# What is the value of cos2x if sinx=1/3 ?

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We know that cos 2x = 1 - 2*(sin x)^2

we have sin x = 1/3

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

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

=> cos 2x = (9 - 2)/9

=> cos 2x = 7/9

**The value of cos 2x = 7/9**

The problem requires the double angle identity:

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

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 - 1/9

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

(cos x)^2 = 8/9

We'll replace the value of (cos x)^2 into the double angle identity:

cos 2x = 2*8/9 - 1

cos 2x = (16-9)/9

cos 2x = 7/9

**The value of cos 2x, if sin x = 1/3, is: cos 2x = 7/9.**