How do you simplify #sin(2cos^-1(4/5))#?

Question

27103 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-15T18:06:00

Use the identity #sin(2A)=2sin(A)cos(A)#
Then use the identity #sin(A) = +-sqrt(1-cos^2(A))#

Given: #sin(2cos^-1(4/5))#

Use the identity #sin(2A)=2sin(A)cos(A)#

#2sin(cos^-1(4/5))cos(cos^-1(4/5))#

The cosine of its inverse yields its argument:

#2sin(cos^-1(4/5))(4/5)#

Perform the multiplication:

#8/5sin(cos^-1(4/5))#

Use the identity #sin(A) = +-sqrt(1-cos^2(A))

#+-8/5sqrt(1-cos^2(cos^-1(4/5)))#

Again, the cosine of its inverse yields its argument:

#+-8/5sqrt(1-(4/5)^2)#

#+-8/5sqrt(25/25-16/25)#

#sin(2cos^-1(4/5)) = +-24/25#

To determine whether to choose the positive or negative value, one would need know whether the angle was in the first or fourth quadrant.