How do you evaluate $arccos ( sin 50)$ ?

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Answer 1 · 2016-05-21T15:01:06

$(360n+-40)^o, n=0,+-1,+-2,+-3,...$

Had it been arc cos (cos a), the answer is that a.

Here, $sin 50^o=cos(360n+-40)^o, n=0,+-1,+-2,+-3,...$ .

So, $arc cos (sin 50^o)=arc cos (cos(360n+-40)^o), n=0,+-1,+-2,+-3,...$

= $(360n+-40)^o, n=0,+-1,+-2,+-3,...$

How do you evaluate arccos ( sin 50) arccos ( sin 50)?