# Page 2 - Math: Additional Topics Study Guide for the SAT Exam

## More Geometry Concepts

### Area, Surface, and Volume

Any closed two-dimensional figure will possess an area. The **area** of a figure is simply the amount of space enclosed by the figure. Because areas are two-dimensional, each area measurement will have **units raised to the second power**.

It is worthwhile to commit the following area formulas to memory:

**Area of a circle**: \(\pi \cdot r^2\), where \(r\) is the radius of the circle

**Area of a triangle**: \(\frac{1}{2}b \cdot h\), where \(b\) is the base, and \(h\) is the height

**Area of a quadrilateral**: \(l \cdot w\), where \(l\) is the length (or base), and \(w\) is the width (or height)

**Area of trapezoid**: \(\frac{1}{2} (b_1 + b_2) \cdot h\), where \(b_1\) and \(b_2\) are the two bases, and \(h\) is the height

In cases where you are asked to find the **area of a figure composed of many different shapes**, break the figure apart into its constituent shapes, calculate the area of each of the shapes, then add each of the areas together to find the total area.

Three-dimensional figures possess both a surface area and a volume. The **surface area** is the sum of the areas of each of the figure’s surfaces. In order to find the surface area of a figure, first, calculate the area of each surface composing the figure and then add the areas together.

The **volume** of a three-dimensional figure is the amount of space enclosed by the figure. In the same way that an area is composed of two dimensions and represented by units squared, a volume is composed of three dimensions and represented by **units cubed**.

It is worthwhile to commit the following volume formulas to memory:

**Volume of a sphere**: \(\frac{4}{3} \cdot \pi \cdot r^3\), where *\(r\) is the radius of the sphere

**Volume of a cylinder**: \(\pi \cdot r^2 \cdot h\), where \(r\) is the radius of the base, and \(h\) is the height

**Volume of a rectangular prism**: \(l \cdot w \cdot h\), where \(l\) is the length, \(w\) is the width, and \(h\) is the height

### Coordinate Geometry

The coordinate plane combines two perpendicular coordinate **axes** to visually represent geometric figures. The **horizontal axis** is commonly used to represent the independent variable (usually \(x\)), and the **vertical axis** is commonly used to represent the dependent variable (usually \(y\), or \(f(x)\)).

The **origin** is the intersection of the \(x\)-axis and the \(y\)-axis and is located at point \((0, 0)\). The intersection also creates four quadrants:

**Quadrant I** contains positive \(x\) and \(y\) values.

**Quadrant II** contains negative \(x\) and positive \(y\) values.

**Quadrant III** contains negative \(x\) and \(y\) values.

**Quadrant IV** contains positive \(x\) and negative \(y\) values.

Points on the \(xy\) coordinate plane are represented as follows: \((x, y)\), where the \(x\) value is the positive or negative distance from \(x = 0\), and the \(y\) value is the positive or negative distance from \(y = 0\).

## Trigonometry

A certain amount of knowledge of right triangle trigonometry and radian measurements will be necessary for some of the SAT exam math questions. Refresh your memory by reviewing these concepts. You will not need to find the value of trigonometric functions that require the use of a calculator.

### Trigonometric Functions

The trigonometric functions **sine**, **cosine**, and **tangent** relate the side lengths of a right triangle to its angles.

In any **right triangle**, there are two legs, two acute angles, a hypotenuse, and a right angle. The trigonometric functions are defined as follows:

Where \(\theta\) is one of the acute angles, \(o\) is the side opposite the angle \(\theta\), \(a\) is the side adjacent to the angle \(\theta\), and \(h\) is the hypotenuse.

Given an angle and any side length, any right triangle can be solved by using the appropriate trigonometric function.

Consider the right triangle with side lengths \(3\), \(4\), and \(5\). Because all side lengths are known, any angle can be found by solving any trig. function for \(\theta\). This is accomplished by evaluating the inverse trig. function:

\(\sin \theta = \frac{3}{5}\), where \(\theta\) is the angle to be solved for, \(3\) is a side length, and \(5\) is the hypotenuse. In order to solve, \(\theta\) must be first liberated from the function; to do so, evaluate the inverse, or \(\arcsin\) of both sides:

\[\arcsin (\sin \theta) = \arcsin (\frac{3}{5})\]\(\theta = \arcsin (\frac{3}{5})\), which can be evaluated using a scientific calculator

The same method can be used to evaluate for an unknown side length when given an angle measurement and a side length.

The expression **SOHCAHTOA** can be used to recall the trig. functions. **SOH** indicates opposite over hypotenuse, **CAH** indicates adjacent over hypotenuse, and **TOA** indicates opposite over adjacent. The first letter of each word part stands for sine, cosine, and tangent.

### Trigonometric Functions

In addition to the sine, cosine, and tangent functions, it is important to develop familiarity with the secant, cosecant, and cotangent functions in addition to the inverse of each of these.

The **secant function** is defined as \(\frac{1}{\cos\theta}\). Given that the cosine function is adjacent side divided by hypotenuse, the secant function is the reciprocal, or hypotenuse divided by adjacent side.

The **cosecant function** is defined as \(\frac{1}{\sin\theta}\). Given that the sine function is the opposite side divided by the hypotenuse, the cosecant function is the reciprocal, or hypotenuse divided by the opposite side.

The **cotangent function** is defined as \(\frac{1}{\tan\theta}\). Given that the tangent function is opposite side over adjacent side, the cotangent function is the reciprocal, or adjacent side divided by opposite side.

The **inverse of each trig. function** can be taken in order to liberate the angle measurement from the trig. function. For example, \(\arcsin {\sin\theta} = \theta\).

### Complementary Angle Relationship

A straight line measures \(180\) degrees. Consider line \(AB\), \(\overleftrightarrow{AB}\), with ray \(XY\), \(\overrightarrow{XY}\), emanating from point \(X\), which is between \(A\) and \(B\). If angle \(a\) (\(\angle a\)) and angle \(b\) (\(\angle b\)) are the angles on either side of ray \(XY\), then \(\angle a + \angle b = 180^{\circ}\).

Angles that combine to \(180\) degrees are known as **supplementary angles**.

In a similar fashion, angles that combine to \(90\) degrees are known as **complementary angles**.

Consider the following: \(\angle X = 3x + 2\) and \(\angle Y = x + 6\). The two angles are complementary. What is the value of \(x\)?

Because the angles are complementary, their sum is equal to \(90\) degrees. Set up an equation and solve for the unknown \(x\):

\[\angle X + \angle Y = 90^{\circ}\] \[3x + 2 + x + 6 = 90\] \[4x + 8 = 90\] \[4x = 82\] \[x = \frac{41}{2}\]### Working with Circles and Trigonometry

To better work with and understand the relationship between circles and trigonometry, it is highly recommended that the unit circle is explored.

The **unit circle** is a circle with a radius of \(1\) that connects the coordinate plane, right triangles, trigonometric functions, degree measurement, and radian measurements. When fully described, it shows a circle divided into a collection of right triangles of various side length measurements, with a vertex lying on the edge of a circle. The coordinates of each vertex along the circle represent cosine, and sine values that satisfy the Pythagorean Formula, \(x^2 + y^2 = 1\), or in trigonometric form, \((\cos x)^2 + (\sin y)^2 = 1\).

The vertices of these triangles also correspond with degree and radian measurements ranging from \(0\) to \(2\pi\) or \(360^{\circ}\). You are encouraged to develop familiarity with the unit circle independently, attempting to understand its design rather than memorizing its form.

## Complex Numbers

Finally, you may encounter some SAT exam questions that contain complex numbers. You may also need to do some arithmetic with these numbers. Go over these procedures and terms so you’ll be ready for this type of question.

### Definition and Standard Form

**Complex numbers** include either a real number, an imaginary number, or the combination of both, in the form of \(a + bi\), where \(a\) is the real portion, and \(bi\) is the complex portion, with \(i = \sqrt{-1}\).

### Addition and Subtraction

In order to add or subtract complex numbers, combine the **real portion** of the complex numbers and the **imaginary portion** of the complex numbers separately. Consider the following problem:

Begin by breaking the problem into the addition of the real portions:

\[3 + (-4) = -1\]Then combine the imaginary portion:

\[-4i + 5i = i\]Finally, combine the real and imaginary portions:

\[-1 + i\]### Multiplication

Multiplication of complex numbers entails the **distribution** of one complex number (both the real and imaginary portion) through the other complex number. Consider the following multiplication problem:

Apply the distributive property and multiply the first term through the second parenthesis:

\[3 \cdot (-1) + 3 \cdot (-5i) = -3 -15i\]Next distribute the imaginary portion through the second parenthesis:

\[2i \cdot (-1) + 2i \cdot (-5i) = -2i - 10i^2 = -2i - 10(-1) = -2i + 10\]Notice that the \(i^2\) was replaced with \(-1\), this is because \(i = \sqrt{-1}\) so \(i^2 = (\sqrt{-1})^2 = -1\).

Finally, combining all like terms:

\[-3 - 15i - 2i + 10 = 7 - 17i\]### Conjugate

The conjugate of a complex number is another complex number that is equivalent except for the sign of the complex portion. For example, the **complex conjugate** of the complex number \(6 + 2i\) is \(6 - 2i\).

Complex conjugates are used to rationalize complex expressions.

### Division

Dividing complex numbers entails utilizing the complex conjugate in order to rationalize the denominator. Consider the complex fraction:

\[\frac{2 - 2i}{3 + 5i}\]By multiplying numerator and denominator by the conjugate of the denominator, the denominator can be converted to a real number:

\[\frac{2 - 2i}{3 + 5i} \cdot \frac{3 - 5i}{3 - 5i} = \frac{6 - 10i - 6i - 10i^2}{9 - 15i + 15i - 25i^2} = \frac{16 - 16i}{34}\]