Question

How do you evaluate $ {\cos ^{ - 1}}(\dfrac{{ - 1}}{2}) $ ?

Answer 1

Hint: Trigonometry functions tell us the relation between the sides and the angles of a right-angled triangle. Sine, cosine and tangent are the main functions of trigonometry. They are simply the ratio of two sides of a right-angled triangle. Inverse trigonometric functions are the inverse of the trigonometric functions, that is we know the numerical value and we have to find the angle. We know the cosine value of some basic angles like $ 0,\dfrac{\pi }{6},\dfrac{\pi }{4} $ etc. So finding out the angle whose cosine is equal to $ \dfrac{{ - 1}}{2} $ , we can find the correct answer.

Complete step-by-step answer:
We have to find out $ {\cos ^{ - 1}}(\dfrac{{ - 1}}{2}) $ , we know that $ \cos \dfrac{\pi }{3} = \dfrac{1}{2} $
 $ \Rightarrow - \cos \dfrac{\pi }{3} = - \dfrac{1}{2} $
Now, we know that cosine is positive in the first and fourth quadrant but is negative in the second and the third quadrant, so –
 $
   - \cos \dfrac{\pi }{3} = \cos (\pi - \dfrac{\pi }{3}),\, - \cos \dfrac{\pi }{3} = \cos (\pi + \dfrac{\pi }{3}) \\
   \Rightarrow - \cos \dfrac{\pi }{3} = \cos \dfrac{{2\pi }}{3},\, - \cos \dfrac{\pi }{3} = \cos \dfrac{{4\pi }}{3} \\
  $
So, $ {\cos ^{ - 1}}(\cos \dfrac{{2\pi }}{3}) = \dfrac{{2\pi }}{3},\,{\cos ^{ - 1}}(\cos \dfrac{{4\pi }}{3}) = \dfrac{{4\pi }}{3} $
Hence, $ {\cos ^{ - 1}}( - \dfrac{1}{2}) $ is equal to $ \dfrac{{2\pi }}{3} $ or $ \dfrac{{4\pi }}{3} $ .
So, the correct answer is “$ \dfrac{{2\pi }}{3} $ or $ \dfrac{{4\pi }}{3} $”.

Note: All the trigonometric and inverse trigonometric functions can be plotted on a graph, a graph is divided into four quadrants, and the signs of the trigonometric functions are different in different quadrants. All the functions have a positive value in the first quadrant; in the second quadrant, only sine functions have positive value while the cosine and tangent functions are negative; tangent functions are positive in the third quadrant and the other two functions are negative and in the fourth quadrant only cosine functions are positive. The same is true for the reciprocal of these functions. Following this system of signs, we rewrite $ - \cos \dfrac{\pi }{3} $ . In the third quadrant, $ \dfrac{\pi }{2} < x < \pi $ so we subtract $ \dfrac{\pi }{3} $ from $ \pi $ and in the fourth quadrant $ \pi < x < \dfrac{{3\pi }}{2} $ so we add $ \dfrac{\pi }{3} $ to $ \pi $ .