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STEP Siklos Booklet Question 6

Hi there,

I am having trouble understanding the rationale behind the answer to question 6:

A) Show that, if tan2θ=2tanθ+1, then tan2θ=−1.

B) Find all solutions of the equation
tan θ = 2 + tan 3θ
which satisfy 0 < θ < 2π, expressing your answers as rational multiples of π .

C) Find all solutions of the equation
cot φ = 2 + cot 3φ
which satisfy −3π < φ < π , expressing your answers as rational multiples of π.

For the part B of the question (find all solutions of tan...), the answer booklet arrives at one solution t=1 and concludes that one set of roots is given by:

θ = nπ + π/4

Why is this??

Thanks for any help!


Kelsey
Reply 1
Avatar for Smaug123
Smaug123
15
Original post by kelsey.phillips

For the part B of the question (find all solutions of tan...), the answer booklet arrives at one solution t=1 and concludes that one set of roots is given by:

θ = nπ + π/4

tan⁡(θ)=1\tan(\theta) = 1 iff θ=π4+nπ\theta = \frac{\pi}{4} + n \pi, n∈Zn \in \mathbb{Z}.
Original post by kelsey.phillips

For the part B of the question (find all solutions of tan...), the answer booklet arrives at one solution t=1 and concludes that one set of roots is given by:

θ = nπ + π/4

Why is this??


I haven't looked at Siklos' solution but for this question you should find tan⁡3θ\tan{3\theta} in terms of tan⁡θ\tan{\theta} and then solve the resulting cubic equation for tt. As you know solving cubic equations usually involve first spotting an obvious root by inspection (in this case t=1t=1).

You should learn sine, cosine and tangent of π/3\pi/3, π/4\pi/4 and π/6\pi/6 or at least know how to derive them from the relevant triangles. Anyway, tan⁡π4=1\tan{\frac{\pi}{4}}=1 so using the fact that tangent is periodic πn\pi n (that is, tan⁡(θ+πn)=tan⁡θ\tan(\theta +\pi n)=\tan{\theta}), tan⁡θ=1  ⟺  θ=π4+πn\tan{\theta}=1 \iff \theta=\frac{\pi}{4}+\pi n where nn is an integer.

It is probably worth learning the following identities too:

[br]sin⁡A=sin⁡B  ⟺  A=(−1)nB+πn[br]cos⁡A=cos⁡B  ⟺  A=±B+2πn[br]tan⁡A=tan⁡B  ⟺  A=B+πn[br] [br]\sin{A}=\sin{B} \iff A=(-1)^n B+\pi n[br]\cos{A}=\cos{B} \iff A=\pm B + 2\pi n[br]\tan{A}=\tan{B} \iff A=B+\pi n[br]
where nn is an integer.
Original post by kelsey.phillips
Hi there,

I am having trouble understanding the rationale behind the answer to question 6:

A) Show that, if tan2θ=2tanθ+1, then tan2θ=−1.

B) Find all solutions of the equation
tan θ = 2 + tan 3θ
which satisfy 0 < θ < 2π, expressing your answers as rational multiples of π .

C) Find all solutions of the equation
cot φ = 2 + cot 3φ
which satisfy −3π < φ < π , expressing your answers as rational multiples of π.

For the part B of the question (find all solutions of tan...), the answer booklet arrives at one solution t=1 and concludes that one set of roots is given by:

θ = nπ + π/4

Why is this??

Thanks for any help!


Kelsey


Clearly (I hope) tan⁡ϕ=1⇒ϕ=π4 \tan \phi = 1 \Rightarrow \phi = \frac{\pi}{4} is one solution.
Other solutions in the given range will lie in first and third quadrants, hence π4+nπ \frac{\pi}{4}+n\pi
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Avatar for Alpha95
Alpha95
OP
1
Original post by brianeverit
Clearly (I hope) tan⁡ϕ=1⇒ϕ=π4 \tan \phi = 1 \Rightarrow \phi = \frac{\pi}{4} is one solution.
Other solutions in the given range will lie in first and third quadrants, hence π4+nπ \frac{\pi}{4}+n\pi


Hi, thank you very much for your response!

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