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

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### 2 Answers

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)

Adding (1) and (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**

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****}.**