How do you find #sin^-1 (cos(pi/6))#?

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Answer 1 · 2018-04-18T04:58:36

#color(blue)(sin^-1 (cos (pi/6)) = pi/3 " or " 60^@#

www.math-only-math.com/images/trigonometrical-ratios-table.png

#"Let " theta = sin^-1 (cos (pi/6))#

From the table above, #cos (pi/6) = (sqrt3/2)#

#:. theta = sin^-1 (sqrt3/2)#

#sin theta = sqrt3/2#

But #sin 60 = sqrt3/2# as seen in the table.

Hence #color(blue)(theta = sin ^-1 (sin (60)) = 60^@ " or " pi/3#