Given pi/2<x<pi and sinx=3/5 calculate cotx.

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have sin x = 3/5.

Use the property (cos x)^2 = 1 - (sin x)^2

=> (cos x)^2 = 1 - (3/5)^2

=> (25 - 9)/25

=> 16/25

For pi/2 < x < pi , the value of cos x is negative

=> cos x = -4/5

cot x = (cos x)/(sin x)

=> (-4/5)/(3/5)

=> -4/3

The value of cot x = -4/3

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that the angle x is located in the 2nd quadrant, therefore the values of cotangent function are negative.

We know that cot x = cos x/sin x

We'll determine cos x, applying the Pythagorean identity:

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

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

(cos x)^2 = 1 - 9/25

(cos x)^2 = (25 - 9)/25

(cos x)^2 = 16/25

cos x = -4/5

We'll keep only the negative value for cos x, since x is in the second quadrant and cosine function is negative.

cot x = (-4/5)/(3/5)

cot x = -4/3

The value of cotangent function is: cot x = -4/3.

We’ve answered 318,959 questions. We can answer yours, too.

Ask a question