Question

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 formula

B. Yes he can write tan 3θ=(2θ+θ) and use the sum formula followed by the double angle formula

C. no. Because 3 is not a multiple of 2, he will have a 3/2 θ term that cannot be simplified

D. no. Because tan 2θ=(2tanθ)/(1-tan^2θ) there is no way to multiply the denominator by 3 to obtain 1-3tan^2θ

Answer 1

1. B, E

2. B, E

3. B

4. B

5. D

6. A

7. D

8. A

9. B

10. B

11. A

Explanation:

Answer 2

We already know that:

`tan \ 2x = (2 \ tan \ x)/(1 - \ tan^2 \ x) \\ and \\ tan \ (x+y) = (tan \ x + \ tan \ y)/(1 \ - \ tan \ x \ \ tan \ y)`


`tan \ 3 \theta = tan \ (2\theta + \theta) = (tan \ 2\theta + \ tan \ \theta)/(1 \ - \ tan \ 2\theta \ \ tan \ \theta)`

`= \ ((2 \ tan \theta)/(1 - \ tan^2 \theta) + \ tan \ \theta)/(1 \ - (2 \ tan \ \theta)/(1 - \ tan^2 \ 2\theta) \ \ tan \ \theta)`
by multiplying numerator and denominator
`* (1 \ - \ tan^2 \theta)`
and summing the similar variables

∴
`tan \ 3 \theta = (3 \ tan \ \theta - \ tan^3 \ \theta )/(1 \ - \ 3 \ tan^2 \ \theta)`


∴ The correct statement is (B)
B. Yes he can write tan 3θ=(2θ+θ) and use the sum formula followed by the double angle formula