How do you evaluate #cos^-1 (-sqrt2/2)#?

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Answer 1 · 2018-06-26T13:22:27

#cos^-1(-sqrt2/2)=(3pi)/4#

Let ,

#A=cos^-1(-sqrt2/2)#

#=>A=cos^-1(-1/sqrt2)#

#=>A=pi-cos^-1(1/sqrt2)tocolor(red)([becausecos^-1(- x)#=#color(red)(pi-cos^-1x)]#

#=>A=pi-cos^-1(cos(pi/4))...to[becausecolor(blue)(cos(pi/4)=1/sqrt2) ]#

#=>A=pi-pi/4#

#=>A=(3pi)/4#