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Question
Prove that sin 4α = `4 tan alpha (1 - tan^2alpha)/(1 + tan^2 alpha)^2`
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Solution
sin 4α = sin 2(2α)
= `(2tan(2alpha))/(1 + tan^2 (2alpha))`
= `(2((2tan alpha)/(1 - tan^2 alpha)))/(1 + ((2 tan alpha)/(1 - tan^2 alpha))^2`
= `((4 tan alpha)/(1 - tan^2 alpha))/(((1 - tan^2 alpha)^2 + 4 tan^2 alpha)/(1 - tan^2 alpha)^2)`
= `(4 tan alpha (1 - tan^2 alpha))/(1 + tan^2 alpha - 2 tan^2 alpha + 4 tan^2 alpha)`
= `(4 tan alpha (1 - tan^2 alpha))/(1 + 2 tan^2 alpha + tan^4 alpha)`
sin 4α = `(4 tan alpha (1 - tan^2 alpha))/(1 + tan^2 alpha)^2`
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