# What is sin x if tan x=2/3 and x is in the set (0,pi)?

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The range (0,pi) covers the first and the second quadrant where the values of the sine function are positive.

We'll apply, for the beginning, the Pythagorean identity:

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

We'll divide the formula with the value (sin x)^2:

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

But the ratio sin x /cos x= tan x and cos x/sin x=1/tan x

The formula will become:

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

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

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

sin x= 1/sqrt(1+9/4)

sin x = 2/sqrt13 => sin x = 2sqrt13/13

**The requested value for sin x is : sin x = 2sqrt13/13**

tanx is > 0 for o<x<pi/2 in (0.pi).

tanx = 2/3.

we know that sinxx = tanx/sqrt(1+tan^2x). We put tanx = 2/3

sinxx = (2/3)/sqrt(1+(2/3)^2)

sinx = 2/sqrt(3^2+2^2)

**Sinx= 2/sqrt13.**