International Baccalaureate · DP 2: Analysis and Approaches HL

Revision: Geometry and Trigonometry

⌚ ~51 min ▢ 6 blocks ✓ 17 exercises ➣ Prerequisites : Further Trigonometry, Non-Right-Angled Triangle Trigonometry

I Toolbox: Arc Length and Sector Area

Recappreviously studied — click to reread

Proposition — Arc Length

On a circle of radius $r$, an angle of $\theta$ radians, with $\theta > 0$, subtends an arc of length $\ell = r\,\theta$. Equivalently, $\theta = \dfrac{\ell}{r}$ and $r = \dfrac{\ell}{\theta}$.

Proposition — Area of a Sector in Radians

A sector is the part of a disc bounded by two radii and the arc between them. On a circle of radius $r$, the sector of angle $\theta$ radians, with $0 < \theta \leq 2\pi$ — the sector whose arc is the one the angle $\theta$ sweeps — has area $A = \dfrac{1}{2}\,r^2\,\theta$. With its arc length $\ell = r\,\theta$, this is also $A = \dfrac{1}{2}\,\ell\,r$.

Skills to practice

Finding Exact Arc Lengths and Sector Areas

II Toolbox: The Double Angle Formulas

Recappreviously studied — click to reread

Proposition — Double Angle Formulas

For every real number $A$: $$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A,$$ $$\sin 2A = 2\sin A\cos A.$$ They are obtained by taking $B = A$ in the formulas for $\cos(A + B)$ and $\sin(A + B)$, which gives $\sin 2A$ and $\cos^2 A - \sin^2 A$, then using $\cos^2 A + \sin^2 A = 1$ for the other two forms of $\cos 2A$.

Example

Compute the exact value of $\cos\dfrac{\pi}{12}$ in two ways:

with an addition formula, writing $\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$;
with a double angle formula, since $2 \times \dfrac{\pi}{12} = \dfrac{\pi}{6}$.

Answer

With the cosine of a difference: $$\begin{aligned}[t] \cos\frac{\pi}{12} &= \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) &&\\ &= \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4} && \text{(cosine of a difference)}\\ &= \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} && \text{(special values)}\\ &= \frac{\sqrt{2} + \sqrt{6}}{4}. \end{aligned}$$
With $A = \dfrac{\pi}{12}$, the formula $\cos 2A = 2\cos^2 A - 1$ gives $\cos\dfrac{\pi}{6} = 2\cos^2\dfrac{\pi}{12} - 1$, so $$\begin{aligned}[t] \cos^2\frac{\pi}{12} &= \frac{1}{2}\left(1 + \cos\frac{\pi}{6}\right) &&\\ &= \frac{1}{2}\left(1 + \frac{\sqrt{3}}{2}\right) = \frac{2 + \sqrt{3}}{4}. \end{aligned}$$ Since $\dfrac{\pi}{12} \in \left[0, \dfrac{\pi}{2}\right]$, $\cos\dfrac{\pi}{12} \ge 0$, hence $\cos\dfrac{\pi}{12} = \dfrac{\sqrt{2 + \sqrt{3}}}{2}$.

The two results are the same number: $\left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right)^2 = \dfrac{2 + 2\sqrt{12} + 6}{16} = \dfrac{8 + 4\sqrt{3}}{16} = \dfrac{2 + \sqrt{3}}{4}$, and both are positive.

Skills to practice

Using the Double Angle Formulas

III Exam-Style Questions

Skills to practice

Working without Technology: Geometry and Trigonometry
Working with Technology: Geometry and Trigonometry
Working without Technology: Further Geometry and Trigonometry
Working with Technology: Further Geometry and Trigonometry