Question

If \[\cos \theta =\dfrac{-1}{2}\] and \[{{0}^{0}}<\theta <{{360}^{0}}\] then the values of $\theta $ are
A. \[{{120}^{0}}\] and \[{{300}^{0}}\].
B. ${{60}^{0}}$ and \[{{120}^{0}}\]
C. \[{{120}^{0}}\] and \[{{240}^{0}}\]
D. ${{60}^{0}}$ and \[{{240}^{0}}\]

Answer 1

Hint: To find the value of $\theta $, we will use the trigonometric quadrants of functions and trigonometric table of values. In the first quadrant the angle is less than or equal to ${{90}^{0}}$ depicted as $\dfrac{\pi }{2}-\theta $ and all the functions are positive. In the second quadrant only function sin and cosec is positive and the angle is greater than ${{90}^{0}}$ but less than ${{180}^{0}}$ that is $\dfrac{\pi }{2}+\theta ,\pi -\theta $. In the third quadrant, only function tan and cot is positive and the angle is greater than ${{180}^{0}}$ but less than ${{270}^{0}}$ depicted as $\pi +\theta ,\dfrac{3\pi }{2}-\theta $.In the fourth quadrant, only function cos and sec is positive and the angle is greater than ${{270}^{0}}$ but less than ${{360}^{0}}$ depicted as $\dfrac{3\pi }{2}+\theta ,2\pi -\theta $.

Formula Used: We are given a trigonometric equation \[\cos \theta =\dfrac{-1}{2}\] where $-\pi <\theta <\pi $ and we have to derive the value of $\theta $.
We know from the trigonometric table of values that $\cos {{60}^{0}}=\dfrac{1}{2}$ so the given equation will be,
$\begin{align}
  & \cos \theta =-\dfrac{1}{2} \\
 & \cos \theta =-\cos {{60}^{0}} \\
\end{align}$
As cos is positive in first and fourth quadrant we will represent the derived equation in second quadrant $\cos ({{180}^{0}}-\theta )=-\cos \theta $and third $\cos ({{180}^{0}}+\theta )=-\cos \theta $. So,
$\begin{align}
  & \cos \theta =\cos ({{180}^{0}}-{{60}^{0}}) \\
 & \cos \theta =\cos {{120}^{0}} \\
 & \theta ={{120}^{0}}
\end{align}$
$\begin{align}
  & \cos \theta =\cos ({{180}^{0}}+{{60}^{0}}) \\
 & \cos \theta =\cos {{240}^{0}} \\
 & \theta ={{240}^{0}}
\end{align}$
Therefore the value of $\theta $ is $\theta ={{120}^{0}},{{240}^{0}}$.
The value of $\theta $ for the trigonometric equation \[\cos \theta =\dfrac{-1}{2}\] when \[{{0}^{0}}<\theta <{{360}^{0}}\] is $\theta ={{120}^{0}},{{240}^{0}}$

Complete step by step solution: We are given a trigonometric equation \[\cos \theta =\dfrac{-1}{2}\] where $-\pi <\theta <\pi $ and we have to derive the value of $\theta $.
We know from the trigonometric table of values that $\cos {{60}^{0}}=\dfrac{1}{2}$ so the given equation will be,
$\begin{align}
  & \cos \theta =-\dfrac{1}{2} \\
 & \cos \theta =-\cos {{60}^{0}} \\
\end{align}$
As cos is positive in first and fourth quadrant we will represent the derived equation in second quadrant $\cos ({{180}^{0}}-\theta )=-\cos \theta $and third $\cos ({{180}^{0}}+\theta )=-\cos \theta $. So,
$\begin{align}
  & \cos \theta =\cos ({{180}^{0}}-{{60}^{0}}) \\
 & \cos \theta =\cos {{120}^{0}} \\
 & \theta ={{120}^{0}}
\end{align}$
$\begin{align}
  & \cos \theta =\cos ({{180}^{0}}+{{60}^{0}}) \\
 & \cos \theta =\cos {{240}^{0}} \\
 & \theta ={{240}^{0}}
\end{align}$
Therefore the value of $\theta $ is $\theta ={{120}^{0}},{{240}^{0}}$.
The value of $\theta $ for the trigonometric equation \[\cos \theta =\dfrac{-1}{2}\] when \[{{0}^{0}}<\theta <{{360}^{0}}\] is $\theta ={{120}^{0}},{{240}^{0}}$

Option ‘C’ is correct

Note: In the four quadrants of trigonometry, only in the first and second quadrant the function changes into other function along with signs that is when functions are depicted with angle $\dfrac{\pi }{2}-\theta ,\dfrac{\pi }{2}+\theta $ because the function changes at angle $\dfrac{\pi }{2}$. The function sin changes into cos, sec into cosec, tan into cot and vice versa.