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

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:

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tan(?) = x/4 ? would desire to be in Quad. III considering the fact that tan(?) is helpful and attitude lies in -?/2 < ? < ?/2. sin(?) = -x/?(x^2 + sixteen) cos(?) = -4/?(x^2 + sixteen) sin(2?) = 2sin(?)cos(?) = 8x/(x^2 + sixteen)

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tan(θ) = x/6

(sin(θ) + cos(θ))² = 1 + sin(2θ)

sin(θ) = x/√(36 - x²)

cos(θ) = 6/√(36 - x²)

(x + 6)²/(36 - x²) - 1 = sin(2θ)

= (x² + 12x + 36 - 36 + x²)/(36 - x²) = sin(2θ)

= (2x(x + 6))/(6 + x)(6 - x) = sin(2θ)

= 2x/(6 - x) = sin(2θ)

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