How do you do simplify #sin(cos^(-1)(1/5))#?

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Answer 1 · 2015-11-17T23:58:58

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

Consider a right angled triangle with hypotenuse #5# and one leg #1#. Then the other leg will have length #sqrt(5^2-1^2) = sqrt(24) = 2sqrt(6)#.

So if #cos(theta) = 1/5# then #sin(theta) = (2sqrt(6))/5#

So:

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

Another way of calculating this is to start with:

#cos^2(theta) + sin^2(theta) = 1#

Let #theta = cos^(-1)(1/5)#

Then #cos(theta) = 1/5# and we find:

#sin(theta) = sqrt(1-cos^2(theta)) = sqrt(1-(1/5)^2)) = sqrt(24/25) = (2sqrt(6))/5#