# What is sinx if x is in the interval (0,90) an cot x = 1/3?

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### 2 Answers

For x lying in the interval (0, 90), sin x and cos x are positive.

cot x = 1/3

=> cos x/ sin x = 1/3

=> sqrt (1 - (sin x)^2)/(sin x) = 1/3

let y = sin x

=> sqrt(1-y^2)/y = 1/3

square both the sides

=> (1 - y^2)/y^2 = 1/9

=> 9 - 9y^2 = y^2

=> y^2 = 9/10

=> y = 3/sqrt 10

**The required value of sin x = 3/sqrt 10**

We'll recall the definition of the cotangent function:

cot x = cos x/sin x

We know, from enunciation, that cot x = 1/3

We'll apply the Pythagorean identity:

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

(sin x)^2*[(cot x)^2 + 1] = 1

(sin x)^2 = 1/[(cot x)^2 + 1]

sin x = +1/sqrt[(cot x)^2 + 1] or sin x = -1/sqrt[(cot x)^2 + 1]

sin x = 1/sqrt[(1/3)^2 + 1]

sin x = 1/sqrt [(1+9)/9] => sin x = 1/sqrt (10/9) => sin x = 3/sqrt 10 or sin x = -3/sqrt 10

**We notice that x belongs to the 1st quadrant, therefore, the value of the sine function is positive:sin x = 3*sqrt 10/10.**