How do you evaluate #arccos(1/(2*sqrt(2)))#?

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Answer 1 · 2017-03-25T17:50:38

#arc cos(1/(2sqrt2))=69.3^@#

Let #arc cos(1/(2sqrt2))=A#

then #cosA=1/(2sqrt2)=sqrt2/4=0.35355#

and using tables or scientific calculator

#A=69.3^@#

Answer 2 · 2017-03-25T22:21:21

#x = +- 69^@30 + k360^2#

#cos x = 1/(2sqrt2) = sqrt2/4 = 0.3535#
Calculator and unit circle give -->
#x = +- 69^@30 + k360^@#