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

justaguide | Certified Educator

We have to determine sin x* cos x given that sin x/cos x = 1/5

(sin x)/(cos x) = 1/5 ...(1)

(cos x)/(sin x) = 5 ...(2)

(sin x)/(cos x) + (cos x)/(sin x) = 5 + 1/5

=> [(sin x)^2 + (cos x)^2]/(cos x)(sin x) = (25 + 1)/5

=> 1/(cos x)(sin x) = 26/5

=> (cos x)(sin x) = 5/26

The required value of (cos x)(sin x) = 5/26

giorgiana1976 | Student

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).

(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}.