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If tan(θ) = x/6 for − π/2 < θ < π/2 , find an expression for sin(2θ) in terms of x. Someone show how?

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Start with the basic definition of tangent - opposite over adjacent. So x is opposite, 6 is adjacent. Since the hypotenuse = sqrt(a^2 + b^2), the hypotenuse is sqrt(x^2 + 36). Now we can define sin(θ) and cos(θ) as

sin(θ) = x / sqrt(x^2 + 36)

cos(θ) = 6 / sqrt(x^2 + 36)

The sin((2θ) = 2 sin(θ) cos(θ) so

sin(2θ) = 2 * [x/sqrt(x^2 + 36)] [6/sqrt(x^2 + 36)]

= 12x / (x^2 + 36)

I've verified this solution using Excel.

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there's a formula: sin(2w)= 2tan(w)/1+tan(w)^2. just do that! we know tan(w)=x/6. plug in:

(x/3)/(1+x^2/36)=12x/(36+x^2) thats the answer