How do you evaluate #cos A = 0.5878#?

Question

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

4251 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-02T22:29:58

If by #0.5878# you mean

#\sqrt{{5-\sqrt{5}}/8}=0.5877852523#

(to ten decimal places), then the answer is #A={3\pi}/{10}# radians or #54°#.

#cos(\pi/5)={\sqrt{5}+1}/4#.

Then

#sin({3\pi}/{10})={\sqrt{5}+1}/4#

because #{3\pi}/{10}+\pi/5=\pi/2#.

Then

#cos^2({3\pi}/{10})=1-sin^2({3\pi}/{10})=1-({\sqrt{5}+1}/4)^2={5-\sqrt{5}}/8#.