How do you evaluate cos^-1(cos ((9pi)/8))cos−1(cos(9π8))?

1 Answer
Bdub
Oct 12, 2016

cos^-1(cos((9pi)/8))=cos^-1(cos((7pi)/8))=(7pi)/8cos−1(cos(9π8))=cos−1(cos(7π8))=7π8

Explanation:

Since the restriction for cos^-1xcos−1x is [0,pi], (9pi)/8[0,π],9π8 is in quadrant three and cosine is negative in quadrant three. So we need to find the reference angle which is (9pi)/8-pi=pi/89π8−π=π8. Hence the argument x in quadrant II is (7pi)/8 7π8 since cosine is negative in quadrant two from the restriction. Lastly, use the property f^-1(f(x)=xf−1(f(x)=x so

cos^-1(cos((9pi)/8))=cos^-1(cos((7pi)/8))=(7pi)/8cos−1(cos(9π8))=cos−1(cos(7π8))=7π8

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