How do you evaluate #arccos(1)# without a calculator?

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Answer 1 · 2018-04-10T16:33:41

#arccos(1)=0#

We know that,

#color(red)(cos^-1(costheta)=theta, thetain[0,pi]...to(I)#

We have,

#arccos(1)=cos^-1(1)#

But, #color(blue)(cos0=1#

#arccos(1)=color(red)(cos^-1(cos0)#, where, #color(red)(0 in [0,pi]#

Using #(I),# we get

#arccos(1)=0#