Geometry and Trigonometry

Geometry and trigonometry form a significant part of the Leaving Certificate syllabus, particularly Paper 2. This topic covers coordinate geometry, trigonometric functions, identities, and geometric Theorems and proofs.

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

This is the Pythagorean theorem applied to the horizontal and vertical displacements.

The midpoint $M$ of $AB$:

$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

The slope of the line through $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$M = \frac{y_2 - y_1}{x_2 - x_1}$$

A vertical line has undefined slope. A horizontal line has slope $0$.

Point-slope form:

$$Y - y_1 = m(x - x_1)$$

General form:

$$Ax + by + c = 0$$

Slope-intercept form:

$$Y = mx + c$$

Where $m$ is the slope and $c$ is the $y$-intercept.

• Parallel lines have equal slopes: $m_1 = m_2$.
• Perpendicular lines: $m_1 \cdot m_2 = -1$.

Proof of the perpendicular condition. If two lines with slopes $m_1$ and $m_2$ are Perpendicular, then the angle between them is $90^\circ$. Using the tangent addition formula: $\tan(\alpha + \beta) = \frac{m_1 + m_2}{1 - m_1 m_2}$. Setting $\alpha + \beta = 90^\circ$: $\tan 90^\circ$ Is undefined, so $1 - m_1 m_2 = 0$ Giving $m_1 m_2 = -1$.

Example (OL): Find the equation of the line through $(1, 3)$ perpendicular to $y = 2x + 1$.

The slope of the given line is $m_1 = 2$ So $m_2 = -\frac{1}{2}$.

$$Y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{7}{2}$$

Example (HL): Find the equation of the perpendicular bisector of the segment joining $A(2, 5)$ And $B(6, 1)$.

Midpoint: $M = \left(\frac{2+6}{2}, \frac{5+1}{2}\right) = (4, 3)$.

Slope of $AB$: $m = \frac{1 - 5}{6 - 2} = -1$.

Slope of perpendicular bisector: $m_{\perp} = 1$.

Equation: $y - 3 = 1(x - 4) \implies y = x - 1$.

The perpendicular distance from $(x_0, y_0)$ to $ax + by + c = 0$:

$$D = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

Example (HL): Find the distance from $(3, 2)$ to $2x + y - 5 = 0$.

$$D = \frac{|6 + 2 - 5|}{\sqrt{4 + 1}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$$

Proof sketch. Let $P(x_0, y_0)$ be the point and $ax + by + c = 0$ the line. The closest point $Q$ on the line to $P$ lies along the perpendicular. The line through $P$ perpendicular to $ax + by + c = 0$ has equation $b(x - x_0) - a(y - y_0) = 0$. Solving the two equations Simultaneously gives $Q$ And the distance $PQ$ simplifies to the formula above.

The area of a triangle with vertices (x_1, y_1)$$(x_2, y_2)$$(x_3, y_3):

$$A = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

This is derived from the shoelace formula (also known as Gauss’s area formula).

Example (HL): Find the area of the triangle with vertices (1, 2)$$(4, 6)$$(3, -1).

$$A = \frac{1}{2}|1(6 - (-1)) + 4((-1) - 2) + 3(2 - 6)| = \frac{1}{2}|7 - 12 - 12| = \frac{1}{2}|-17| = 8.5$$

Two lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$: lines intersect at a unique point.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$: lines are parallel (no intersection).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$: lines are coincident (infinitely many intersections).

Centre-radius form: $(x - h)^2 + (y - k)^2 = r^2$ with centre $(h, k)$ and radius $r$.

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$ with centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.

The circle exists only if $g^2 + f^2 - c > 0$.

Example (OL): Find the centre and radius of $x^2 + y^2 - 4x + 6y - 3 = 0$.

Completing the square:

$$(x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \implies (x - 2)^2 + (y + 3)^2 = 16$$

Centre $(2, -3)$ Radius $4$.

The tangent at a point $(x_1, y_1)$ on the circle $x^2 + y^2 = r^2$ has equation:

$$X_1 x + y_1 y = r^2$$

Proof. The radius to $(x_1, y_1)$ has slope $y_1/x_1$. The tangent is perpendicular, so its Slope is $-x_1/y_1$. Using point-slope form: $y - y_1 = -\frac{x_1}{y_1}(x - x_1)$ Which simplifies to $x_1 x + y_1 y = x_1^2 + y_1^2 = r^2$.

Example (HL): Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.

$$3x + 4y = 25$$

Example (HL): Show that the line $3x - 4y + 25 = 0$ is a tangent to $x^2 + y^2 = 25$.

Substitute $y = \frac{3x + 25}{4}$ into $x^2 + y^2 = 25$:

$$X^2 + \left(\frac{3x + 25}{4}\right)^2 = 25$$ $$16x^2 + 9x^2 + 150x + 625 = 400$$ $$25x^2 + 150x + 225 = 0$$ $$X^2 + 6x + 9 = 0 \implies (x + 3)^2 = 0$$

The discriminant is $\Delta = 0$ Confirming a tangent. The point of tangency is $x = -3$ $y = \frac{-9 + 25}{4} = 4$.

Substitute the line into the circle equation. The discriminant of the resulting quadratic tells you:

• $\Delta > 0$: two intersection points
• $\Delta = 0$: tangent (one intersection point)
• $\Delta < 0$: no intersection

Example (HL): Find the equation of the circle through (0, 0)$$(4, 0) And $(0, 4)$.

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.

Substituting $(0,0)$: $c = 0$.

Substituting $(4,0)$: $16 + 8g = 0 \implies g = -2$.

Substituting $(0,4)$: $16 + 8f = 0 \implies f = -2$.

The circle is $x^2 + y^2 - 4x - 4y = 0$ With centre $(2, 2)$ and radius $\sqrt{4+4} = 2\sqrt{2}$.

For a right-angled triangle with angle $\theta$:

On the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$.

Key values:

Pythagorean identities:

$$\sin^2\theta + \cos^2\theta = 1$$ $$1 + \tan^2\theta = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$

Proof of the second identity. Divide $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$: $\tan^2\theta + 1 = \sec^2\theta$.

Example (HL): Given $\sin\theta = \frac{3}{5}$ and $\theta$ is in the second quadrant, find $\cos\theta$ and $\tan\theta$.

$$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$

Since $\theta$ is in the second quadrant, $\cos\theta < 0$ So $\cos\theta = -\frac{4}{5}$.

$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$$$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$ $$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$

Example (HL): Find the exact value of $\sin 75^\circ$.

$$\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Example (HL): Find the exact value of $\tan 15^\circ$.

$$\tan 15° = \tan(45° - 30°) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$$$\sin 2A = 2\sin A \cos A$$ $$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$ $$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$

Proof of $\cos 2A = 2\cos^2 A - 1$. Using the compound angle formula:

$$\cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A$$

Since $\sin^2 A = 1 - \cos^2 A$:

$$\cos 2A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1$$

Proof that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$:

$$\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta$$ $$= 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta$$ $$= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta$$ $$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta$$

Similarly, $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$:

$$\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$$ $$= (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta$$ $$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta$$ $$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta = 4\cos^3\theta - 3\cos\theta$$$$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$ $$\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$ $$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$ $$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$

Example (HL): Evaluate $\sin 75° - \sin 15^\circ$.

$$\sin 75° - \sin 15° = 2\cos\frac{90°}{2}\sin\frac{60°}{2} = 2\cos 45°\sin 30° = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}$$

Example (OL): Solve $\sin\theta = \frac{1}{2}$ for $0 \leq \theta \leq 2\pi$.

$$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$

Example (HL): Solve $2\cos^2\theta + 3\cos\theta - 2 = 0$ for $0 \leq \theta \leq 2\pi$.

Let $u = \cos\theta$: $2u^2 + 3u - 2 = 0 \implies (2u - 1)(u + 2) = 0$.

$u = \frac{1}{2}$ or $u = -2$ (rejected since $|\cos\theta| \leq 1$).

$\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}, \frac{5\pi}{3}$.

Example (HL): Solve $\sin 2\theta = \sin \theta$ for $0 \le \theta \lt 2\pi$.

$$2\sin\theta\cos\theta = \sin\theta$$ $$\sin\theta(2\cos\theta - 1) = 0$$

$\sin\theta = 0$: $\theta = 0, \pi$.

$2\cos\theta - 1 = 0$: $\cos\theta = \frac{1}{2}$ So $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$.

Solutions: $0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$.

• Geometry Trig
• Irish LC Mathematics