Question

cos^-1(cos \(\frac { 7π }{ 6 }\)) is equal to
(A) \(\frac { 7π }{ 6 }\)
(B) \(\frac { 5π }{ 6 }\)
(C) \(\frac { π }{ 5 }\)
(D) \(\frac { π }{ 6 }\)

Answer 1

(B) \(\frac { 5π }{ 6 }\)

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cos^-1(cos \(\frac { 7π }{ 6 }\)) = cos^-1cos ( π + \(\frac { π }{ 6 }\) )
= cos^-1(- cos\(\frac { π }{ 6 }\) )
= cos^-1[cos ( π – \(\frac { π }{ 6 }\) )
= cos^-1cos\(\frac { 5π }{ 6 }\)
= \(\frac { 5π }{ 6 }\).
Note that cos^-1(cos \(\frac { 7π }{ 6 }\) ) ≠ \(\frac { 7π }{ 6 }\) since
the range of principal value branch of cos^-1 is [0, π].
∴ cos^-1(cos \(\frac { 7π }{ 6 }\)) = cos^-1cos \(\frac { 5π }{ 6 }\) = \(\frac { 5π }{ 6 }\).