How do you evaluate #Sin[Arccos(-1/5)]#?

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Answer 1 · 2015-06-29T18:33:07

#sin(arccos(-1/5)) = (2sqrt(6))/5#

Let #alpha = arccos(-1/5)#

By definition of #arccos#, we have #0 <= alpha <= pi#

More specifically, since #cos(alpha) = -1/5 < 0#,

we have #pi/2 < alpha <= pi#

Since #alpha# is in this quadrant, #sin(alpha) >= 0#

Now #cos^2(alpha) + sin^2(alpha) = 1#

So:

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

(must be positive square root since #sin(alpha) >= 0#)

#=sqrt(1-(-1/5)^2))#

#=sqrt(1-1/25)#

#=sqrt(24/25)#

#=sqrt(6*4/25)#

#=sqrt(6)sqrt(4/25)#

#=(2sqrt(6))/5#