What is cosx*sinx if cosx-sinx=1/3?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find cos x*sin x given that cos x - sin x = 1/3

cos x - sin x = 1/3

square both the sides

=> (cos x - sin x)^2 = 1/9

=> (cos x)^2 + (sin x)^2 - 2*sin x* cos x = 1/9

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

=> 1 - 2*sin x* cos x = 1/9

=> 2*sin x * cos x = 1 - 1/9

=> 2*sin x * cos x = 8/9

=> sin x* cos x = 4/9

The required value of sin x* cos x = 4/9

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll raise to square both sides, the given difference:

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

We'll expand the binomial:

(cos x)^2 - 2sin x*cos x + (sin x)^2 = 1/9

We'll use the Pythagorean identity, such as (sin x)^2 + (cos x)^2=1.

1 - 2sin x*cos x = 1/9

We'll isolate 2sin x*cos x to the left side:

-2sin x*cos x = 1/9 - 1

-2sin x*cos x = -8/9

sin x*cos x = 8/2*9

sin x*cos x = 4/9

The requested value of the product sin x*cos x = 4/9.

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