amir wants to show that tan3θ = (3tanθ-tan^3θ)/(1-3tan^2θ). He says he should use the double angle formula for the tangent function to do this. Is he correct?A. Yes he can write tan 3θ= tan 2(3/2 θ) and use the double angle formulaB. Yes he can write tan 3θ=(2θ+θ) and use the sum formula followed by the double angle formulaC. no. Because 3 is not a multiple of 2, he will have a 3/2 θ term that cannot be simplifiedD. no. Because tan 2θ=(2tanθ)/(1-tan^2θ) there is no way to multiply the denominator by 3 to obtain 1-3tan^2θ
Accepted Solution
A:
We already know that: [tex]tan \ 2x = \frac{2 \ tan \ x}{1 - \ tan^2 \ x} \\ and \\ tan \ (x+y) = \frac{tan \ x + \ tan \ y}{1 \ - \ tan \ x \ \ tan \ y} [/tex]
[tex]tan \ 3 \theta = tan \ (2\theta + \theta) = \frac{tan \ 2\theta + \ tan \ \theta}{1 \ - \ tan \ 2\theta \ \ tan \ \theta} [/tex] [tex] = \ \frac{\frac{2 \ tan \theta}{1 - \ tan^2 \theta} + \ tan \ \theta}{1 \ - \frac{2 \ tan \ \theta}{1 - \ tan^2 \ 2\theta} \ \ tan \ \theta}
[/tex] by multiplying
numerator and denominator [tex]* (1 \ - \ tan^2 \theta) [/tex] and summing the similar variables