Meditaliano IMAT Prep

Lesson 3: Trigonometry, Geometry & Analytic Geometry

Lesson 3: Trigonometry & Geometry

In this lesson, we explore two cornerstones of mathematics: Trigonometry, the study of the relationships between angles and side lengths of triangles, and Plane Geometry, which deals with figures on a flat surface. We will build a strong foundation with trigonometric ratios, master the all-important Unit Circle, and then apply these concepts to geometric figures, with a special focus on circles.

Part 1: Trigonometry

Trigonometry is essential for modeling periodic phenomena and solving geometric problems. It begins with the simple right-angled triangle but extends to all angles via the unit circle.

Angles: Degrees and Radians

Angles can be measured in two main units:

  • Degrees (°): A full circle is divided into 360 degrees.
  • Radians (rad): A full circle is $2\pi$ radians.

Angle Conversion Formulas

The key relationship is $180^\circ = \pi$ radians.

  • Radians = Degrees $\times \frac{\pi}{180^\circ}$
  • Degrees = Radians $\times \frac{180^\circ}{\pi}$

Trigonometric Ratios and Identities

Ratios in a Right-Angled Triangle

For any acute angle $\theta$ in a right-angled triangle, we define the three basic trigonometric ratios using the lengths of the sides: Opposite (O), Adjacent (A), and Hypotenuse (H).

θ Adjacent (A) Opposite (O) Hypotenuse (H)

SOH CAH TOA

  • $\sin(\theta) = \frac{\color{#ef4444}{\text{Opposite}}}{\color{#16a34a}{\text{Hypotenuse}}}$
  • $\cos(\theta) = \frac{\color{#7e22ce}{\text{Adjacent}}}{\color{#16a34a}{\text{Hypotenuse}}}$
  • $\tan(\theta) = \frac{\color{#ef4444}{\text{Opposite}}}{\color{#7e22ce}{\text{Adjacent}}}$

Fundamental Identities

Trigonometric identities are equations that are true for all values of the variables. They are essential for simplifying expressions and solving equations.

Pythagorean Identities

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

Ratio Identity

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

IMAT Challenge

Challenge an IMAT Question!

Official Paper: 2024 - Q54

In a right triangle, let $a$ and $b$ represent the legs and $c$ the hypotenuse. If $\alpha$ is the angle opposite $a$, which of the following relations is true? *Note: Conceptually similar to Q41.*

The Unit Circle

The unit circle (radius 1, center at origin) extends trigonometry to all angles. For any point $(x,y)$ on the circle at angle $\theta$: $x = \cos(\theta)$ and $y = \sin(\theta)$.

30° $(\frac{\sqrt{3}}{2}, \frac{1}{2})$ 45° $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ 60° $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ 90° (0,1) 120° (-$\frac{1}{2}, \frac{\sqrt{3}}{2}$) 180° (-1,0) 270° (0,-1) 0° (1,0)

The Unit Circle, showing coordinates $(\cos\theta, \sin\theta)$ for key angles.

Graphs of Trigonometric Functions

Plotting the values from the unit circle reveals the periodic (repeating) nature of trigonometric functions.

$y=\sin(x)$

$\pi/2$$\pi$$3\pi/2$$2\pi$1-1

$y=\cos(x)$

$\pi/2$$\pi$$3\pi/2$$2\pi$1-1

$y = \tan(x)$

$\\frac{\\pi}{2}$ $\\frac{3\\pi}{2}$ $\\pi$ $2\\pi$

Advanced Trigonometric Formulas & Phase Shifts

General Sinusoidal Transformation Form

$$ y = A \sin(B(x - C)) + D \quad \text{or} \quad y = A \cos(B(x - C)) + D $$

Where:

  • $|A|$: Amplitude (vertical stretch/compression).
  • $B$: Determines the Period ($T = \frac{2\pi}{B}$ for sine/cosine, and $T = \frac{\pi}{B}$ for tangent).
  • $C$: Phase Shift (horizontal translation; shift right if $C > 0$, left if $C < 0$).
  • $D$: Vertical Shift (vertical translation; shifts the midline of the wave).

Trigonometric Phase Shift & Symmetry Identities

These identities allow you to shift the input angle or reflect it, which is crucial for solving equations and identifying wave shapes.

Co-function Phase Shifts ($\pi/2$ or $90^\circ$)

Shifting by $\pi/2$ converts sine to cosine and vice-versa:

  • $$ \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) $$
  • $$ \cos\left(\theta + \frac{\pi}{2}\right) = -\sin(\theta) $$
  • $$ \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta) $$
  • $$ \cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta) $$

Periodicity & Half-Period Shifts ($\pi$ or $180^\circ$)

Shifting by $\pi$ changes the sign of sine and cosine:

  • $$ \sin(\theta \pm \pi) = -\sin(\theta) $$
  • $$ \cos(\theta \pm \pi) = -\cos(\theta) $$
  • $$ \tan(\theta \pm \pi) = \tan(\theta) \quad (\text{since period is } \pi) $$

Even and Odd (Symmetry) Identities

Cosine is an **even** function (symmetric about y-axis), while sine and tangent are **odd** functions (symmetric about origin):

  • $$ \sin(-\theta) = -\sin(\theta) $$
  • $$ \cos(-\theta) = \cos(\theta) $$
  • $$ \tan(-\theta) = -\tan(\theta) $$

Detailed Mathematical Derivations

1. Derivation of the Pythagorean Identity: $\sin^2\theta + \cos^2\theta = 1$

Consider a right-angled triangle with legs $a$ (opposite), $b$ (adjacent), and hypotenuse $c$. By the Pythagorean Theorem:

$$ a^2 + b^2 = c^2 $$

Dividing both sides of the equation by $c^2$:

$$ \frac{a^2}{c^2} + \frac{b^2}{c^2} = 1 \quad \Rightarrow \quad \left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1 $$

Since by definition $\sin\theta = \frac{a}{c}$ and $\cos\theta = \frac{b}{c}$, we substitute these in to get:

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

2. Derivation of Angle Addition Formulas: $\cos(A - B)$

We can derive the angle addition/subtraction formulas using unit vectors. Let $\vec{u}$ and $\vec{v}$ be two unit vectors on the Cartesian plane making angles $A$ and $B$ with the positive x-axis respectively.

Their coordinates are $\vec{u} = (\cos A, \sin A)$ and $\vec{v} = (\cos B, \sin B)$.

The angle between the two vectors is $A - B$. We can calculate their dot product $\vec{u} \cdot \vec{v}$ in two ways:

  • Using coordinates: $\vec{u} \cdot \vec{v} = \cos A\cos B + \sin A\sin B$
  • Using geometric definition: $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos(A - B) = (1)(1)\cos(A - B) = \cos(A - B)$

Equating the two expressions yields the angle subtraction formula for cosine:

$$ \cos(A - B) = \cos A\cos B + \sin A\sin B $$

To find $\cos(A + B)$, replace $B$ with $-B$ and use the symmetry identities ($\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$):

$$ \cos(A + B) = \cos A\cos(-B) + \sin A\sin(-B) = \cos A\cos B - \sin A\sin B $$

3. Derivation of Double-Angle Formulas

Double-angle formulas are directly derived by setting $B = A = \theta$ in the sum formulas:

  • • Sine Double-Angle:

    $\sin(A + B) = \sin A\cos B + \cos A\sin B \quad \xrightarrow{B=A=\theta} \quad \sin(2\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta$

    $$ \sin(2\theta) = 2\sin\theta\cos\theta $$

  • • Cosine Double-Angle:

    $\cos(A + B) = \cos A\cos B - \sin A\sin B \quad \xrightarrow{B=A=\theta} \quad \cos(2\theta) = \cos^2\theta - \sin^2\theta$

    Using $\sin^2\theta + \cos^2\theta = 1$, we can substitute $\sin^2\theta = 1 - \cos^2\theta$ to get:

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

    Or substitute $\cos^2\theta = 1 - \sin^2\theta$ to get:

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

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

4. Derivation of Half-Angle Formulas

Half-angle formulas are derived by starting with the cosine double-angle formulas and solving for the single-angle terms.

  • • Sine Half-Angle:

    Start with $\cos(2\alpha) = 1 - 2\sin^2\alpha$. Let $\alpha = \frac{\theta}{2}$, then $2\alpha = \theta$:

    $$ \cos\theta = 1 - 2\sin^2\left(\frac{\theta}{2}\right) \quad \Rightarrow \quad 2\sin^2\left(\frac{\theta}{2}\right) = 1 - \cos\theta $$

    $$ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{2} $$

  • • Cosine Half-Angle:

    Start with $\cos(2\alpha) = 2\cos^2\alpha - 1$. Let $\alpha = \frac{\theta}{2}$, then $2\alpha = \theta$:

    $$ \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \quad \Rightarrow \quad 2\cos^2\left(\frac{\theta}{2}\right) = 1 + \cos\theta $$

    $$ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos\theta}{2} $$

Solving Triangles

BACacb

The Law of Sines

Used in any triangle to relate the sides to the sines of their opposite angles. Useful when you know an angle and its opposite side.

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

The Law of Cosines

Used in any triangle to find a side length if you know two sides and the angle between them, or to find an angle if you know all three sides.

$c^2 = a^2 + b^2 - 2ab\cos(C)$

IMAT Challenge

Challenge an IMAT Question!

Official Paper: 2021 - Q56

PQR and PRS are right-angled triangles. $\angle PRS = \angle PQR = 90^{\circ}$. RS=5 cm. Area of PQR is $14\text{ cm}^2$ and $\tan(PRQ)=1/7$. What is the value of $ \cos(RPS) $?
IMAT Challenge

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Official Paper: 2016 - Q53

A right-angled triangle has an area of $18\text{ cm}^2$. One of the two shorter sides is twice the length of the other one. What is the length of the hypotenuse?

Solving Trigonometric Equations & Inequalities

These are solved by finding base solutions (usually in $0 \le \theta < 2\pi$) and then generalizing by adding the period $n \cdot 2\pi$ or $n \cdot \pi$.

Example Equation: Solve $\sin(\theta) = \frac{1}{2}$

1. Find base solutions (Unit Circle): $\sin$ is positive in Q1 and Q2. The angles are $\theta = \frac{\pi}{6}$ (Q1) and $\theta = \frac{5\pi}{6}$ (Q2).

2. Generalize: Add the period $n \cdot 2\pi$ to each solution: $\theta = \frac{\pi}{6} + 2n\pi$ or $\theta = \frac{5\pi}{6} + 2n\pi$

Example Inequality: Solve $\cos(\theta) > \frac{1}{2}$ for $0 \le \theta < 2\pi$

1. Solve for equality: $\cos(\theta) = \frac{1}{2}$ at $\theta = \frac{\pi}{3}$ (Q1) and $\theta = \frac{5\pi}{3}$ (Q4).

2. Test intervals on Unit Circle: We want where the x-coordinate is greater than $\frac{1}{2}$. This is the region to the right of $x = 1/2$.

3. Solution: $0 \le \theta < \frac{\pi}{3}$ or $\frac{5\pi}{3} < \theta < 2\pi$

Part 2: Plane & Analytic Geometry

Triangle Properties

  • Congruence: Two triangles are congruent if they have the same size and shape. Conditions: SSS, SAS, ASA.
  • Similarity: Two triangles are similar if they have the same shape but possibly different sizes (corresponding angles are equal). Conditions: AA, SSS similarity, SAS similarity.

Lines in the Plane

Internal and External Division

The coordinates of a point P that divides the line segment from $A(x_1, y_1)$ to $B(x_2, y_2)$ in the ratio $m:n$ are:

Internal Division ($m:n$): $P = \left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n} \right)$

External Division ($m:n$): $P = \left( \frac{-nx_1 + mx_2}{m-n}, \frac{-ny_1 + my_2}{m-n} \right)$

Finding the Equation of a Line

  • Slope-Intercept Form: $y = mx + c$ (where $m$ is slope, $c$ is y-intercept).
  • Point-Slope Form: $y - y_1 = m(x - x_1)$ (given point $(x_1, y_1)$ and slope $m$).
  • Two-Point Form: $y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1)$ (given points $(x_1, y_1)$ and $(x_2, y_2)$).

Parallel and Perpendicular Conditions

For two lines $L_1$ with slope $m_1$ and $L_2$ with slope $m_2$:

Parallel Condition: $L_1 \parallel L_2 \iff m_1 = m_2$

Perpendicular Condition: $L_1 \perp L_2 \iff m_1 \cdot m_2 = -1$ (or $m_2 = -\frac{1}{m_1}$)

IMAT Challenge

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Official Paper: 2022 - Q53

Find the area enclosed by the three lines: $$y=0,\quad y=2x-4,\quad y=11-x$$
IMAT Challenge

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Official Paper: 2017 - Q56

What is the gradient of the straight line passing through the points with coordinates $(2, -3)$ and $(-1, 6)$?
IMAT Challenge

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Official Paper: 2014 - Q55

What is the equation of the straight line which passes through $(-6, 2)$ and is perpendicular to $4y+3x=8$?
IMAT Challenge

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Official Paper: 2012 - Q79

The line L has equation $y=2x-1$. Four of the following five points are the same distance from L. Which one is at a different distance?
IMAT Challenge

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Official Paper: 2011 - Q79

The graph shows the line joining A $(2,1)$ and B $(6,3)$, and its perpendicular bisector. Which is the equation of the bisector?

Circles

Equation of a Circle

The standard equation of a circle with center $(h,k)$ and radius $r$ is:

$(x-h)^2 + (y-k)^2 = r^2$

Circle Theorems

  • A tangent is perpendicular to the radius at the point of tangency.
  • Inscribed Angle Theorem: An angle $\theta$ inscribed in a circle is half of the central angle $2\theta$ that subtends the same arc.
  • An angle inscribed in a semicircle is always a right angle ($90^\circ$).
  • Tangent-Chord Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
$2\theta$ $\theta$ $\alpha$ $\alpha$

Intersections and Regions

Circle and Line Intersections

To find the intersection of a circle (e.g., $x^2 + y^2 = r^2$) and a line (e.g., $y = mx + c$):

  1. Substitute the line equation into the circle equation. (e.g., $x^2 + (mx+c)^2 = r^2$)
  2. This creates a quadratic equation in $x$. Solve it.
  3. The discriminant ($\Delta$) of this quadratic tells you the number of intersection points:
    • $\Delta > 0$: Two points (secant line)
    • $\Delta = 0$: One point (tangent line)
    • $\Delta < 0$: No points (line misses the circle)

Locus and Regions

A locus is a set of points satisfying a geometric condition. An inequality describes a region.

  • Locus Example: "The locus of points equidistant from A and B" is the perpendicular bisector of the segment AB.
  • Region Example: $y > 2x + 1$ is the half-plane above the line $y = 2x + 1$.
  • Region Example: $(x-h)^2 + (y-k)^2 < r^2$ describes the region *inside* the circle.
IMAT Challenge

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Official Paper: 2024 - Q52

Let $\phi$ be the acute angle formed between the tangent at point A to a circle and one of its chords, AB. Considering any point D on the larger of the arcs AB, denoted by $\theta$ as the angle ADB, what relationship exists between the angles $\phi$ and $\theta$?
IMAT Challenge

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Official Paper: 2016 - Q56

The straight-line graph given by $$x/4+y/6=1$$ intersects the x-axis at A $(a,0)$ and the y-axis at B $(0,b)$. A circle passes through A and B and has a diameter AB. What are the coordinates of the centre?
IMAT Challenge

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Official Paper: 2013 - Q56

Which of the following is the equation of the circle with centre $(-1.5, 0.5)$ and radius 3?
IMAT Challenge

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Official Paper: 2012 - Q78

ABCDE is a regular pentagon. R is a rotation mapping A$\to$B, B$\to$C... S is reflection in y-axis. Which sequence (left to right) does NOT leave vertex D in the same position?

Part 3: Advanced Function Topics (Review)

Quadratic Functions: Maximum and Minimum

The vertex of a parabola $y = ax^2 + bx + c$ is the key to finding its maximum or minimum value. The vertex occurs at $x = -\frac{b}{2a}$.

  • If $a > 0$ (parabola opens up), the vertex is a minimum.
  • If $a < 0$ (parabola opens down), the vertex is a maximum.

Example: Find the minimum value of $y = 2x^2 - 8x + 5$.

1. $a = 2, b = -8$. The parabola opens up, so it has a minimum.

2. The vertex is at $x = -\frac{-8}{2(2)} = \frac{8}{4} = 2$.

3. The minimum value is the y-value at the vertex: $y = 2(2)^2 - 8(2) + 5 = 2(4) - 16 + 5 = 8 - 16 + 5 = -3$.

Absolute Value Functions (Advanced)

The graph of $y = |f(x)|$ is created by reflecting any part of $y=f(x)$ that is below the x-axis ($y<0$) to be above the x-axis.

To solve equations or inequalities with absolute values, you must consider two cases:

Case 1: The expression inside is non-negative. $|A| \to A$

Case 2: The expression inside is negative. $|A| \to -A$

Example: Solve $|x - 3| < 2$

This is a "sandwich" inequality: "The distance from $x$ to 3 is less than 2."

$-2 < x - 3 < 2$

Add 3 to all parts: $1 < x < 5$

Exponential & Logarithmic Equations & Inequalities

This is a review from Lesson 2, as it's a critical skill.

Exponential

Strategy: Get a common base. $a^M = a^N \implies M = N$.

Equation: Solve $2^{x+1} = 8$

$2^{x+1} = 2^3 \implies x+1 = 3 \implies x = 2$

Logarithmic

Strategy 1: Combine logs. $\log_a M = \log_a N \implies M = N$.

Strategy 2: Convert to exponential form. $\log_a M = N \implies a^N = M$.

CRITICAL: Always check your domain! The argument of a log must be positive ($M > 0$).

Inequality: Solve $\log_2(x-1) < 3$

1. Domain: $x-1 > 0 \implies x > 1$.

2. Solve: Since base is 2 ($> 1$), keep the sign. $x-1 < 2^3 \implies x-1 < 8 \implies x < 9$.

3. Intersection: We need $x > 1$ AND $x < 9$.

Final Solution: $1 < x < 9$

Part 4: Practice Problems