R-Method

A2 · P3 · 3 min

Any expression of the form asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta is a single sine or cosine wave with amplitude R=a2+b2R = \sqrt{a^2 + b^2} and a phase shift α\alpha. Writing it that way makes equations solvable in one step and reveals maximum and minimum values instantly.

The four forms

Key result
asin⁡θ+bcos⁡θ≡Rsin⁡(θ+α),asin⁡θ−bcos⁡θ≡Rsin⁡(θ−α)a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha), \qquad a\sin\theta - b\cos\theta \equiv R\sin(\theta - \alpha)acos⁡θ+bsin⁡θ≡Rcos⁡(θ−α),acos⁡θ−bsin⁡θ≡Rcos⁡(θ+α)a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha), \qquad a\cos\theta - b\sin\theta \equiv R\cos(\theta + \alpha)

with R=a2+b2>0R = \sqrt{a^2 + b^2} > 0 and tan⁡α=ba\tan\alpha = \dfrac{b}{a}, 0<α<90∘0 < \alpha < 90^\circ (for a,b>0a, b > 0).

Method
  1. Expand the target form with the compound-angle formula: e.g. Rsin⁡(θ+α)=Rcos⁡αsin⁡θ+Rsin⁡αcos⁡θR\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta.
  2. Compare coefficients: Rcos⁡α=aR\cos\alpha = a, Rsin⁡α=bR\sin\alpha = b.
  3. Square and add for RR; divide for tan⁡α\tan\alpha.
  4. Check that both cos⁡α\cos\alpha and sin⁡α\sin\alpha have the signs the equations require.
Writing in R-form

Express 3sin⁡θ+4cos⁡θ3\sin\theta + 4\cos\theta in the form Rsin⁡(θ+α)R\sin(\theta + \alpha) with R>0R > 0 and 0∘<α<90∘0^\circ < \alpha < 90^\circ.

Solution

Rcos⁡α=3R\cos\alpha = 3, Rsin⁡α=4R\sin\alpha = 4. R=9+16=5R = \sqrt{9 + 16} = 5; tan⁡α=43\tan\alpha = \tfrac{4}{3}, α=53.13∘\alpha = 53.13^\circ.

3sin⁡θ+4cos⁡θ=5sin⁡(θ+53.13∘)3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13^\circ).

Solving an equation

Hence solve 3sin⁡θ+4cos⁡θ=23\sin\theta + 4\cos\theta = 2 for 0∘≤θ≤360∘0^\circ \leq \theta \leq 360^\circ.

Solution

5sin⁡(θ+53.13∘)=2⇒sin⁡(θ+53.13∘)=0.45\sin(\theta + 53.13^\circ) = 2 \Rightarrow \sin(\theta + 53.13^\circ) = 0.4.

Let u=θ+53.13∘u = \theta + 53.13^\circ, 53.13∘≤u≤413.13∘53.13^\circ \leq u \leq 413.13^\circ. sin⁡u=0.4\sin u = 0.4: u=23.58∘u = 23.58^\circ (too small), 156.42∘156.42^\circ, 383.58∘383.58^\circ.

θ=103.3∘,330.5∘\theta = 103.3^\circ, 330.5^\circ.

Maximum and minimum values

Because −1≤sin⁡(θ+α)≤1-1 \leq \sin(\theta + \alpha) \leq 1, the expression Rsin⁡(θ+α)R\sin(\theta + \alpha) has maximum RR and minimum −R-R.

Greatest and least values

Find the maximum value of 3sin⁡θ+4cos⁡θ3\sin\theta + 4\cos\theta and the smallest positive θ\theta at which it occurs. Also find the minimum value of 103sin⁡θ+4cos⁡θ+7\dfrac{10}{3\sin\theta + 4\cos\theta + 7}.

Solution

Maximum 55, when sin⁡(θ+53.13∘)=1\sin(\theta + 53.13^\circ) = 1, i.e. θ+53.13∘=90∘\theta + 53.13^\circ = 90^\circ, θ=36.9∘\theta = 36.9^\circ.

The denominator 3sin⁡θ+4cos⁡θ+73\sin\theta + 4\cos\theta + 7 has maximum 1212, so the fraction has minimum 1012=56\tfrac{10}{12} = \tfrac{5}{6}.

Cosine form in radians

Express 2cos⁡θ−5sin⁡θ2\cos\theta - \sqrt{5}\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta + \alpha) and solve 2cos⁡θ−5sin⁡θ=12\cos\theta - \sqrt{5}\sin\theta = 1 for 0≤θ≤2π0 \leq \theta \leq 2\pi.

Solution

Rcos⁡(θ+α)=Rcos⁡αcos⁡θ−Rsin⁡αsin⁡θR\cos(\theta + \alpha) = R\cos\alpha\cos\theta - R\sin\alpha\sin\theta, so Rcos⁡α=2R\cos\alpha = 2, Rsin⁡α=5R\sin\alpha = \sqrt{5}: R=3R = 3, tan⁡α=52\tan\alpha = \tfrac{\sqrt{5}}{2}, α=0.8411\alpha = 0.8411.

3cos⁡(θ+0.8411)=1⇒cos⁡(θ+0.8411)=133\cos(\theta + 0.8411) = 1 \Rightarrow \cos(\theta + 0.8411) = \tfrac{1}{3}. With u=θ+0.8411u = \theta + 0.8411 in [0.8411,7.124][0.8411, 7.124]: u=1.2310,5.0522u = 1.2310, 5.0522 (and 1.2310+2π=7.5141.2310 + 2\pi = 7.514 is too large).

θ=0.390,4.21\theta = 0.390, 4.21 (3 s.f.).

Sketching

y=Rsin⁡(θ+α)y = R\sin(\theta + \alpha) is the sine curve with amplitude RR, translated α\alpha to the left. Mark the maximum at θ=90∘−α\theta = 90^\circ - \alpha and the yy-intercept Rsin⁡α=bR\sin\alpha = b.

Watch out

When solving after converting, the interval for u=θ+αu = \theta + \alpha shifts too. Solutions for uu near the lower end may correspond to negative θ\theta; solutions just beyond 360∘+α360^\circ + \alpha are also valid for θ\theta. Always work out the interval for uu first.

Exam tip

The angle α\alpha should be kept to at least 2 decimal places during the working (or stored on the calculator), then the final answers rounded to 1 decimal place. Rounding α\alpha early can shift the final answer by 0.1∘0.1^\circ and cost the accuracy mark.

Practice

Question
  1. Express 5sin⁡θ−12cos⁡θ5\sin\theta - 12\cos\theta as Rsin⁡(θ−α)R\sin(\theta - \alpha).
  2. Solve 5sin⁡θ−12cos⁡θ=6.55\sin\theta - 12\cos\theta = 6.5 for 0∘≤θ≤360∘0^\circ \leq \theta \leq 360^\circ.
  3. Find the maximum value of 3cos⁡θ+sin⁡θ\sqrt{3}\cos\theta + \sin\theta and the value of θ\theta in [0,2π][0, 2\pi] where it occurs.
  4. Find the range of f(θ)=8cos⁡θ−6sin⁡θ+3f(\theta) = 8\cos\theta - 6\sin\theta + 3.
Answers
  1. 13sin⁡(θ−67.38∘)13\sin(\theta - 67.38^\circ).
  2. sin⁡(θ−67.38∘)=0.5\sin(\theta - 67.38^\circ) = 0.5: θ−67.38∘=30∘,150∘\theta - 67.38^\circ = 30^\circ, 150^\circ; θ=97.4∘,217.4∘\theta = 97.4^\circ, 217.4^\circ.
  3. 2cos⁡(θ−π6)2\cos\left(\theta - \tfrac{\pi}{6}\right): maximum 22 at θ=π6\theta = \tfrac{\pi}{6}.
  4. R=10R = 10: −7≤f(θ)≤13-7 \leq f(\theta) \leq 13.

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