Maths4U
« Back to Subchapter

Trigonometry – Expressing a sinθ + b cosθ in the form R sin(θ ± α) or R cos(θ ± α)

📘 Notes

🧮 A Level Maths (9709): Expressing \( a\sin\theta + b\cos\theta \) in the form \( R\sin(\theta \pm \alpha) \) or \( R\cos(\theta \pm \alpha) \)

This technique is often used to simplify trigonometric expressions and to solve equations involving both sine and cosine terms.

1. 📌 The Key Idea

Any expression of the form \[ a\sin\theta + b\cos\theta \] can be written as \[ R\sin(\theta + \alpha) \quad\text{or}\quad R\cos(\theta - \alpha), \] where \[ R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \frac{b}{a}, \quad 0^\circ \le \alpha \le 90^\circ. \]

Similarly, for \[ a\cos\theta + b\sin\theta \] we can write \[ R\cos(\theta - \alpha) \quad\text{or}\quad R\sin(\theta + \alpha). \]

2. 🧠 Derivation of the Formula

Start with \(\displaystyle a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)\).

Expand the right-hand side using the sine addition formula: \[ R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha. \]

Compare coefficients of \(\sin\theta\) and \(\cos\theta\): \[ a = R\cos\alpha, \qquad b = R\sin\alpha. \]

Then \(\displaystyle R = \sqrt{a^2 + b^2}\), and dividing gives \(\displaystyle \tan\alpha = \frac{b}{a}\).

✅ Final result: \(\displaystyle a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)\), where \(\displaystyle R = \sqrt{a^2 + b^2}\), \(\displaystyle \alpha = \arctan\left(\frac{b}{a}\right)\).

3. 📝 Worked Examples

Example 1: Express \( 3\sin\theta + 4\cos\theta \) in the form \( R\sin(\theta+\alpha) \).

\( a = 3 \), \( b = 4 \). \(\displaystyle R = \sqrt{3^2+4^2} = \sqrt{25} = 5.\)

\(\displaystyle \tan\alpha = \frac{b}{a} = \frac{4}{3}\). \(\displaystyle \alpha = \arctan\left(\frac{4}{3}\right) \approx 53.1^\circ\).

\(\displaystyle 3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.1^\circ)\).

Example 2: Express \( 5\cos\theta + 12\sin\theta \) in the form \( R\cos(\theta - \alpha) \).

\( a = 5 \) (cos coefficient), \( b = 12 \) (sin coefficient). \(\displaystyle R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).

\(\displaystyle \tan\alpha = \frac{12}{5}\), \(\displaystyle \alpha = \arctan\left(\frac{12}{5}\right) \approx 67.4^\circ\).

\(\displaystyle 5\cos\theta + 12\sin\theta = 13\cos(\theta - 67.4^\circ)\).

Example 3: Solve \( 3\sin\theta + 4\cos\theta = 2 \) for \( 0^\circ \le \theta \le 360^\circ \).

From Example 1: \[ 3\sin\theta + 4\cos\theta = 5\sin(\theta+53.1^\circ). \] So \[ 5\sin(\theta+53.1^\circ)=2 \quad\Rightarrow\quad \sin(\theta+53.1^\circ)=\frac{2}{5}=0.4. \]

\(\displaystyle \theta+53.1^\circ = \arcsin(0.4) \approx 23.6^\circ\) or \( 180^\circ - 23.6^\circ = 156.4^\circ \).

\(\displaystyle \theta = -29.5^\circ \) (add 360° → \( 330.5^\circ \)) or \( 103.3^\circ \).

✅ Final answers: \(\displaystyle \theta \approx 103.3^\circ, \; 330.5^\circ\).

📌 Key Points to Remember

  • Always find \( R \) first using \( R = \sqrt{a^2 + b^2} \).
  • Use \( \tan\alpha = \dfrac{b}{a} \) to find the angle \( \alpha \).
  • You can choose to express the result as either:
    • \( R\sin(\theta + \alpha) \) when starting with \( a\sin\theta + b\cos\theta \), or
    • \( R\cos(\theta - \alpha) \) when starting with \( a\cos\theta + b\sin\theta \).
  • This method is often used to solve equations and find maximum or minimum values of trig expressions.
  • For A Level exams, your answer for \( \alpha \) should usually be given to 1 decimal place (unless exact values are expected).