How do you find the value of #sin(arccos1)#?

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Answer 1 · 2016-12-01T11:34:15

#sin(arccos(1)) = 0#

#y=arccos(x)# is defined as the value of #y in [0,pi)# for which #cos y =x#

In this interval #cosx =1# for #x=0#, so:

#arccos(1) = 0#

and then:

#sin(arccos(1)) = sin 0 = 0#

In general, for #x in [-1,1]# as #arccosx# is the inverse function of #cos x#:

#cos(arccos x) = x#

and as for every #theta#

#sin theta = sqrt (1-cos^2 theta)#

#sin(arccos x) = sqrt (1-x^2 )#

which obviously brings to the same result.