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Question

Mathematics Question on Inverse Trigonometric Functions

The value of cos−1(cos(7π4))cos^{-1}(cos(\dfrac{7\pi}{4})) is

A

00

B

π2 \dfrac{\pi}{2}

C

π3 \dfrac{\pi}{3}

D

π4 \dfrac{\pi}{4}

E

π6 \dfrac{\pi}{6}

Answer

π4 \dfrac{\pi}{4}

Explanation

Solution

Given that

cos−1(cos(7π4))cos^{-1}(cos(\dfrac{7\pi}{4}))

So. let's first simplify the bracketed expression

cos(7π4) cos(\dfrac{7\pi}{4})

=cos(2π−π4)= cos(2\pi-\dfrac{\pi}{4})

=cos(2π)×cos(π4)+sin(2π)×sin(π4)= cos(2\pi)×cos(\dfrac{\pi}{4}) + sin(2\pi)×sin(\dfrac{\pi}{4})

=1×cos(π4)+0=1×cos(\dfrac{\pi}{4}) +0

=cos(π4)=cos(\dfrac{\pi}{4})

Now form the parent expression we can write

cos−1(cos(7π4))=cos−1cos(π4)cos^{-1}(cos(\dfrac{7\pi}{4}))= cos^{-1}cos(\dfrac{\pi}{4})

cos−1(cos(7π4))=π4cos^{-1}(cos(\dfrac{7\pi}{4}))= \dfrac{\pi}{4} (_Ans.)