Radian measure, arc length and sector area, exact values, the Pythagorean and addition identities, reciprocal and inverse functions, and solving trigonometric equations.

What this dot point is asking

Edexcel wants you to work in radians, use the arc length and sector area formulae, recall exact trigonometric values, apply the Pythagorean and addition identities and the double angle formulae, use the reciprocal functions ($\sec$, $\csc$, $\cot$) and inverse functions, express $a\cos\theta + b\sin\theta$ in the form $R\cos(\theta \pm \alpha)$, and solve trigonometric equations over a given interval.

The answer

Radians, arc length and sector area

Identities

The reciprocal functions are $\sec\theta = \dfrac{1}{\cos\theta}$, $\csc\theta = \dfrac{1}{\sin\theta}$ and $\cot\theta = \dfrac{1}{\tan\theta}$. Exact values worth memorising include $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$ and $\tan\dfrac{\pi}{3} = \sqrt{3}$.

The R form

You can write $a\sin\theta + b\cos\theta$ as $R\sin(\theta + \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$. This is useful for finding maxima and minima and for solving equations. Because $\sin$ and $\cos$ range between $-1$ and $1$, the combined expression ranges between $-R$ and $R$, so the maximum value of $a\sin\theta + b\cos\theta$ is $R$ and the minimum is $-R$. The maximum occurs when the bracket equals $\dfrac{\pi}{2}$, which is what makes the R form the standard tool for optimisation questions.

Solving equations

Examples in context

Try this

Q1. A sector has radius $6$ cm and angle $\tfrac{\pi}{3}$ radians. Find its arc length and area. [3 marks]

• Cue. Arc $= 6 \times \tfrac{\pi}{3} = 2\pi$ cm; area $= \tfrac{1}{2}(36)\tfrac{\pi}{3} = 6\pi$ cm squared.

Q2. Solve $\cos 2x = \tfrac{1}{2}$ for $0 \le x \le \pi$. [4 marks]

• Cue. $2x = \tfrac{\pi}{3}$ or $\tfrac{5\pi}{3}$, so $x = \tfrac{\pi}{6}$ or $x = \tfrac{5\pi}{6}$.