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Problem 8: Β Trig. equations (βœ“ βœ“) 1997 Paper II

Show that, if tan2πœƒ = 2tanπœƒ + 1, then tan2πœƒ = βˆ’1.
Find all solutions of the equation tanπœƒ = 2 + tan3πœƒ

which satisfy 0 < πœƒ < 2Ο€, expressing your answers as rational multiples of Ο€.

Find all solutions of the equation cotΟ• = 2 + cot3Ο•

which satisfy βˆ’3Ο€ 2 < Ο• < Ο€ 2 , Β expressing your answers as rational multiples of Ο€.


There are three distinct parts. It is pretty certain that they are related, but it is not obvious what the relationship is. Part (i) must surely help with part (ii) in some way that will only become apparent once part (ii) is under way.

In the absence of any other good ideas, it looks right to start part (ii) by expressing the double and triple angle tans in terms of single angle tans. You should remember the formula

tan(A + B) = tanA + tanB 1 βˆ’ tanAtanB.

You can use this for tan3πœƒ and hence (later on) for cot3πœƒ. If you ever forget the tan(A + B) formula, you can quickly work it out from the corresponding sin and cos formulae:

tan(A + B) = sin(A + B) cos(A + B) = sinAcosB + cosAsinB cosAcosB βˆ’ sinAsinB.

You are are not expected to remember the more complicated triple angle formulae (I certainly don’t).

You may well find yourself trying to solve cubic equations at some stage in this question; no need to panic β€” there is sure to be one easily spottable root, in which case you can reduce the cubic to a quadratic.

Interesting, isn’t it, that the range of Ο• for part (iii) is not the obvious 0 < πœƒ < 2Ο€? Maybe that is significant.

Solution to problem 8

We will write t for tanπœƒ (or tanΟ•) throughout.

(i) tan2πœƒ = 2t 1 βˆ’ t2 = βˆ’1 (since 2t = t2 βˆ’ 1 by minor rearrangement of the given equation).

(ii) We first work out tan3πœƒ. We have

tan3πœƒ = tan(πœƒ + 2πœƒ) = tanπœƒ + tan2πœƒ 1 βˆ’ tanπœƒtan2πœƒ = t + 2t 1βˆ’t2 1 βˆ’ t 2t 1βˆ’t2 = 3t βˆ’ t3 1 βˆ’ 3t2,

so the equation becomes

3t βˆ’ t3 1 βˆ’ 3t2 = t βˆ’ 2,i.e.t3 βˆ’ 3t2 + t + 1 = 0.

One solution (by inspection) is t = 1. Thus one set of roots is given by πœƒ = nΟ€ + 1 4Ο€.

There are no other obvious integer roots, but we can reduce the cubic equation to a quadratic equation by dividing out the known factor (t βˆ’ 1). I would start by writing t3 βˆ’ 3t2 + t + 1 ≑ (t βˆ’ 1)(t2 + at βˆ’ 1) since the coefficients of t2 and of t0 in the quadratic bracket are obvious. Then I would multiply out the brackets to find that a = βˆ’2. Now we see the connection with part (i): t2 + at βˆ’ 1 = 0 β‡’ tan2πœƒ = βˆ’1, and hence 2πœƒ = nΟ€ βˆ’1 4Ο€.

The roots are therefore πœƒ = nΟ€ + 1 4Ο€ and πœƒ = 1 2nΟ€ βˆ’1 8Ο€. The multiples of Ο€ in the given range are {1 4, 3 8, 7 8, 5 4, 11 8 , 15 8 }.

(iii) Β For the last part, we could set cotΟ• = 1 tanΟ• Β and cot3Ο• = 1 tan3Ο• Β in the given equation, thereby obtaining

1 t = 2 + 1 βˆ’ 3t2 3t βˆ’ t3 .

This simplifies to the cubic equation t3 + t2 βˆ’ 3t + 1 = 0. There is an integer root t = 1, and the remaining quadratic is t2 + 2t βˆ’ 1 = 0. Learning from the first part, we write this as 2t 1 βˆ’ t2 = 1, which means that tan2Ο• = nΟ€ + 1 4Ο€. Proceeding as before gives (noting the different range) the following multiples of Ο€: {1 4, 1 8,βˆ’3 8,βˆ’3 4,βˆ’7 8,βˆ’11 8 }.


There was a small but worthwhile notational point in this question: it is often possible to use the abbreviation t for tan (or s for sin, etc), which can save a great deal of writing.

There are two other points worth recalling. First is the way that part (i) fed into part (ii), but had to be mildly adapted for part (iii). This is a typical device used in STEP questions aimed to see how well you learn new ideas. Second is what to do when faced with a cubic equation. There is a formula for the roots of a cubic, but no one knows it nowadays. Instead, you have to find at least one root by inspection. Having found one root, you have a quick look to see if there are any other obvious roots and, if not, then divide out the known factor to obtain a quadratic equation.

The detectives amongst you will have worked out the reason for the peculiar choice βˆ’3 2Ο€ < Ο• < 1 2Ο€ for part (iii). The reciprocal relation between tan and cot we used at the start of part (iii) is not the only way to relate these two trigonometric function. We could have instead used cotA = tan(1 2Ο€ βˆ’ A).

The equation of part (ii) transforms exactly into the equation of part (iii) if we set Ο• = 1 2Ο€ βˆ’ πœƒ. Furthermore, the given range of Ο• corresponds exactly to the range of πœƒ given in part (ii). We can therefore write down the solutions for part (iii) directly from the solutions for part (ii).