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

1. $\sqrt{9 + 16} = \sqrt{25} = 5$.
2. $\arctan 1 = \dfrac{\pi}{4}$.
3. $\cos 0 = 1$.

If question 1 was hard, revise Pythagoras before you continue.

$R = \sqrt{a^2 + b^2}$ with $a = 3$ and $b = 4$. Then $\tan\alpha = \dfrac{b}{a}$.

$R = \sqrt{9 + 16} = 5$ and $\tan\alpha = \dfrac{4}{3}$, so $\alpha \approx 0.9273$. The greatest value is $R$.

The greatest value is $5$, reached when $\theta = 0.9273$.

Here $a = 5$ and $b = -12$. The negative $b$ puts $\alpha$ in the fourth quadrant.

$R = \sqrt{25 + 144} = 13$. Also $\tan\alpha = \dfrac{-12}{5}$, so the related acute angle is about $1.176$ and $\alpha \approx -1.176$.

The greatest value is $13$.

$a = -1$, so cosine is negative, and $b = \sqrt{3}$ is positive. That puts $\alpha$ in the second quadrant.

$R = 2$. The related acute angle is $\arctan\sqrt{3} = \dfrac{\pi}{3}$, so $\alpha = \pi - \dfrac{\pi}{3}$.

Replace the left-hand side with $5\cos(\theta - 0.9273)$ from question 1.

$\cos(\theta - 0.9273) = 0.4$. The principal value is $1.1593$ radians, and cosine is positive at the negative of that value as well.

The second is outside the interval, so add $2\pi$ to get $\theta = 6.0512$. The solutions are $\theta \approx 2.087$ and $\theta \approx 6.051$.

The greatest height is $R = \sqrt{a^2 + b^2}$.

$R = \sqrt{144 + 25} = \sqrt{169}$.

The greatest height is $13$ m. The tide is never as high as $12 + 5 = 17$ m, because the two waves do not peak at the same angle.

For the sine form, $\sin\beta = \dfrac{a}{R}$ and $\cos\beta = \dfrac{b}{R}$, where $a$ is the cosine coefficient and $b$ the sine coefficient.

$R = 5$. Here the cosine coefficient is $3$ and the sine coefficient is $4$, so $\sin\beta = \dfrac{3}{5}$ and $\cos\beta = \dfrac{4}{5}$.

where $\beta \approx 0.6435$ radians, because $\sin\beta = \dfrac{3}{5}$.