Consider the lengths $DE$ and $AD$ in the following diagram to prove the addition formulae for $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$.
$AB = \cos\beta\cos\alpha$ <br> $BC = \cos\beta\sin\alpha$ <br> $FE = \sin\beta\cos\alpha$ <br> $FC = \sin\beta\sin\alpha$ <br> $AD = AB - FC$ <br> $DE = FE + BC$
Consider the lengths $FC$ and $AF$ in the following rectangle to prove the subtraction formulae for $\sin(\alpha-\beta)$ and $\cos(\alpha-\beta)$.
$\angle BAD = 90 - \alpha$ <br> $CE = \sin\beta\cos\alpha$ <br> $BE = \sin\beta\sin\alpha$ <br> $BD = \cos\beta\cos\alpha$ <br> $AD = \cos\beta\sin\alpha$ <br> $FC = AD - CE$ <br> $AF = BD + BE$
Write the following in the form $\sin(x + \alpha)$:
- $\dfrac{1}{\sqrt{2}}(\sin x + \cos x)$
- $\dfrac{1}{2}(\sin x + \sqrt{3}\cos x)$
- $\sin(x+45^{\circ})$
- $\sin(x+60^{\circ})$
Given $\tan\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}$, find the exact value of $\tan x$
$8-5\sqrt{3}$
Given $180^{\circ} < \alpha < 270^{\circ}$, $\beta$ is obtuse, and $$\sin \alpha = -\frac{3}{5} \quad \cos \beta = -\frac{12}{13}$$ find:
- $\cos(\alpha-\beta)$
- $\tan(\alpha+\beta)$
- $\dfrac{33}{65}$
- $\dfrac{16}{63}$
Given $\alpha$ is reflex, $\beta$ is obtuse, and $$\tan \alpha = \frac{7}{24} \quad \sin \beta = \frac{5}{13}$$ find:
- $\sin(\alpha+\beta)$
- $\tan(\alpha-\beta)$
- $-\dfrac{36}{325}$
- $\dfrac{204}{253}$
Find exact values for:
- $\tan 75^{\circ}$
- $\cos 105^{\circ}$
- $2+\sqrt{3}$
- $\dfrac{\sqrt{2}-\sqrt{6}}{4}$
Given that $x=\sin\theta$, $y=\sin2\theta$, and that $\theta$ is acute, find an expression for $y$ in terms of $x$.
$2x\sqrt{1-x^2}$
Given that $\tan\theta = \dfrac{3}{4}$ and that $\theta$ is acute, find the value of $\sin4\theta$.
$\dfrac{336}{625}$
Solve, in the interval $0 \leqslant \theta < 360^{\circ}$:
- $3\cos\theta = 2\sin(\theta+60^{\circ})$
- $\sin(\theta+30^{\circ})+2\sin\theta=0$
- $51.7^{\circ}, 232^{\circ}$
- $170^{\circ}, 350^{\circ}$
Solve the following equations in the interval $0 \leqslant \theta < 2\pi$:
- $3\cos2\theta = 2\cos^2\theta$
- $2\sin2\theta=3\tan\theta$
- $\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
- $0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}, 2\pi$
Solve, in the interval $0 \leqslant x < \pi$:
- $\cos^2x-\sin2x=\sin^2x$
- $\sin4x=\cos2x$
- $\dfrac{\pi}{8}, \dfrac{5\pi}{8}$
- $\dfrac{\pi}{12}, \dfrac{\pi}{4}, \dfrac{5\pi}{12}, \dfrac{3\pi}{4}$
Solve, in the interval $0 \leqslant x < 2\pi$:
- $5\sin2x+4\sin x = 0$
- $2\cos^2\frac{x}{2}-4\sin^2\frac{x}{2}=-3$
- $0, 1.98, \pi, 4.30$
- $2.30, 3.98$
By expanding $R\sin(x+\alpha)$, where $R>0$ and $\alpha$ is acute, solve, in the interval $0 \leqslant \theta < 360^{\circ}$:
- $6\sin x + 8\cos x = 5\sqrt{3}$
- $8\cos x + 15\sin x = 10$
- $6.87^{\circ}, 66.9^{\circ}$
- $7.96^{\circ}, 116^{\circ}$
By expanding $R\sin(x-\alpha)$, where $R>0$ and $\alpha$ is acute, solve, in the interval $0 \leqslant \theta < \pi$: $$5\sin\tfrac{\theta}{2}-12\cos\tfrac{\theta}{2}=-6.5$$
$1.30$
By expanding $R\sin(x-\alpha)$, where $R<0$ and $\alpha$ is acute, solve, in the interval $0 \leqslant \theta < \pi$: $$2\cos3\theta-3\sin3\theta = -1$$
$0.290, 1.15, 2.38$
Solve, in the interval $0\leq\theta\leq2\pi$: $$\sqrt{2}\cos\left(\theta-\dfrac{\pi}{4}\right)+(\sqrt{3}-1)\sin\theta=2$$
$\dfrac{\pi}{3}$
The height, $H$ metres, of water at a harbour is modelled by $$H = 5 + 4\sin(30t)^{\circ} + 3\cos(30t)^{\circ}$$ where $t$ is the time in hours after midnight. Boats can leave the harbour when the height of the water is at least 4 m.
- Find the maximum height of water at the harbour.
- What is the latest time before 9am, to the nearest minute, that a boat can leave the harbour?
- $10$
- $5:09$ am
Given that $\cos\theta = \dfrac{3}{5}$ and that $\dfrac{3\pi}{2} \leq \theta \leq 2\pi$, find the exact value of $\cos3\theta$.
$-\dfrac{117}{125}$
The temperature of a swimming pool is given by $$25 + 4\cos (12t)^{\circ} + 2\sin (12t)^{\circ}$$ where $t$ is the time in hours after the pool opens. What percentage of the time that the pool is open for is the temperature below $26$?
$57.2\%$
Solve, in the interval $0 \leqslant x < 2\pi$: $$4\tan\tfrac{1}{2}x=\tan x$$
$0, 1.23, 5.05$
Solve, in the interval $0 \leqslant x < 2\pi$: $$3\cos x - \sin\frac{x}{2}-1=0$$
$\dfrac{\pi}{3}, \dfrac{5\pi}{3}$
Find the minimum value of $$\dfrac{18}{50+9\cos\theta+40\sin\theta}$$ and the smallest positive value of $\theta$ at which this occurs.
$(1.35, \frac{18}{91})$