Trigonometric Functions — Worked Solutions (HSC Maths Advanced)

These two formulas, $\ell = r\theta$ and $A = \tfrac{1}{2}r^2\theta$, only work when the angle $\theta$ is measured in radians — which it is here, $\tfrac{\pi}{3}$.

For the arc length we just multiply the radius by the angle: $\ell = 6 \times \tfrac{\pi}{3} = 2\pi$ cm.

For the area of the sector: $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}\times 36 \times \tfrac{\pi}{3}$. Half of $36$ is $18$, and $18 \times \tfrac{\pi}{3} = 6\pi$ cm$^2$.

So the arc length is $2\pi$ cm and the sector area is $6\pi$ cm$^2$. Leaving answers in terms of $\pi$ keeps them exact, which is what "exact" asks for.

First let's get the sine on its own by dividing by $2$: $\sin x = \tfrac{\sqrt{3}}{2}$.

The angle whose sine is $\tfrac{\sqrt{3}}{2}$ is $\tfrac{\pi}{3}$ — that's our reference (related) angle. Now, where is sine positive? In the first and second quadrants, because $y$-coordinates are positive there.

So one solution is the reference angle itself, $x = \tfrac{\pi}{3}$ (quadrant 1). For quadrant 2 we use $\pi$ minus the reference angle: $x = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$.

Both fall inside $0 \le x \le 2\pi$, so the solutions are $x = \tfrac{\pi}{3}$ and $x = \tfrac{2\pi}{3}$.

A function of the form $y = a\cos(bx)$ has two key features. The amplitude — how far it rises above and falls below its centre line — is $|a|$. The period — how long before the wave repeats — is $\tfrac{2\pi}{b}$, because the larger $b$ is, the faster the wave cycles.

Here $a = 3$ and $b = 2$.

So the amplitude is $|3| = 3$, and the period is $\tfrac{2\pi}{2} = \pi$.

We differentiate each term separately. For $\sin(3x)$ we use the chain rule: the derivative of sine is cosine, and we multiply by the derivative of the inside, $3x$, which is $3$. So $\tfrac{d}{dx}\sin(3x) = 3\cos(3x)$.

For $\cos x$, the derivative of cosine is $-\sin x$ — note the negative sign, which often catches people out.

Adding them: $y' = 3\cos(3x) - \sin x$.

Let's find the primitive first. Integrating $\cos(kx)$ gives $\tfrac{1}{k}\sin(kx)$ — the $\tfrac{1}{k}$ appears to undo the chain-rule factor that differentiation would introduce. So the primitive of $\cos(2x)$ is $\tfrac{1}{2}\sin(2x)$.

Now apply the limits using the fundamental theorem of calculus: $\displaystyle\int_0^{\pi/2}\cos(2x)\,dx = \Big[\tfrac{1}{2}\sin(2x)\Big]_0^{\pi/2}$.

At the upper limit $x = \tfrac{\pi}{2}$: $2x = \pi$, and $\sin\pi = 0$, so this term is $0$. At the lower limit $x = 0$: $\sin 0 = 0$, so this term is also $0$.

Subtracting, $0 - 0 = 0$. The integral equals $0$ — which makes sense, since over $[0,\tfrac{\pi}{2}]$ the curve $\cos(2x)$ has equal areas above and below the axis.