5.9 The forms r cos(theta +/- alpha) and r sin(theta +/- alpha) — Practise

Work through these on your own. Find $R$ before you do anything else, then place $\alpha$ in the correct quadrant. Give angles in radians to three decimal places unless the question asks for an exact value.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Write $3\cos\theta + 4\sin\theta$ in the form $R\cos(\theta - \alpha)$ , and state the greatest value.
Write $5\cos\theta - 12\sin\theta$ in the form $R\cos(\theta - \alpha)$ , and state the greatest value.
Write $-\cos\theta + \sqrt{3}\sin\theta$ in the form $R\cos(\theta - \alpha)$ , in exact form.
State the greatest value of $3\cos\theta + 4\sin\theta$ .
State the least value of $5\cos\theta - 12\sin\theta$ .
Write $4\sin\theta + 3\cos\theta$ in the form $R\sin(\theta + \beta)$ , with $\beta$ acute.
Solve $3\cos\theta + 4\sin\theta = 2$ for the smallest positive value of $\theta$ , to three decimal places.
The tide height is $h = 12\cos\theta + 5\sin\theta$ metres. State the greatest height.
For $12\cos\theta + 5\sin\theta$ , find the phase $\alpha$ in radians to three decimal places.
Solve $12\cos\theta + 5\sin\theta = 10$ for the smallest positive value of $\theta$ , to three decimal places.

Answers and concise solutions

1. The single wave and its greatest value

Answer: $R = \sqrt{3^2 + 4^2} = 5$ and $\tan\alpha = \dfrac{4}{3}$ , so $\alpha \approx 0.927$ . The single wave is $5\cos(\theta - 0.927)$ , and its greatest value is $5$ .

2. The single wave and its greatest value

Answer: $R = \sqrt{25 + 144} = 13$ . Here $a = 5$ and $b = -12$ , so $\alpha \approx -1.176$ and the wave is $13\cos(\theta + 1.176)$ . The greatest value is $13$ .

3. The single wave

Answer: $R = \sqrt{1 + 3} = 2$ . The coefficient of cosine is negative and the coefficient of sine is positive, so $\alpha$ is in the second quadrant. The related acute angle is $\dfrac{\pi}{3}$ , so $\alpha = \dfrac{2\pi}{3}$ and the wave is $2\cos\left(\theta - \dfrac{2\pi}{3}\right)$ .

4. The greatest value

Answer: The greatest value is $R = \sqrt{3^2 + 4^2} = 5$ . It is reached when $\theta - \alpha = 0$ , that is when $\theta \approx 0.927$ .

5. The least value

Answer: The least value is $-R = -\sqrt{5^2 + 12^2} = -13$ . It is reached when $\cos(\theta - \alpha) = -1$ , that is at the opposite phase to the peak.

6. The single wave

Answer: $R = \sqrt{4^2 + 3^2} = 5$ . For the sine form, $\cos\beta = \dfrac{4}{5}$ and $\sin\beta = \dfrac{3}{5}$ , so $\beta \approx 0.644$ and the wave is $5\sin(\theta + 0.644)$ .

7. The smallest positive solution

Answer: The left-hand side is $5\cos(\theta - 0.927)$ . So $\cos(\theta - 0.927) = 0.4$ , giving $\theta - 0.927 = \pm 1.159$ . The positive solution is $\theta \approx 0.927 + 1.159 = 2.087$ radians.

8. The greatest height

Answer: The greatest height is $R = \sqrt{144 + 25} = 13$ m. The height is never $17$ m, because the cosine and sine terms do not reach their peaks at the same angle.

9. The phase

Answer: $\tan\alpha = \dfrac{5}{12}$ , so $\alpha \approx 0.395$ radians. Both coefficients are positive, so $\alpha$ is acute and no quadrant adjustment is needed.

10. The smallest positive solution

Answer: Write the left-hand side as $13\cos(\theta - 0.395)$ . Then $\cos(\theta - 0.395) = \dfrac{10}{13} = 0.7692$ , so $\theta - 0.395 \approx 0.693$ . The smallest positive solution is $\theta \approx 0.395 + 0.693 = 1.088$ radians.