Question

Solve: $ {{\operatorname{cosec}}^{-1}}\left( \cos x \right) $ is real if; \[\]
A. $ x\in \left[ -1,1 \right] $ \[\]
B. $ x\in R $ \[\]
C. $ x $ is an odd multiple of $ \dfrac{\pi }{2} $ \[\]
D. $ x $ is multiple of $ \pi $ \[\]

Answer 1

Hint: We recall the domain and range of cosec inverse function as $ {{\operatorname{cosec}}^{-1}}x:\left( -\infty ,1 \right]\bigcup \left[ 1,\infty \right)\to \left[ \dfrac{-\pi }{2},0 \right)\bigcup \left( 0,\dfrac{\pi }{2} \right] $ and the domain and range of cosine function as $ \cos x:R\to \left[ -1,1 \right] $ . We use these domain and ranges to see which option is correct. \[\]

Complete step by step answer:
We are given in the question a composite function $ {{\operatorname{cosec}}^{-1}}\left( \cos x \right) $ where the cosec inverse functions $ {{\operatorname{cosec}}^{-1}}x $ takes all the values returned by cosine function $ \cos x $ inside the bracket. We know from domain of cosec inverse function can take values as input which do not belong to the interval $ \left( -1,1 \right) $ .\[\]
A. We are given $ x\in \left[ -1,1 \right] $ . So cosine function $ \cos x $ c will take map this interval $ \left[ -1,1 \right] $ to the interval $ \left[ 0,\cos \left( 1 \right) \right] $ but we know that range of cosine function is $ \left[ -1,1 \right] $ . So we have $ -1<0<\cos \left( 1 \right)<1 $ . So the domain for cosec inverse function is $ \left[ 0,\cos \left( 1 \right) \right] $ and cosec inverse function cannot take values from $ \left( -1,1 \right) $ . So option A is not correct.\[\]
B. We are given $ x\in R $ . We have already prove that $ {{\operatorname{cosec}}^{-1}}\left( \cos x \right) $ is not real if $ x\in \left[ -1,1 \right] $ . So $ x $ is may not be real for all $ x\in R $ . Hence option B is incorrect.\[\]
C. $ x $ is an odd multiple of $ \dfrac{\pi }{2} $ which means for some $ n\in \mathsf{\mathbb{Z}} $ we have $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ for $ n=0 $ we have $ x=\left( 2\cdot 0+1 \right)\dfrac{\pi }{2}=\dfrac{\pi }{2} $ . Then we have $ \cos x=\cos \left( \dfrac{\pi }{2} \right)=0 $ but inverse cosec function is not defined for $ x=0 $ since $ 0\in \left[ -1,1 \right] $ . So option $ C $ is not correct.\[\]
D. We are given $ x $ is multiple of $ \pi $ which means $ x=n\pi ,n\in \mathsf{\mathbb{Z}} $ . If $ n $ is odd we get $ \cos x=-1 $ and if $ n $ is even we get $ \cos x=1 $ . The cosec inverse is well defined for $ x=-1,1 $ . So option D is only correct option.\[\]

Note:
We can alternatively solve by observing that cosec inverse function can take only those values at the intersection of domain of cosec inverse function and range cosine inverse function that is $ \left( \left( -\infty ,1 \right]\bigcup \left[ 1,\infty \right) \right)\bigcap \left[ -1,1 \right]=\left\{ -1,1 \right\} $ . We can find for what values of $ x $ the value of cosine $ \cos x=-1,1 $ . We note that mapping of $ \left[ -1,1 \right] $ on $ \left[ 0,1 \right] $ comes from the fact that $ \cos \left( -x \right)=\cos x,-\dfrac{\pi }{2} < -1 < 1 < \dfrac{\pi }{2} $ and $ \cos x\in \left[ 0,1 \right] $ for all $ x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] $ .