Question

How do you evaluate $ \arccos \left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ ?

Answer 1

Hint: We know that $ \arccos $ is the another way of writing $ {{\cos }^{-1}} $ so we can write the given expression as $ {{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ . Now, equate this expression to $ \theta $ and then take cosine on both sides and hence solve.

Complete step by step answer:
The expression given in the above problem is as follows:
 $ \arccos \left( \cos \left( \dfrac{5\pi }{4} \right) \right) $
Rewriting the above expression we get,
 $ {{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{4} \right) \right) $
Equating the above expression to $ \theta $ we get,
 $ \theta ={{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{4} \right) \right) $
Taking cosine on both the sides of the above equation we get,
 $ \cos \theta =\cos \left( {{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{4} \right) \right) \right) $
In the expression, written on the R.H.S of the above equation the expression $ \cos \left( {{\cos }^{-1}} \right) $ becomes 1 and we get,
 $ \cos \theta =\left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ ……… Eq. (1)
Now, the above equation is written in the following form:
 $ \cos \theta =\cos \alpha $
And the general solution of the above equation is:
 $ \theta =2n\pi \pm \alpha $
The expression written on the R.H.S of the above equation contains a variable “n”. The values that “n” can take is 0, 1, 2, 3…….
Using the above general formula relation in solving eq. (1) we get,
In eq. (1), $ \theta $ is $ \theta $ but $ \alpha $ is $ \dfrac{5\pi }{4} $ then the general solution for eq. (1) is:
 $ \theta =2n\pi \pm \dfrac{5\pi }{4} $
As we have assumed that $ \arccos \left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ is $ \theta $ so the value of $ \arccos \left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ is equal to:
 $ 2n\pi \pm \dfrac{5\pi }{4} $
In the above expression, where “n” can take values as 0, 1, 2, 3……..

Note:
 In the above solution, the blunder that could be possible is:
In the expression given in the above problem i.e. $ {{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{4} \right) \right) $ , you might think that the value of $ {{\cos }^{-1}}\left( \cos \right) $ is 1 and you will get the value of the given expression as $ \dfrac{5\pi }{4} $ . This is wrong, you can write $ \cos \left( {{\cos }^{-1}} \right) $ as 1 but you cannot write $ {{\cos }^{-1}}\left( \cos \right) $ as 1.
Hence, make sure you won’t make this mistake in the examination.