How do you evaluate sin(arccos 3/4)?

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Answer 1 · 2018-06-15T05:27:29

#sin(arccos(3/4))=sqrt7/4#

We know that,

#color(orange)((A) sin ( arc sinx)=x # , where , #x in [-1,1]#

Let ,

#m=sin(color(blue)(arccos(3/4)))...to(1)#

We know that,

#color(red)(arccosx=arcsin(sqrt(1-x^2))#

#:.arccos(3/4)=arcsin(sqrt(1-9/16))#

#=>color(blue)(arccos(3/4)=arcsin(sqrt7/4)#

From #(1)# . we have

#m=sin(color(blue)(arcsin(sqrt7/4)))...tocolor(orange)( Apply(A)#

#=>m=sqrt7/4~~0.66#

#i.e.sin(arccos(3/4))=sqrt7/4#

Answer 2 · 2018-06-16T11:06:04

#+- 0.66#

cos x = 3/4
Calculator and unit circle -->
arc #x = +- 41^@41#

a. sin (41^@41) = 0.66
b. sin (-41^@41) = - 0.66