Solve sin(2cos^-1(4/5))?

Question

All i remember is that you have to draw a triangle...totally forgot all my trig over the summer. Any help would be great!

forgot to add to put it in terms of radians w/o a calc

Answer 1

sin [ 2 arc cos^(4/5) ] . . . . . . arc cos^(4/5) = 36deg 52'12"
sin [ 73deg 44' 23" ] = 0.96

Answer 2

If the right answer is 24/25 then & only then read my method.!!!

now

let cos^1(4/5) = x then
cos(x) = 4/5
therefore sin(x)= 3/5
{Because u draw a triangle with hypotaneous is 5 & one side is 4 then the other side would be 3 therefore sin(x) = 3/5 }
therefore x= sin^-1(3/5)
therefore cos^-1(4/5) = sin^-1(3/5) eq no 1.
sin(2cos^-1(4/5)) = 2sin(cos^-1(4/5))cos(cos^-1(4/5))
=2sin(sin^-1(3/5))cos(cos^-1(4/5))
{using eq no 1}
=2*(4/5)*(3/5)
Answer is =24/25

Answer 3

sin 2t = 2 sin t cos t
sin(2cos^-1(4/5)) = 2(3/5)(4/5) = 24/25

Answer 4

let cos^-1(4/5)=x
cosx=4/5
opposite side=sqrt(5^2-4^2)
=3
sinx=3/5
x=sin^-1(3/5)
question becomes sin(2sin^-1(3/5) )
this is of the form sin(2y) which is equal to 2siny *cosy
here y=sin^-1(3/5)
therefore sin(2sin^-1(3/5))
=2sin (sin^-1(3/5))*cos(sin^-1(3/5))
it is proved at the top that sin^-1(3/5)=cos^-1(4/5)
we know that sin^-1(sin(x))=x ,cos^-(cos(x))=x
=2[ 3/5*4/5]
ans is 24/25