Porisma

Worked solution

Find a phase shift and then bound the shifted expression

Express 5sin⁡θ+12cos⁡θ5\sin\theta+12\cos\theta in the form rcos⁡(θ−α)r\cos(\theta-\alpha). Show that −6≤5sin⁡θ+12cos⁡θ+7≤20-6\le5\sin\theta+12\cos\theta+7\le20.

CJC · O Level revision · Example 23

Combine the two trigonometric terms

Answer

13cos⁡(θ−α)13\cos(\theta-\alpha), where α=tan⁡−1(5/12)≈22.6∘\alpha=\tan^{-1}(5/12)\approx22.6^\circ.

Match the cosine and sine coefficients

12cos⁡θ+5sin⁡θ=rcos⁡(θ−α),r=122+52=13,α=tan⁡−1(5/12)≈22.6∘.\begin{aligned}12\cos\theta+5\sin\theta&=r\cos(\theta-\alpha),\\r&=\sqrt{12^2+5^2}=13,\\\alpha&=\tan^{-1}(5/12)\approx22.6^\circ.\end{aligned}

The source reorders the sum as 12cosθ+5sinθ. Its printed phase 22.6° is rounded; the exact arctangent specifies the identity in consistent angle units.

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Scale the cosine range and add seven

Answer

−6≤5sin⁡θ+12cos⁡θ+7≤20-6\le5\sin\theta+12\cos\theta+7\le20.

Retain both bounds at every stage

−1≤cos⁡(θ−α)≤1,−13≤13cos⁡(θ−α)≤13,−13≤5sin⁡θ+12cos⁡θ≤13,−13+7≤5sin⁡θ+12cos⁡θ+7≤13+7,−6≤5sin⁡θ+12cos⁡θ+7≤20.\begin{aligned}-1&\le\cos(\theta-\alpha)\le1,\\-13&\le13\cos(\theta-\alpha)\le13,\\-13&\le5\sin\theta+12\cos\theta\le13,\\-13+7&\le5\sin\theta+12\cos\theta+7\le13+7,\\-6&\le5\sin\theta+12\cos\theta+7\le20.\end{aligned}

The source callout supplies the cosine extrema 1 and −1. Scaling by positive 13 preserves the inequalities; adding 7 shifts both endpoints.

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