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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)
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