Geometry and Trigonometry | Highers

Geometry and Trigonometry

Higher Trigonometry

Trigonometric Functions

The three primary trigonometric functions for an angle $\theta$ in a right-angled triangle are:

\sin\theta = \frac{\mathrm{opposite}{\mathrm{hypotenuse}, \quad \cos\theta = \frac{\mathrm{adjacent}{\mathrm{hypotenuse}, \quad \tan\theta = \frac{\mathrm{opposite}{\mathrm{adjacent}

On the unit circle (radius 1), the point at angle $\theta$ from the positive $x$ -axis has Coordinates $(\cos\theta, \sin\theta)$ . This definition extends the trig functions to all real Angles, not just those in $[0, \pi/2]$ .

Key Identity:

\sin^2\theta + \cos^2\theta = 1

Proof (geometric). In the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$ . By the Pythagorean theorem, the distance from the origin is $\sqrt{\cos^2\theta + \sin^2\theta} = 1$ So $\cos^2\theta + \sin^2\theta = 1$ .

Dividing through by $\cos^2\theta$ :

1 + \tan^2\theta = \sec^2\theta

Dividing through by $\sin^2\theta$ :

1 + \cot^2\theta = \cosec^2\theta

Radians

Angles can be measured in radians. One full revolution is $2\pi$ radians.

\pi \mathrm{ radians = 180°

Why radians? In calculus, the derivative formula $\frac{d}{dx}[\sin x] = \cos x$ holds only when $x$ is in radians. If $x$ is in degrees, you get an extra factor of $\frac{\pi}{180}$ . Radians arise because the arc length subtended by angle $\theta$ on a unit circle is exactly $\theta$ .

Arc Length:

S = r\theta

Where $s$ is arc length, $r$ is radius, and $\theta$ is in radians.

Sector Area:

A = \frac{1}{2}r^2\theta

Segment Area:

A = \frac{1}{2}r^2(\theta - \sin\theta)

Example: Find the length of the arc and the area of the sector for a circle of radius 8 cm with An angle of $\dfrac{5\pi}{6}$ radians.

Arc length: $s = 8 \times \dfrac{5\pi}{6} = \dfrac{20\pi}{3} \approx 20.94 \mathrm{ cm$ .

Sector area: $A = \dfrac{1}{2} \times 64 \times \dfrac{5\pi}{6} = \dfrac{160\pi}{6} = \dfrac{80\pi}{3} \approx 83.78 \mathrm{ cm^2$ .

Example: A sector of a circle of radius 6 cm has an area of $24\pi \mathrm{ cm^2$ . Find the Perimeter of the sector.

\frac{1}{2} \times 36 \times \theta = 24\pi \implies \theta = \frac{48\pi}{36} = \frac{4\pi}{3}

Arc length: $s = 6 \times \frac{4\pi}{3} = 8\pi \mathrm{ cm$ .

Perimeter = $2r + s = 12 + 8\pi \mathrm{ cm$ .

Trigonometric Identities

Addition Formulae:

\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}

Proof of $\cos(A + B)$ . Consider two points on the unit circle: $P$ at angle $A$ with Coordinates $(\cos A, \sin A)$ And $Q$ at angle $-(A+B)$ with coordinates $(\cos(A+B), -\sin(A+B))$ . Rotating the entire figure by angle $A$ maps $P$ to $(1, 0)$ and $Q$ to The point at angle $-B$ Namely $(\cos B, -\sin B)$ . Since rotation preserves distances:

[\cos(A+B) - \cos A]^2 + [-\sin(A+B) - \sin A]^2 = (\cos B - 1)^2 + (-\sin B)^2

Expanding and simplifying using $\cos^2 A + \sin^2 A = 1$ yields $\cos(A+B) = \cos A \cos B - \sin A \sin B$ .

Double Angle Formulae:

\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}

The three forms of $\cos 2A$ are all useful in different contexts. Use $\cos 2A = 2\cos^2 A - 1$ When everything is in terms of $\cos$ And $\cos 2A = 1 - 2\sin^2 A$ when everything is in terms of $\sin$ .

Proof that $\sin 2A = 2\sin A \cos A$ .

\sin 2A = \sin(A + A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A

$\blacksquare$

Example: Express $\cos 3\theta$ in terms of $\cos\theta$ .

\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\theta \cos\theta \sin\theta

= 2\cos^3\theta - \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

Example: Prove 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

$\blacksquare$

Solving Trigonometric Equations

When solving trig equations in a given interval, always check for all solutions. The periodicity of Trig functions means there are multiple solutions.

Function equals zero separately. Dividing by $\cos x$ loses the solutions where $\cos x = 0$ .

Example: Solve $\sin 2x = \cos x$ for $0 \leq x < 2\pi$ .

2\sin x \cos x = \cos x

2\sin x \cos x - \cos x = 0

\cos x(2\sin x - 1) = 0

Either $\cos x = 0$ or $\sin x = \dfrac{1}{2}$ .

$\cos x = 0$ : $x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$ .

$\sin x = \dfrac{1}{2}$ : $x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$ .

Solutions: $x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$ .

Example: Solve $3\cos^2 x - \cos x - 2 = 0$ for $0 \leq x < 2\pi$ .

Let $u = \cos x$ . Then $3u^2 - u - 2 = 0$ .

$(3u + 2)(u - 1) = 0$

$u = -\dfrac{2}{3}$ or $u = 1$ .

$\cos x = 1$ : $x = 0$ .

$\cos x = -\dfrac{2}{3}$ : $x = \arccos\left(-\dfrac{2}{3}\right) \approx 2.301$ or $x = 2\pi - 2.301 \approx 3.982$ .

Example: Solve $\cos 2x = 1 - 3\sin x$ for $0 \leq x < 2\pi$ .

Use $\cos 2x = 1 - 2\sin^2 x$ :

1 - 2\sin^2 x = 1 - 3\sin x

2\sin^2 x - 3\sin x = 0

\sin x(2\sin x - 3) = 0

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

$2\sin x - 3 = 0$ : $\sin x = 1.5$ Which has no solution since $|\sin x| \le 1$ .

Solutions: $x = 0, \pi$ .

Example: Solve $2\sin^2 x + 3\cos x - 3 = 0$ for $0 \le x \lt 2\pi$ .

Use $\sin^2 x = 1 - \cos^2 x$ :

2(1 - \cos^2 x) + 3\cos x - 3 = 0

-2\cos^2 x + 3\cos x - 1 = 0

2\cos^2 x - 3\cos x + 1 = 0

$(2\cos x - 1)(\cos x - 1) = 0$

$\cos x = \frac{1}{2}$ : $x = \frac{\pi}{3}, \frac{5\pi}{3}$ .

$\cos x = 1$ : $x = 0$ .

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

The Sine and Cosine Rules

Sine Rule:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

Where $R$ is the circumradius of the triangle.

Use the sine rule when you know an angle and its opposite side, or two angles and one side.

Cosine Rule:

A^2 = b^2 + c^2 - 2bc\cos A

Use the cosine rule when you know all three sides (to find an angle) or two sides and the included Angle (to find the third side).

Area of a triangle:

A = \frac{1}{2}ab\sin C

Example: In triangle $ABC$ , $a = 8$ , $b = 5$ , $C = 60^\circ$ . Find $c$ .

C^2 = 64 + 25 - 2(8)(5)\cos 60° = 89 - 40 = 49

$c = 7$ .

Example: In triangle $ABC$ , $a = 7$ , $b = 9$ , $B = 55^\circ$ . Find angle $A$ .

By the sine rule:

\frac{\sin A}{7} = \frac{\sin 55°}{9}

\sin A = \frac{7\sin 55°}{9} \approx \frac{7 \times 0.8192}{9} \approx 0.6372

A = \arcsin(0.6372) \approx 39.6° \quad \mathrm{or \quad A = 180° - 39.6° = 140.4°

Both values are valid since $A + B = 39.6° + 55° = 94.6° < 180^\circ$ and $A + B = 140.4° + 55° = 195.4° > 180^\circ$ . Only $A \approx 39.6^\circ$ is valid (since the sum of angles Must be less than $180^\circ$ ).

Wave Function (R-Addition Formula)

An expression of the form $a\sin x + b\cos x$ can be written as $R\sin(x + \alpha)$ or $R\cos(x - \alpha)$ Where:

R = \sqrt{a^2 + b^2}, \quad \alpha = \arctan\left(\frac{b}{a}\right)

Derivation. We want $a\sin x + b\cos x = R\sin(x + \alpha) = R\sin x \cos\alpha + R\cos x \sin\alpha$ . Matching Coefficients: $R\cos\alpha = a$ and $R\sin\alpha = b$ . Squaring and adding: $R^2 = a^2 + b^2$ So $R = \sqrt{a^2 + b^2}$ . Dividing: $\tan\alpha = b/a$ .

Applications: The maximum value is $R$ and the minimum is $-R$ .

Example: Express $3\sin x + 4\cos x$ in the form $R\sin(x + \alpha)$ .

$R = \sqrt{9 + 16} = 5$

\alpha = \arctan\left(\frac{4}{3}\right)

3\sin x + 4\cos x = 5\sin\left(x + \arctan\left(\frac{4}{3}\right)\right)

Maximum value is $5$ Occurring when $\sin(x + \alpha) = 1$ .

Example: Find the maximum value of $2\sin\theta - \sqrt{3}\cos\theta$ and the smallest positive Value of $\theta$ at which it occurs.

R = \sqrt{4 + 3} = \sqrt{7}

2\sin\theta - \sqrt{3}\cos\theta = \sqrt{7}\sin(\theta + \alpha)

Where $\tan\alpha = \dfrac{-\sqrt{3}}{2}$ So $\alpha = -\arctan\left(\dfrac{\sqrt{3}}{2}\right) \approx -0.714 \mathrm{ rad$ .

Maximum value is $\sqrt{7}$ Occurring when $\sin(\theta + \alpha) = 1$ I.e., $\theta + \alpha = \dfrac{\pi}{2}$ So $\theta = \dfrac{\pi}{2} + \arctan\left(\dfrac{\sqrt{3}}{2}\right) \approx 2.285 \mathrm{ rad$ .

Example: Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$ and find its Maximum value.

R = \sqrt{25 + 144} = \sqrt{169} = 13

5\sin\theta - 12\cos\theta = 13\sin(\theta - \alpha)

Where $\tan\alpha = \dfrac{12}{5}$ So $\alpha = \arctan\!\left(\dfrac{12}{5}\right)$ .

Maximum value is $13$ .

Solving Equations Using the Wave Function

The wave function technique is especially powerful for solving equations of the form $a\sin x + b\cos x = c$ .

Example: Solve $3\sin x + 4\cos x = 5$ for $0 \le x \lt 2\pi$ .

Since $R = 5$ We have $5\sin(x + \alpha) = 5$ So $\sin(x + \alpha) = 1$ .

X + \alpha = \frac{\pi}{2} + 2k\pi

X = \frac{\pi}{2} - \alpha + 2k\pi = \frac{\pi}{2} - \arctan\frac{4}{3} + 2k\pi

For $k = 0$ : $x \approx 1.571 - 0.927 = 0.644$ rad. For $k = 1$ : $x \approx 0.644 + 2\pi \approx 6.927$ (outside range).

There is exactly one solution in $[0, 2\pi)$ .

Note that if $|c| > R = \sqrt{a^2 + b^2}$ The equation has no real solutions, because the maximum Of $a\sin x + b\cos x$ is $R$ .

Coordinate Geometry

The Straight Line

The equation of a straight line passing through $(x_1, y_1)$ with gradient $m$ :

Y - y_1 = m(x - x_1)

Gradient between two points:

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

Example: Find the equation of the perpendicular bisector of the line segment joining $A(2, 5)$ And $B(8, 3)$ .

Midpoint: $M = \left(\dfrac{2 + 8}{2}, \dfrac{5 + 3}{2}\right) = (5, 4)$ .

Gradient of AB: $m_{AB} = \dfrac{3 - 5}{8 - 2} = -\dfrac{1}{3}$ .

Gradient of perpendicular bisector: $m = 3$ .

Equation: $y - 4 = 3(x - 5)$ I.e., $y = 3x - 11$ .

Circles

The general equation of a circle with centre $(a, b)$ and radius $r$ :

(x - a)^2 + (y - b)^2 = r^2

Expanded form: $x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - r^2) = 0$ .

Given the expanded form $x^2 + y^2 + 2gx + 2fy + c = 0$ The centre is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$ (provided $g^2 + f^2 - c > 0$ ).

Example: Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ .

Complete the square:

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

$(x - 3)^2 + (y + 2)^2 = 25$

Centre $(3, -2)$ Radius $5$ .

Example: Find the equation of the circle with centre $(2, -3)$ that passes through $(5, 1)$ .

R^2 = (5 - 2)^2 + (1 + 3)^2 = 9 + 16 = 25

$(x - 2)^2 + (y + 3)^2 = 25$

Tangent to a Circle:

The tangent at a point on the circle is perpendicular to the radius at that point.

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

X_1 x + y_1 y = r^2

Example: Find the equation of the tangent to $(x - 2)^2 + (y + 1)^2 = 25$ at the point $(5, 3)$ .

Verify $(5, 3)$ lies on the circle: $(5-2)^2 + (3+1)^2 = 9 + 16 = 25$ . Confirmed.

Gradient of radius from $(2, -1)$ to $(5, 3)$ : $m_r = \dfrac{3 - (-1)}{5 - 2} = \dfrac{4}{3}$ .

Gradient of tangent: $m_t = -\dfrac{3}{4}$ .

Equation: $y - 3 = -\dfrac{3}{4}(x - 5)$ I.e., $4y - 12 = -3x + 15$ Or $3x + 4y - 27 = 0$ .

Example: Find the equation of the tangent to $x^2 + y^2 + 4x - 6y + 9 = 0$ at the point $(-2, 3)$ .

Complete the square: $(x + 2)^2 + (y - 3)^2 = 4$ . Centre $(-2, 3)$ Radius $2$ .

Since $(-2, 3)$ is the centre, not a point on the circle, we must check: $(-2+2)^2 + (3-3)^2 = 0 \ne 4$ . The point $(-2, 3)$ is inside the circle, so there is no tangent From this point to the circle. The point must lie on the circle for a tangent to exist.

Intersection of Line and Circle

Substitute the line equation into the circle equation and solve the resulting quadratic. The Discriminant of the resulting quadratic tells you:

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

Example: Find where the line $y = 2x + 1$ intersects the circle $x^2 + y^2 = 10$ .

X^2 + (2x + 1)^2 = 10

X^2 + 4x^2 + 4x + 1 = 10

5x^2 + 4x - 9 = 0

$(5x + 9)(x - 1) = 0$

$x = -\frac{9}{5} \mathrm{ or x = 1$

When $x = 1$ : $y = 3$ . When $x = -\dfrac{9}{5}$ : $y = -\dfrac{13}{5}$ .

Points of intersection: $(1, 3)$ and $\left(-\dfrac{9}{5}, -\dfrac{13}{5}\right)$ .

Distance from a Point to a Line

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

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

Proof. Let $P = (x_0, y_0)$ and let $Q$ be the foot of the perpendicular from $P$ to the line. The line through $P$ perpendicular to $ax + by + c = 0$ has direction $(a, b)$ So its parametric Form is $(x_0 + at, y_0 + bt)$ . Substituting into the line equation: $a(x_0 + at) + b(y_0 + bt) + c = 0$ Giving $t = -\frac{ax_0 + by_0 + c}{a^2 + b^2}$ . The distance Is $|t|\sqrt{a^2 + b^2} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ . $\blacksquare$

Example: Find the distance from $(3, 2)$ to the line $4x + 3y - 5 = 0$ .

D = \frac{|12 + 6 - 5|}{\sqrt{16 + 9}} = \frac{13}{5}

3D Coordinate Geometry (Advanced Higher)

Equations of Lines in 3D

A line through point $\mathbf{a} = (a_1, a_2, a_3)$ with direction vector $\mathbf{d} = (d_1, d_2, d_3)$ has parametric equations:

X = a_1 + td_1, \quad y = a_2 + td_2, \quad z = a_3 + td_3

In vector form: $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ .

Example: Find the equation of the line through $(1, 2, -1)$ in the direction $(3, -1, 4)$ .

\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}

Parametrically: $x = 1 + 3t$$y = 2 - t$$z = -1 + 4t$ .

Skew, Parallel, and Intersecting Lines

Two lines in 3D can be:

Parallel: direction vectors are scalar multiples
Intersecting: there exists a common point (same values of $s$ and $t$ satisfy all three coordinate equations simultaneously)
Skew: neither parallel nor intersecting

Example: Determine whether the following lines intersect:

$L_1$ : $\mathbf{r} = (1, 0, 2) + s(2, 1, -1)$

$L_2$ : $\mathbf{r} = (3, 1, -1) + t(1, -1, 3)$

Equate coordinates:

$1 + 2s = 3 + t \quad (1)$

$s = 1 - t \quad (2)$

$2 - s = -1 + 3t \quad (3)$

From (2): $s = 1 - t$ . Substitute into (1): $1 + 2(1 - t) = 3 + t$ So $3 - 2t = 3 + t$ Giving $t = 0$ , $s = 1$ .

Check (3): $2 - 1 = -1 + 0$ I.e., $1 = -1$ . This is false, so the lines are skew.

Distance from a Point to a Line in 3D

The shortest distance from point $P$ to the line through $A$ with direction $\mathbf{d}$ is:

D = \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|}

Example: Find the perpendicular distance from the point $(1, 2, 3)$ to the line $\mathbf{r} = (0, 1, -1) + t(2, -1, 3)$ .

$\overrightarrow{AP} = (1, 1, 4)$ , $\mathbf{d} = (2, -1, 3)$ .

\overrightarrow{AP} \times \mathbf{d} = \begin{pmatrix} 1 \cdot 3 - 4 \cdot (-1) \\ 4 \cdot 2 - 1 \cdot 3 \\ 1 \cdot (-1) - 1 \cdot 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \\ -3 \end{pmatrix}

|\overrightarrow{AP} \times \mathbf{d}| = \sqrt{49 + 25 + 9} = \sqrt{83}

|\mathbf{d}| = \sqrt{4 + 1 + 9} = \sqrt{14}

D = \frac{\sqrt{83}}{\sqrt{14}} = \sqrt{\frac{83}{14}} \approx 2.435

Distance Between Two Skew Lines

The shortest distance between two skew lines $\mathbf{r}_1 = \mathbf{a}_1 + t\mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{a}_2 + s\mathbf{d}_2$ is:

D = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}

Example: Find the shortest distance between the skew lines $\mathbf{r} = (1, 0, 0) + s(1, 2, 0)$ And $\mathbf{r} = (0, 0, 1) + t(0, 1, 1)$ .

$\mathbf{a}_2 - \mathbf{a}_1 = (-1, 0, 1)$ .

$\mathbf{d}_1 = (1, 2, 0)$ , $\mathbf{d}_2 = (0, 1, 1)$ .

\mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \cdot 1 - 0 \cdot 1 \\ 0 \cdot 0 - 1 \cdot 1 \\ 1 \cdot 1 - 2 \cdot 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}

(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) = (-1)(2) + 0(-1) + 1(1) = -1

|\mathbf{d}_1 \times \mathbf{d}_2| = \sqrt{4 + 1 + 1} = \sqrt{6}

D = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}

Intuition

Trigonometry is the mathematics of triangles and circles — it connects the angles of a triangle to the lengths of its sides through the sine, cosine, and tangent ratios. The unit circle is the master key: it extends trigonometric functions beyond right-angled triangles to any angle, revealing the periodic, wave-like nature of sine and cosine. Trigonometric identities are like algebraic simplifications — they let you rewrite complex expressions in simpler forms. The practical applications are everywhere: GPS uses trigonometry to calculate positions, audio engineers use it to analyse sound waves, and architects use it to design structures.

Worked Examples

See the examples integrated throughout the sections above.

Common Pitfalls

Degrees vs radians: Always check which units are being used. Calculus requires radians. If a question gives angles in degrees, convert before differentiating or integrating.
Forgetting to check the domain: When solving $\cos x = -\dfrac{2}{3}$ There are two solutions in $[0, 2\pi)$ : one in the second quadrant and one in the third quadrant.
Sign errors in the wave function: When writing $a\sin x + b\cos x = R\sin(x + \alpha)$ ensure $\alpha$ has the correct sign. The quadrant of $\alpha$ depends on the signs of $a$ and $b$ .
Incorrectly completing the square for circles: Remember to add the constant terms to both sides. For $x^2 + y^2 - 6x + 4y - 12 = 0$ You add 9 and 4 to both sides.
Assuming lines in 3D always intersect: Always check all three coordinates when testing for intersection. Even if two coordinates match, the third may not.
Dividing by zero in trig equations: When you factor and divide by $\cos x$ , $\sin x$ Or $\tan x$ You lose solutions. Always consider the case where the factor equals zero separately.
Using the wrong form of $\cos 2A$ : All three forms are equivalent, but using the wrong one for the given context makes the algebra much harder.
Confusing the ambiguous case of the sine rule: When $\sin A = k$ where $0 < k < 1$ There are two possible angles ( $A$ and $180° - A$ ). Both may or may not be valid in the triangle. Always check the sum of angles.
Forgetting that $R$ in the wave function is always positive: $R = \sqrt{a^2 + b^2}$ is defined as the positive square root. The maximum of $a\sin x + b\cos x$ is $R$ and the minimum is $-R$ .

Practice Questions

Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$ and find its maximum value.
Solve $\cos 2x = 1 - 3\sin x$ for $0 \leq x < 2\pi$ .
Find the equation of the circle with centre $(2, -3)$ that passes through $(5, 1)$ .
Find the equation of the tangent to $x^2 + y^2 + 4x - 6y + 9 = 0$ at the point $(-2, 3)$ .
A sector of a circle of radius 6 cm has an area of $24\pi \mathrm{ cm^2$ . Find the perimeter of the sector.
Prove that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .
Determine whether the lines $\mathbf{r} = (1, 2, 0) + s(1, -1, 2)$ and $\mathbf{r} = (3, 0, 4) + t(2, 1, -1)$ intersect, are parallel, or are skew.
Find the minimum value of $3\cos x + 4\sin x$ and the smallest positive value of $x$ at which it occurs.
Find the perpendicular distance from the point $(1, 2, 3)$ to the line $\mathbf{r} = (0, 1, -1) + t(2, -1, 3)$ .
In triangle $ABC$ , $a = 7$ , $b = 9$ , $B = 55^\circ$ . Find angle $A$ (there may be two solutions).
Solve $2\sin^2 x + 3\cos x - 3 = 0$ for $0 \le x \lt 2\pi$ .
Find the shortest distance between the skew lines $\mathbf{r} = (1, 0, 0) + s(1, 2, 0)$ and $\mathbf{r} = (0, 0, 1) + t(0, 1, 1)$ .
Find the angle between the lines $\mathbf{r} = (0, 0, 0) + s(1, 2, -1)$ and $\mathbf{r} = (1, 1, 0) + t(2, -1, 3)$ .
Find the area of triangle $ABC$ given $a = 10$ , $b = 8$ , $c = 6$ .
The line $y = mx + 7$ is tangent to the circle $x^2 + y^2 - 4x + 2y - 20 = 0$ . Find the possible values of $m$ .
Express $\cos 4\theta$ in terms of $\cos\theta$ using double angle formulae.

flowchart TD
    A[2_Trigonometry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to geometry and trigonometry, including key theorems, methods, and problem-solving approaches.

Key concepts include:

sine, cosine, and tangent functions
trigonometric identities
solving trigonometric equations
the sine and cosine rules
radian measure and arc length

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Cross-References

Algebra and Functions — Trigonometric functions are transcendental functions that extend the concept of function beyond polynomials.
Calculus — Differentiation and integration of trigonometric functions are essential techniques in calculus.
Vectors — Trigonometry is used to resolve vectors into components and find angles between vectors.

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