What is sin x*cos x if sin x/cos x=1/5 and x is in the interval (0,pi)

The interval (0,pi) covers the first and the second
quadrant where the values of the sine function are positive and the values of cosine
function are both positive (1st quadrant) and negative (2nd
quadrant).

We'll start with the Pythagorean
identity:

(sin x)^2 + (cos
x)^2=1

We'll divide the formula by  (cos
x)^2:

(sin x)^2/ (cos x)^2 + (cos x)^2/(cos x)^2 = 1 / (cos
x)^2

But the ratio sin x /cos x= tan x =>(sin x)^2/
(cos x)^2 = (tan x)^2 = 1/25

The formula will
become:

(tan x)^2 + 1 = 1/(cos
x)^2

cos x = +1/sqrt[1+(tan x)^2] or cos x = -1/sqrt[1+(tan
x)^2]

cos x =
1/sqrt[1+(1/5)^2]

cos x=
1/sqrt(1+1/25)

cos x = 5/sqrt26 or cos x =
-5/sqrt26

But sin x/cos x =1/5 => sin x = cos x/5
=> sin x = 1/sqrt26

sin x*cos x = 5/26 or sin x*cos
x = -5/26

The requested values of the product
sin x*cos x, in the interval (0,pi), are : {-5/26 ;
5/26}.