2}$
Now let us find the value of tan.
Now we know that $\tan \left( \theta \right)=\dfrac{\sin \theta }{\cos \theta }$
Hence we get,
 $\Rightarrow \tan \left( {{120}^{\circ }} \right)=\dfrac{\sin \left( {{120}^{\circ }} \right)}{\cos \left( {{120}^{\circ }} \right)}=\dfrac{\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}}=-\sqrt{3}$
Hence we get, $\tan \left( {{120}^{\circ }} \right)=-\sqrt{3}$ .
Now we know that for remaining functions we have $\cot \theta =\dfrac{1}{\tan \theta }$ , $\sec \theta =\dfrac{1}{\cos \theta }$ and $\cos ec\theta =\dfrac{1}{\sin \theta }$ .
Hence we get,
$\cot \left( {{120}^{\circ }} \right)=\dfrac{-1}{\sqrt{3}}$ , $\sec \left( {{120}^{\circ }} \right)=\dfrac{1}{\dfrac{-1}{2}}=2$ and $\cos ec\left( {{120}^{\circ }} \right)=\dfrac{1}{\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}}$
Hence the values of trigonometric ratios are,
$\begin{align}
  & \sin \left( {{120}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} \\
 & \cos \left( {{120}^{\circ }} \right)=\dfrac{-1}{2} \\
 & \tan \left( {{120}^{\circ }} \right)=-\sqrt{3} \\
 & \cot \left( {{120}^{\circ }} \right)=\dfrac{-1}{\sqrt{3}} \\
 & \sec \left( {{120}^{\circ }} \right)=-2 \\
 & \cos ec\left( {{120}^{\circ }} \right)=\dfrac{2}{\sqrt{3}} \\
\end{align}$

Note: Now note that we know $\sin \left( \pi -x \right)=\sin x$ and $\cos \left( \pi -x \right)=-\cos x$ . Hence using this we can easily write $\sin \left( {{120}^{\circ }} \right)=\sin \left( 180-60 \right)=\sin \left( {{60}^{\circ }} \right)$ and $\cos \left( {{120}^{\circ }} \right)=\cos \left( 180-60 \right)=-\cos \left( {{60}^{\circ }} \right)$ . Hence we can easily find the value of sin and cos. Now for the rest of the functions we will use the same method and find their values. Also we can use $\left( \pi -\theta \right)$ formula for the rest of functions as well.