Solving a trig equation with the t-formulae

t-formulae

A-level Further Maths (9FM0) · Further Pure 1

ORIGINAL

Substitute t = tan(θ/2), solve in t — then check θ = π, the root the substitution can never find.

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Question

Use the substitution

t = \tan\dfrac{\theta}{2}

to solve

3\sin\theta - 4\cos\theta = 4

for

0 \le \theta < 2\pi

. Give any non-exact answers to 3 significant figures.

Every step worked, with the reasoning.

1
$\sin\theta = \dfrac{2t}{1 + t^2}$ , $\ \cos\theta = \dfrac{1 - t^2}{1 + t^2}$
The t-formulae, with $t = \tan\tfrac{\theta}{2}$ .
2
$\dfrac{6t}{1 + t^2} - \dfrac{4(1 - t^2)}{1 + t^2} = 4$
Substitute into the equation.
3
$6t - 4 + 4t^2 = 4 + 4t^2 \Rightarrow 6t = 8 \Rightarrow t = \dfrac43$
Multiply through by $1 + t^2$ (always positive); the $t^2$ terms cancel.
4
$\tan\dfrac{\theta}{2} = \dfrac43$ with $0 \le \dfrac{\theta}{2} < \pi$ , so $\dfrac{\theta}{2} = 0.9273\ldots$ and $\theta = 1.85$ (3 s.f.)
$\tan$ has period $\pi$ , so exactly one value of $\tfrac{\theta}{2}$ lies in $[0, \pi)$ ; double it for $\theta$ .
5
Check $\theta = \pi$ : $3\sin\pi - 4\cos\pi = 0 + 4 = 4$ , so $\theta = \pi$ is also a solution.
At $\theta = \pi$ , $t = \tan\tfrac{\pi}{2}$ is undefined, so the substitution can never produce this root — always test it separately.

Answer: $\theta = 1.85$ (3 s.f.) or $\theta = \pi$