Amir wants to show that tangent 3 theta equals start fraction 3 tangent theta minus tangent cubed theta over 1 minus 3 tangent squared theta end fraction. He says he should use the double-angle formula for the tangent function to do this. Is he correct?

Question

(1 point)
Responses

Yes. He can write tangent 3 theta equals tangent 2 left parenthesis start fraction 3 over 2 end fraction theta right parenthesis and use the double-angle formula.
Yes. He can write Image with alt text: tangent 3 theta equals tangent 2 left parenthesis start fraction 3 over 2 end fraction theta right parenthesis and use the double-angle formula.

Yes. He can write tangent 3 theta equals tangent left parenthesis 2 theta plus theta right parenthesis and use the sum formula followed by the double-angle formula.
Yes. He can write Image with alt text: tangent 3 theta equals tangent left parenthesis 2 theta plus theta right parenthesis and use the sum formula followed by the double-angle formula.

No. Because 3 is not a multiple of 2, he will have a start fraction 3 over 2 end fraction theta term that cannot be simplified.
No. Because 3 is not a multiple of 2, he will have a Image with alt text: start fraction 3 over 2 end fraction theta term that cannot be simplified.

No. Because tangent 2 theta equals start fraction 2 tangent theta over 1 minus tangent squared theta end fraction, there is no way to multiply the denominator by 3 to obtain 1 minus 3 tangent squared theta.
No. Because Image with alt text: tangent 2 theta equals start fraction 2 tangent theta over 1 minus tangent squared theta end fraction , there is no way to multiply the denominator by 3 to obtain Image with alt text: 1 minus 3 tangent squared theta .

Answer 1

To show that \(\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}\), Amir is correct in that he can utilize the sum and double-angle formulas for tangent.

The correct approach would be:

• First, recognize that \(3\theta\) can be expressed as \(2\theta + \theta\).
• Then, use the sum formula for tangent: \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\).
• Set \(a = 2\theta\) and \(b = \theta\), and then apply the double angle formula for \(\tan 2\theta\): \(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}\).

Therefore, the correct response is: Yes. He can write \(\tan 3\theta = \tan(2\theta + \theta)\) and use the sum formula followed by the double-angle formula.