Quick fp2 question

QP
MS

I'm stuck at question 6 part c. At this point, I've established that tan3(theta)=1 and that the roots (theta) are tan(pi/12), tan(5pi/12), tan(9pi/12).

I don't understand the "hence" because the marks cheme shows tan(pi/12)+tan(5pi/12)+tan(9pi/12)=3 but shouldn't it be tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)=3

Yours Sincerely,
A confused student

No, because the cubic corresponds to the $\tan(3\theta)$
identity in a (iii).

$x=\tan\theta$

No, because the cubic corresponds to the $\tan(3\theta)$
identity in a (iii).

$x=\tan\theta$

Actually wait,

Now how does tan(pi/12)+tan(5pi/12)+tan(9pi/12)=1 ?

No, each of those are the theta for which

$\tan 3\theta = 1$

By the tan graph, your equation equals 3 and the sum with thrice those values also equals 3.

I understand that tan(theta)=1 i.e. tan(3pi/12)+tan(15pi/12)+tan(27pi/12)=3

But how could I "hence" show that the sum of the thetas with a third of the value also equal 3?

Because I'm pretty sure (tan3A+tan3B)/(tanA+tanB)=1 is an identity :P

I'm sure it would suffice to observe the identity you quoted.

Alternatively, you may subtract multiples of pi to see why it works.

QP
MS

I'm stuck at question 6 part c. At this point, I've established that tan3(theta)=1 and that the roots (theta) are tan(pi/12), tan(5pi/12), tan(9pi/12).

I don't understand the "hence" because the marks cheme shows tan(pi/12)+tan(5pi/12)+tan(9pi/12)=3 but shouldn't it be tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)=3

Yours Sincerely,
A confused student

I think you've confused yourself here!

The first part of the question lets you write tan(3X) in terms of t where t=tan X.

If you take $X = \pi/12$ then tan(3X) = 1, so if $t = tan(X) = tan(\pi/4)$ then t satisfies the cubic equation $t^3 - 3t^2 - 3t + 1 = 0$

We can used this logic to find 2 other distinct roots of the cubic. We know that if $X = 5\pi/12, 3X = 5\pi/4, tan(3X) = 1$
and
$X = 9\pi/12, 3X = 9\pi/4, tan(3X) = 1$

Therefore, the roots of the cubic are $a= tan(\pi/12), b = tan(5\pi/12), c = tan(9\pi/12)$

Now factorize the cubic as (t-a)(t-b)(t-c) and compare coefficients of t^2:
-(a+b+c) = -3
so
$a+b+c = 3$
or
$tan(\pi/12) + tan(5\pi/12) + tan(9\pi/12) = 3$

But $tan(9\pi/12) = tan(3\pi/4) = -1$

and the required result follows.

Oh thank you! It never occurred to me that the topic of roots of polynomials would be thrown in there!