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Pre-Calculus – Study Guide

Functions and Their Graphs

A function is relationship between two variables such that each independent variable has only one dependent variable.

Function Notation

$y=f(x)$

Where $f(x)$ is the value of the function, y is the dependent variable, and x is the independent variable.

To find the function value given the value of the function, plug in the value of the function given for x.

Example: Let $f(x)=x^{2}+5x+2$ , find the function values of $f(2)$ .

To solve this we will substitute x = 2 into our function.
$f(2)=2^{2}+5\cdot 2+2=4+10+2=16$

The Vertical Line Test

You can tell if an equation is a function by performing the vertical line test. Look at the graph of the equation. Imagine moving a vertical line back and forth across the graph (You can use your pencil or the edge of a ruler for this.). If there is any point at which the vertical line passes through the graph of the equation more than once, then it is not a function.

Domain and Range

The domain of a function is the set of all possible inputs, or independent variables, for the function.

The range of a function is the set of all possible outputs, or dependent variables, for the function.

Domain and range can be found algebraically or by graphing.

Example: Find the domain and range for the function $f(x)=\frac{3}{x}$

To solve for the domain algebraically we need to consider what x values are possible for this function. The x is in the denominator which means some values of x will not be possible. If x = 0 then the function would be undefined because 0 cannot be in the denominator. Therefore, x = 0 is not a possible value for the function. This is the only value for x that is not possible so the domain is $(-\inf,\ 0)(0,\ \inf)$ .
To solve for the range we must consider all of the y values that are possible for the function. Once again we must take into consideration that the x is in the denominator. Therefore, there is not a possible way to every achieve y = 0 because 3 divided by anything will never by 0. The range is then $(-\inf,\ 0)(0,\ \inf)$ .

Now that we have solved the domain and range algebraically we can verify our answers using a graphical approach. The function $f(x)=\frac{3}{x}$ can also be written as $y=\frac{3}{x}$ . If we graph this equation we get a graph that looks like:

As we can see from the graph the domain and range are indeed what we found algebraically. The function does not exist at x = 0 or y = 0.

Increasing and Decreasing Functions

A function is increasing on an interval if when analyzing two points $x_{1}<x_{2}$ .
A function is decreasing on an interval if when analyzing two points $x_{1}>x_{2}$ .
A function is constant on an interval if when analyzing two points $x_{1}=x_{2}$ .

Example: For what intervals is the function $f(x)=x^{2}$ increasing, decreasing, or constant.

To solve this we should first graph the equation.

The function is decreasing until it gets to the vertex at $(0, 0)$ . The function is then increasing after the vertex. Therefore, the function is decreasing from $(-\inf,\ 0)$ and increasing from $(0,\ \inf)$ .

Even and Odd Functions

An even function is a function that is symmetric in respect to the y-axis. An odd function is a function that is symmetric with respect to the origin.

Example:

Even Function Odd Function

Zeros of a Function

The zeros of a function are the x-values for which $f(x)=0$ . To find the zeros of a function you can use factoring or the quadratic equation.

Example: Find the zeros of the function $f(x)=2x^{2}+5x+2$ .

First see if the function can factor. The function can factor into $(x+2)(2x+1)=0$ . Set both sets of parenthesis equal to zero and solve for x.
$(x+2)=0$
$x=-2$

$(2x+1)=0$
$2x=-1$
$x=-\frac{1}{2}$

The zeros are $(-2,\ 0)$ and $(-\frac{1}{2},\ 0)$ .

Average Rate of Change

The average rate of change for two points in the slope of the line through these two points.

$Average\ rate\ of\ change\ =\frac{change\ in\ y}{change\ in\ x}$

Example: Find the rate of change for $f(x)=x^{2}+2x+2$ for $x1=0\ and\ x2=3$

First find the y-value for each x-value given. This is found by plugging in the x-value into the function.
$f(0)=0^{2}+2\cdot 0+2=2$
$f(3)=3^{2}+2\cdot 3+2=17$

Now we can use these values to plug into the average rate of change equation.
$Average\ rate\ of\ change\ =\frac{17-2}{3-0}=\frac{15}{3}=5$

Inverse Functions

An inverse function occurs when the coordinate pairs are interchanged. An inverse function is denoted by $f^{-1}(x)$ . To find the inverse of a function simply undo what is done in the function.

Example: Find the inverse of the function $y=\frac{5}{x}$ .
To find the inverse we must undo what is happening in this function. In the function there is division so the division must be undone. The opposite of division is multiplication; therefore our inverse function must be multiplication. $y=5x$

Transformations to Functions

Transformations occur when changes are made to the parent graph.

Shifting graphs can occur in the vertical horizontal direction.

Shift	Result
Vertical shift upward “a” units	$h(x)=f(x)+c$
vertical shift downward “a” units	$h(x)=f(x)-c$
Horizontal shift left “a” units	$h(x)=f(x+c)$
Horizontal shift right “a” units	$h(x)=f(x-c)$

Graphs can be reflected across the x and y-axis.

Reflection	Result
Reflection in the x-axis	$h(x)=-f(x)$
Reflection in the y-axis	$h(x)=f(-x)$

Example: Graph the function $f(x)=-(x-5)^{2}+3$

First start this problem by identifying the parent graph of this function. The parent function is $y=x^2$ . The parent function is a parabola:

Now look at the shifting and reflecting that has been applied to the parent function. The negative in front of the function indicates a reflection in the x-axis. The parabola will open downwards. The $(x-5)$ indicates a horizontal shift to the right 5 units. The +3 added to the parent function indicates a vertical shift upwards 3 units. After making all of these changes to the parent function it looks like this:

Conics

Circle	$(x-h)^{2}+(y-k)^{2}=r^{2}$
	Vertical Axis	Horizontal Axis
Parabola	$(x-h)^{2}=4p(y-k)$	$(y-k)^{2}=4p(x-h)$
Ellipse	$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$	$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
Hyperbola	$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$	$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

* In the table the point (h, k) is the vertex of the parabola and the center of the other conics listed.

Circle

The standard equation for a circle with center (h, k) and radius r is $(x-h)^{2}+(y-k)^{2}=r^{2}$

Example: Find the equation of a circle with center (3, 2) and radius 4.

The center indicates $h=3$ and $k=2$ . The value for r is 4. Plug these values into the equation.

$(x-3)^{2}+(y-2)^{2}=16$

Parabola

The directrix and focus of a parabola are equidistant from the vertex. The directrix is a fixed line on the side of the parabola where there are no points. The focus is a point inside of the parabola.

The standard equation for a parabola with a vertex (h, k) is
$(x-h)^{2}=4p(y-k)$ vertical axis, directrix is $y=k-p$ , focus is $(h,\ k+p)$
$(y-k)^{2}=4p(x-h)$ horizontal axis, directrix is $x=h-p$ , focus is $(h+p,\ k)$

Example: Find the standard form of the equation of the parabola with vertex (1, 2) and focus (1, 5).

First determine if this parabola is in the vertical or horizontal direction. The focus of the parabola is further up the y-axis by 3 units than the vertex. This means the parabola opens upward in the vertical direction. The standard equation $(x-h)^{2}=4p(y-k)$ will be used. The values for hand k are already known because the vertex is defined as (h, k). This means h = 1 and k = 2. The only thing left to find is p. To do this look at the focus because k + p must equal 5. This means p = 3.

Plug in these values to the standard equation.
$(x-1)^{2}=12(y-2)$

Ellipse

The standard equation for an ellipse with center (h, k) is
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ when the major axis is horizontal
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ when the major axis is vertical

The major axis is 2a and the minor axis is 2b.

The distance from the center to the focus is defined as c. The relationship between a, b, and c is $c^{2}=a^{2}-b^{2}$ .

Example: Find the equation of the ellipse with a center at (3, 2) and foci at (5, 2) and (1, 2). The major axis length is 10.

First determine if the major axis is horizontal or vertical. The major axis is horizontal because the foci and center have the same y-value but different x-values. The standard equation $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ will be used. The values of h and k are known because the center of the elllipse is (h, k). Therefore, h = 3, and k = 2. The value for c be found by looking at the foci. The distance between the center of the ellipse and the foci is c. This means that c = 2. The major axis, which is 2a, is given as 10 so a = 5. To find b use $c^{2}=a^{2}-b^{2}$ .

$c^{2}=a^{2}-b^{2}$
$2^{2}=5^{2}-b^{2}$
$4=25-b^{2}$
$-25=-b^{2}$
$21=b^{2}$
$b=\sqrt{21}$

Now that all of the values have been obtained they can be plugged into the standard equation.
$\frac{(x-3)^2}{25}+\frac{(y-2)^2}{21}=1$

Hyperbola

The standard equation for a hyperbola with a center (h, k) is
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ when the transverse axis is horizontal
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$ when the transverse axis is vertical

The vertices are a units from the center. The foci are c units from the center. The relationship between a, b, and c is $c^{2}=a^{2}+b^{2}$ .

Example: Find the equation of the hyperbola with foci (-3, 1) and (5, 1) and vertices (-1, 1) and (3, 1).

First determine if the hyperbola’s transverse axis is horizontal or vertical. The y-value remains constant while the x-value changes for the vertices and foci so the transverse axis is horizontal. The standard equation $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ will be used. The center of the hyperbola is half way between the vertices therefore it must be (1, 1). This means that h = 1 and k = 1. The value of c is the distance between the foci and the center which means c = 4. The value of a is the distance between the vertices and the center which means a = 2. The value of b can be found by using:
$c^{2}=a^{2}+b^{2}$ .
$4^{2}=2^{2}+b^{2}$
$16=4+b^2$
$12=b^2$
$b=\sqrt{12}$

Plug the values into the standard equation.
$\frac{(x-1)^2}{4}+\frac{(y-1)^2}{12}=1$

Logarithmic and Exponential Functions

Exponential Functions

The exponential function is $f(x)=a^x$ where a is greater than 0 and cannot equal 1.
These exponential functions will always have a horizontal asymptote along the x-axis and one y-intercept. The y-intercept will always be (0, 1).

Example: Graph $y=2^x$ and find the domain, range, intercepts, and asymptotes of the function.

Start by graphing the equation. Using a simple t-chart is a quick way to graph these equations.

By looking at the graph we can see this function only has one intercept, a y-intercept. In addition, we can see that the function is approaching the x-axis but will never cross it. The y-intercept can be found by plugging a 0 in for x. This gives $f(x)=2^0$ which is 1. Therefore, the y-intercept will be (0, 1). The horizontal asymptote is the x-axis. The domain is $(-\inf,\ \inf)$ because there is no values of x that will make the function untrue. The range is $(0,\ \inf)$ because of the horizontal asymptote at the x-axis.

Logarithmic Functions

The logarithmic function is $f(x)=log_{a}x$ . This is read as “log base a of x.” Another way to write this same function is $x=a^y$ . Logarithmic and exponential functions are inverse functions. This means that the graph of a logarithmic function will be the reflection of the exponential function across the line $y-x$ . The logarithmic functions will always have a vertical asymptote along the y-axis and a x-intercept. The x-intercept will always be (1, 0).

Example: Graph the function $y=log_{2}x$ and find the domain, range, asymptote, and intercepts.

Start by graphing the equation. Using a simple t-chart is a quick way to graph these equations.

By looking at the graph we can see this function only has one intercept, a x-intercept. In addition, we can see that the function is approaching the y-axis but will never cross it. The x-intercept can be found by plugging a 0 in for y. This give $f(0)=log_{2}x$ or $x=2^0$ which is 1. Therefore, the x-intercept will be (1, 0). The horizontal asymptote is the y-axis. The domain is $(0,\ \inf)$ because there is a vertical asymptote at the y-axis. The range is $(-\inf,\ \inf)$ because of the horizontal asymptote at the x-axis.

Properties of Logarithms

1. $log_{a}1=0$
2. $log_{a}a=1$
3. $log_{a}a^{2}=x\ and\ a^{log_{a}x}=x$
4. $If\ log_{a}x=log_{a}y\ then\ x=y$
5. $log_{a}x=\frac{log_{b}x}{log_{b}a},\ a\neq 1\ and\ b\neq 1$
6. $log_{a}(uv)=log_{a}u+log_{a}v$
7. $log_{a}\frac{u}{v}=log_{a}u-log_{a}v$
8. $log_{a}u^{n}=n\cdot log_{a}u$

Example: Solve $log(3x+7)=log(x+1)$

These two logarithms have the same base so we can apply rule number 4 from above. This means that $3x+7=x+1$ . Combining like terms and solving for x will give x = -3.

Properties of Natural Logarithms

1. $ln(1)=0$
2. $ln(e)=1$
3. $ln(e^{x})=x\ and\ e^{ln(x)}=x$
4. $If\ ln(x)=ln(y),\ then\ x=y$
5. $log_{a}x=\frac{ln(x)}{ln(a)}$
6. $ln(uv)=ln(u)+ln(v)$
7. $ln(\frac{u}{v})=ln(u)-ln(v)$
8. $ln(u^{n})=n\cdot ln(u)$

Example: Simplify $\frac{ln(1)}{4}$ .

To simplify this we first must recognize that $ln(1)=0$ as state in rule 1 above. This now leaves us with $\frac{0}{4}$ which is equal to 0.

Exponential Growth and Decay

Exponential functions can be used to model the growth or decay of a population. The exponential growth model is written as $y=ae^{bx}$ where $b>0$ . The exponential decay model is written as $y=ae^{-bx}$ where $b>0$ .

Example: Assume that flies exhibit exponential growth. At day 1 there are 200 flies and at day 3 there are 400 flies. how many flies are there at day 6.

To solve this we know we will use the model $y=ae^{bx}$ . In the problem we are given two values for x and y but not a value for a or b. We can write two equations with two unknowns and solve for a and b.

$200=ae^{b}\ and\ 400=ae^{3b}$
Solve the first equation in terms of a and substitute it into the second
$a=\frac{200}{e^b}$
$400=(\frac{200}{e^b})(e^{3b})$
Divide each side by 200
$2=\frac{e^{3b}}{e^b}$
Simplify the exponents
$2=e^{2b}$
Take the natural log of each side and simplify
$ln(2)=ln(e^{2b})$
$ln(2)=2b$
$\frac{ln(2)}{2}=b$
$b=0.347$
Substitute this value of b into the first equation and solve for a.
$200=ae^{0.347}$
$200=a(1.415)$
$a=141.34$

Plug the values of a and b into the model
$y=141.34e^{0.347x}$

Plug the value of 6 in for x to find the number of flies at day 6.
$y=141.34e^{0.347(6)}$
$y=1133.61$
Round the number up to the nearest whole number
$y=1134\ flies$

Trigonometry in Triangles

Law of Sines

The Law of Sines is used with oblique triangles. An oblique triangle is a triangle that does not contain a right angle. The Law of Sines is used to solve oblique triangles when given two angles and any side or two sides and an angle opposite one of the sides.

Example: A triangle has angles A = 35° and B = 75° and side length a = 20 feet. What are the remaining sides and angle?

The problem gives 2 angles so we can find the third angle easily. The three agnles of a triangle must add up to 180° therefore angle C will be 70°(35°+75°+70°=180°). To find sides b and c use the Law of Sines.

$\frac{sin(35)}{20}=\frac{sin(75)}{b}=\frac{sin(70)}{c}$

By using cross multiplication the lengths of b and c are found to be b = 33.68 feet and c = 32.77 feet.

Law of Cosines

The Law of Cosines is also used with oblique triangles. The Law of Cosines is used to solve oblique triangles when given three sides or two sides and their included angles.

Example: A triangle has sides b = 10 feet and c = 20 feet and angle A = 100°. Find the remaining side and angles?

The problem gives two sides and an angle so we can use the Law of Cosines to solve this. We are given A, a, and b so we will use $a^{2}=b^{2}+c^{2}-2bccos(A)$ . Plugging in these values we will be able to solve for a.
$a^{2}=10^{2}+20^{2}-2\cdot 10\cdot 20\cos(100)$
$a=23.86$
Now we can use this value and our other side lengths b and c to find angle B.
$b^{2}=a^{2}+c^{2}-2ac\cdot cos(B)$
$10^{2}=23.86^{2}+20^{2}-2\cdot 23.86\cdot 20\cdot cos(B)$
$B=24.35\degree$
We have two angles so we can easily find the third angle, C, Angles of a triangle must add up to 180° so angle C is 55.65° (100°+24.35°+55.65°=180°).

Area of an Oblique Triangle

The area of an oblique triangle can be found by manipulating the Law of Sines. The height of an oblique triangle is $h=b\cdot sin(A)$ . Plugging this into the area equation of a triangle gives:

$Area=\frac{1}{2}bc\cdot sin(A)=\frac{1}{2}ab\cdot sin(C)=\frac{1}{2}ac\cdot sin(B)$ , for two sides with the included angle in between.

Example: Find the area of the triangle with side lengths 4 feet and 5 feet and an included angle of 55°.

Plug these values into the area equation and solve.
$Area=\frac{1}{2}\cdot 4\cdot 5\cdot sin(55)$
$Area=8.19\ feet^{2}$

Trigonometric Functions

Converting between Degrees and Radians

To convert degrees to radians, multiply by $\frac{\pi\ rad}{180\degree}$
To convert radians to degrees, multiply by $\frac{180\degree}{\pi\ rad}$

Example: Convert 145° to radians

$145\degree\cdot\frac{\pi\ rad}{180\degree}=\frac{29\pi}{36}$

Unit Circle

The unit circle is given by the equation $x^{2}+y^{2}=1$ .

Trigonometric functions correspond to points on the unit circle when t is a real number an (x, y) are a point on the circle.
$sin(t)=y$
$cos(t)=x$
$tan(t)=\frac{y}{x}$
$csc(t)=\frac{1}{y}$
$sec(t)=\frac{1}{x}$
$cot(t)=\frac{x}{y}$

Example: Evaluate the 6 trigonometric functions for $t=\frac{n}{3}$ .

First use the unit circle to find the corresponding (x, y) point to $\frac{\pi}{3}$ . It is $(\frac{1}{2},\ \frac{\sqrt{3}}{2})$ . now use these x and y values to plug into the trigonometric functions.
$sin(t)=y=\frac{\sqrt{3}}{2}$
$cos(t)=x=\frac{1}{2}$
$tan(t)=\frac{y}{x}\cdot\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
$csc(t)=\frac{1}{y}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2\sqrt{3}}{3}$
$sec(t)=\frac{1}{x}=\frac{1}{\frac{1}{2}}=2$
$cot(t)=\frac{x}{y}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}$

Graph Sine and Cosine Functions

The sine and function has a domain of all real numbers and a range of [-1, 1]. The sine curve is symmetric with respect to the origin. Sine is an odd function and has a period of 2Π.

$y=a\cdot sin(bx-c)$

The cosine function has a domain of all real numbers and a range of [-1, 1]. The cosine curve is symmetric with respect to the y-axis. Cosine is an even function and has a period of 2Π.

$y=a\cdot cos(bx-c)$

The amplitude of the function is the distance between the maximum and minimum of the function. It is vertical stretching of the graph.

$Amplitude\ =\ |a|$

The period of the function is the horizontal stretching of the graph.

$Period\ =\ \frac{2\pi}{b}$

The phase shift of the function is the horizontal translation of the curve. Finding the phase shift will indicate where the left and right end points have been moved to.

$Phase\ Shift\ =\frac{c}{b}$
$bx-c=0$
$bx-c=2\pi$

Example: Sketch the graph of $y=2sin(x-\frac{\pi}{2})$ .

Start by identifying the amplitude, period, and phase shift. Looking at the equation we can see that $a=2$ , $b=1$ , and $c=\frac{\pi}{2}$ . The amplitude is 2 and the period is $2\pi$ . The phase shift is $\frac{\pi}{2}$ . The end points are normally at (0, 0) and (2Π, 0) but they will now become $(\frac{\pi}{2},\ 0)$ and $(\frac{5\pi}{2},\ 0)$ . Use this information to transform the normal sine graph.

Graph of Other Trigonometric Functions

The tangent function has a domain of all really numbers except $x\neq\frac{\pi}{2}+n\pi$ and the range is all real numbers. The tangent graph is symmetric with respect to the origin and has a period of Π. There are vertical asymptotes at $x=\frac{\pi}{2}+n\pi$ .

$y=a\cdot tan(bx-c)$

The graph of the cotangent function is similar to the tangent function. It has a domain of all real numbers except x ≠ nΠ and the range is all real numbers. The period is Π and there is are vertical asymptotes at x = nΠ.

The graph of cosecant is the reciprocal of the sine function. It has a domain of all real numbers except x ≠ nΠ and the range is $(-\inf,\ 1][1,\ \inf)$ . The graph is symmetric across the origin and there are vertical asymptotes at x = nΠ. The period is 2Π.

The secant graph is the reciprocal of the cosine graph. It has a domain of all real numbers except $x\neq\frac{\pi}{2}+n\pi$ and the range is $(-\inf,\ 1][1,\ \inf)$ . There are vertical asymptotes at $x=\frac{\pi}{2}+n\pi$ and it is symmetric around the y-axis. The period is 2Π.

Find Values of Trigonometric Functions

To find the value of the trigonometric function you must use your knowledge of the unit circle and reference angle. Reference angles are the angles used when the value is not from 0-90°. It is also important to know which functions are positive and negative in each quadrant. A good way to remember this is by saying “All Students Take Calculus.” Each word represents the trig functions (A = all of them, S = sine, T = tangent, C = cosine). Label each quadrant with a letter starting in quadrant I and going in the counterclockwise direction as shown in the left. Therefore in quadrant I all of the trig functions are positive. In quadrant II, sine is positive and cosine and tangent are negative. In quadrant III, tangent is positive and sine and cosine are negative. In quadrant IV, cosine is positive and sine and tangent are negative.

Example: Find the exact value of $sin(\frac{2\pi}{3})$ .

Start by identifying what quadrant this angle is in. $\frac{2\pi}{3}$ is in quadrant II. From the saying “All Students Take Calculus” determine that the sine is positive in this quadrant. Now figure out the reference angle. For quadrant II, the reference angle will be from the x-axis. This means our reference angle will be $\frac{\pi}{3}$ . The sine of $\frac{\pi}{3}$ is $\frac{\sqrt{3}}{2}$ . This means that the sine of $\frac{2\pi}{3}$ is $\frac{\sqrt{3}}{2}$ .

Inverse Trigonometric Functions

Inverse trigonometric functions can occur for the periodic function when the domain is restricted. This will cause the function to pass the Horizontal Line Test which is required for the function to have an inverse function.

Function	Domain	Range
$y=arcsin(x)$	$-1\leq x\leq 1$	$-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$
$y=arccos(x)$	$-1\leq x\leq 1$	$0\leq y\leq\pi$
$y=arctan(x)$	$-\infty\leq x\leq\infty$	$-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$

Example: Find the exact value of $arcsin(\frac{1}{2})$ .

To find the value think about this problem as “what is the angle whose sine is $\frac{1}{2}$ ?” Recognize that the sine of $\frac{\pi}{6}$ is $\frac{1}{2}$ . The range of arcsin restricts the possible answer to $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$ so we only need to look at the first and fourth quadrants. Sine is positive in the first quadrant and negative in the fourth quadrant. Therefore the answer will be in the first quadrant so it will be $\frac{\pi}{6}$ .

Fundamental Trigonometric Identities

Reciprocal Identities:

$sin(u)=\frac{1}{csc(u)}$	$cos(u)=\frac{1}{sec(u)}$	$tan(u)=\frac{1}{cot(u)}$
$csc(u)=\frac{1}{sin(u)}$	$sec(u)=\frac{1}{cos(u)}$	$cot(u)=\frac{1}{tan(u)}$

Quotient Identities:

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

$cot(u)=\frac{cos(u)}{sin(u)}$

Pythagorean Identities:

$sin^{2}(u)+cos^{2}(u)=1$

$1+tan^{2}(u)=sec^{2}(u)$

$1+cot^{2}(u)=csc^{2}(u)$

Cofunction Identities:

$sin(\frac{\pi}{2}-u)=cos(u)$	$cos(\frac{\pi}{2}-u)=sin(u)$
$tan(\frac{\pi}{2}-u)=cot(u)$	$cot(\frac{\pi}{2}-u)=tan(u)$
$sec(\frac{\pi}{2}-u)=csc(u)$	$csc(\frac{\pi}{2}-u)=sec(u)$

Even/Odd Identities

$sin(-u)=-sin(u)$	$cos(-u)=cos(u)$	$tan(-u)=-tan(u)$
$csc(-u)=-csc(u)$	$sec(-u)=sec(u)$	$cot(-u)=-cot(u)$

Example: Verify the identity $\frac{csc^{2}(x)}{cot(x)}=csc(x)sec(x)$ .

To verify this identity we want to transform the left side into the right side by using the trigonometric identities above.
$\frac{csc^{2}(x)}{cot(x)}=csc(x)sec(x)$
First recognize that the $csc^{2}(x)$ can be changed into $1+cot^{2}(x)$ using one of the Pythagorean identities.
$\frac{1+cot^{2}(x)}{cot(x)}=csc(x)sec(x)$
Next separate the function into two separate fractions.
$\frac{1}{cot(x)}+\frac{cot^{2}(x)}{cot(x)}=csc(x)sec(x)$
Change the $\frac{1}{cot(x)}$ to $tan(x)$ using the reciprocal identity. Divide $cot^{2}(x)$ by $cot(x)$ .
$tan(x)+cot(x)=csc(x)sec(x)$
Use the quotient identities to put $tan(x)$ and $cot(x)$ in terms of $sin(x)$ and $cos(x)$ .
$\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)}=csc(x)sec(x)$
Find the common denominator, $sin(x)cos(x)$ , and add the fractions together.
$\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}=csc(x)sec(x)$
Recognize that $sin^{2}(x)+cos^{2}(x)$ can be changed to 1 using one of the Pythagorean identities.
$\frac{1}{sin(x)cos(x)}=csc(x)sec(x)$
Break up the fraction.
$\frac{1}{sin(x)}\cdot\frac{1}{cos(x)}=csc(x)sec(x)$
Use the reciprocal identities.
$csc(x)sec(x)=csc(x)sec(x)$

Sum and Difference Formulas

$sin(u+v)=sin(u)cos(v)+cos(u)sin(v)$
$sin(u-v)=sin(u)cos(v)-cos(u)sin(v)$
$cos(u+v)=cos(u)cos(v)-sin(u)sin(v)$
$cos(u-v)=cos(u)cos(v)+sin(u)sin(v)$
$tan(u+v)=\frac{tan(u)+tan(v)}{1-tan(u)tan(v)}$
$tan(u-v)=\frac{tan(u)-tan(v)}{1+tan(u)tan(v)}$

Example: Find the exact value of sin(105°).

The sin(105°) is not a value found on the unit circle but we can break up 105° into values that are on the unit circle. We can write this problem as sin(60°+45°). Then we can use the sum formula of sin to solve this.
$sin(60\degree +45\degree)=sin(60\degree)cos(45\degree)+cos(60\degree)sin(45\degree)$
$sin(60\degree+45\degree)=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}$
$sin(60\degree+45\degree)=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}$
$sin(60\degree+45\degree)=\frac{\sqrt{6}+\sqrt{2}}{4}$

Other Formulas

Double-Angle Formulas

$sin(2u)=2sin(u)cos(u)$
$tan(2u)=\frac{2tan(u)}{1-tan^{2}(u)}$
$cos(2u)=cos^{2}(y)-sin^{2}(u)=2cos^{2}(u)-1=1-2sin^{2}(u)$

Power-Reducing Formulas

$sin^{2}(u)=\frac{1-cos(2u)}{2}$
$cos^{2}(u)=\frac{1+cos(2u)}{2}$
$tan^{2}(u)=\frac{1-cos(2u)}{1+cos(2u)}$

Half-Angle Formulas

$sin(\frac{u}{2})=\pm\sqrt{\frac{1-cos(u)}{2}}$
$cos(\frac{u}{2})=\pm\sqrt{\frac{1+cos(u)}{2}}$
$tan(\frac{u}{2})=\frac{1-cos(u)}{sin(u)}=\frac{sin(u)}{1+cos(u)}$

Sum-to-Product Formulas

$sin(u)+sin(v)=2sin(\frac{u+v}{2})cos(\frac{u-v}{2})$
$sin(u)-sin(v)=2cos(\frac{u+v}{2})sin(\frac{u-v}{2})$
$cos(u)+cos(v)=2cos(\frac{u+v}{2})cos(\frac{u-v}{2})$
$cos(u)-cos(v)=-2sin(\frac{u+v}{2})sin(\frac{u-v}{2})$

Example: Use the sum-to-product formula to find the exact value of the expression sin(60°) + sin(30°).

Plug the sine values into the appropriate sum-to-product formula and simplify.
$sin(60\degree)+sin(30\degree)=2sin(\frac{60\degree+30\degree}{2})cos(\frac{60\degree-30\degree}{2})$
$sin(60\degree)+sin(30\degree)=2sin(45\degree)cos(15\degree)$
The exact value of sin(45°) can easily be written but cos(15°) cannot therefore a sum and difference formula needs to be used.
$sin(60\degree)+sin(30\degree)=2sin(45\degree)cos(45\degree -30\degree)$
$sin(60\degree)+sin(30\degree)=2sin(45\degree)\cdot [cos(45\degree)cos(30\degree)+sin(45\degree)sin(30\degree)]$
Input exact values and simplify.
$sin(60\degree)+sin(30\degree)=2\cdot\frac{\sqrt{2}}{2}\cdot[\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}]$
$sin(60\degree)+sin(30\degree)=2\cdot\frac{\sqrt{2}}{2}\cdot [\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}]$
$sin(60\degree)+sin(30\degree)=2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{6}+\sqrt{2}}{4}$
$sin(60\degree)+sin(30\degree)=\frac{\sqrt{12}+2}{4}$
$sin(60\degree)+sin(30\degree)=\frac{2\sqrt{3}+2}{4}$
$sin(60\degree)+sin(30\degree)=\frac{\sqrt{3}+1}{2}$

Sequence and Series

Summation Notation

Sequences are represented by summation notation.

$\sum_{i=1}^{n} a_{i} = a_{1}+a_{2}+a_{3}+...+a_{n}$

$i=1$ is the value to start the summation on and $n$ is the value to end on.

Example: Find the sum of $\sum_{i=1}^{4} i^{2}$ .

Plug in the values to the summation.
$\sum_{i=1}^{4} i^{2}=1^{2}+2^{2}+3^{2}+4^{2}$

Simplify and add together.
$\sum_{i=1}^{4} i^{2}=1+4+9+16=30$

Recursive Sequences

Another way to write sequences is to use the recursion formula. A recursion formula defines the nth term of a sequence as a function of a previous term. For this formula you must be given a starting value.

Example: Find the first 4 terms of $a_{n}=a_{n-1}+5$ and $a_{1}=7$ .

We are given the initial term of $a_{1}=7$ . To find the next term, $a_{2}$ , we will use this value because $a_{2-1}=a_1$ .

$a_{2}=a_{1}+5$
$a_{2}=7+5=12$

Now find $a_{3}$ through the same process.
$a_{3}=a_{3-1}+5=a_{2}+5=12+5=17$

Now find $a_{4}=a_{4-1}+5=a_{3}+5=17+5=22$

The first four terms of the sequence are 7, 12, 17, and 22.

Arithmetic Sequence

An arithmetic sequence is a sequence where the next term is different from the term before it by a constant amount. D is the common difference (the number added) between the two terms.
$a_{n}=a_{1}+(n-1)d$

To find the sum of the first n terms of an arithmetic sequence use
$S_{n}=\frac{n}{2}(a_{1}+a_{n})$
where $a_{1}$ is the first term and $a_{n}$ is the nth term.

Example: Find the sum of the first 20 terms of the arithmetic sequence: 4, 10, 16, 11, …

To use the sum formula we need to know $a_{1}$ and $a_{20}$ . We already know $a_{1}$ so we must find $a_{20}$ . Use the formula for an arithmetic sequence by pluggin in what we know. The value for n is 20 and d is 6 because the number added between 4 and 10 is 6.
$a_{20}=4+(20-1)6$
$a_{20}=118$
Now plug in all of the values to the sum formula and solve.
$S_{20}=\frac{20}{2}(4+118)=1220$

Geometric Sequence

A geometric sequence is a sequence where the next term is found by multiplying the previous term by some constant number. The number multiplied each time is called the common ratio, r.
$a_{n}=a_{1}r^{n-1}$

To find the sum of the first n terms of the geometric sequence use
$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

Polar Coordinates and Complex Numbers

Converting between Polar and Rectangular

Polar coordinates are represented by the form (r, θ) where r is the distance from the pole to the point and θ is the directed angle. To convert polar coordinates to rectangular use:

$x=r\cdot cos(\theta)$
$y=r\cdot sin(\theta)$

To convert rectangular coordinates to polar use:

$r=\sqrt{x^{2}+y^{2}}$
$\theta =tan^{-1}(\frac{y}{x})\ when\ x>0$
$\theta =tan^{-1}(\frac{y}{x})+\pi\ when\ x<0$

Example: Find the rectangular coordinates of the polar coordinates $(-2,\ \frac{4\pi}{3})$ .

First using the form of a polar coordinate recognize that $r=-2$ and $\theta=\frac{4\pi}{3}$ . Plug these values into the equations and solve.

$x=-2cos(\frac{4\pi}{3})$
Using the reference angle of $\frac{\pi}{3}$ of the $cos(\frac{4\pi}{3})=-\frac{1}{2}$ .
$x=-2\cdot-\frac{1}{2}=1$
$y=-2sin(\frac{4\pi}{3})$
Using the reference angle of $\frac{\pi}{3}$ of the $sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2}$ .
$y=-2\cdot -\frac{\sqrt{3}}{2}=\sqrt{3}$

The rectangular coordinate is $(1,\ \sqrt{3})$ .

Example: Convert the rectangular equation $(x-5)^{2}+y^{2}=25$ to polar.

To convert a rectangular equation insert $r\cdot cos(\theta)$ for x and $r\cdot sin(\theta)$ for y then simplify.
$(r\ cos\theta-5)^{2}+(r\ sin\theta)^{2}=25$
$(r^{2}cos\theta^{2}-10r\ cos\theta+25)+r^{2}sin^{2}\theta=25$
Subtract 25 from both sides.
$(r^{2}cos\theta^{2}-10r\ cos\theta)+r^{2}sin^{2}\theta=0$
Add $10r\ cos\theta$ to both sides.
$(r^{2}cos\theta^{2})+r^{2}sin^{2}\theta=10r\ cos\theta$
Factor out an $r^{2}$ on the left side.
$r^{2}(cos^{2}\theta+sin^{2}\theta)=10r\ cos\theta$
Use the trigonometric identity $cos^{2}\theat+sin^{2}\theta=1$
$r^{2}=10r\ cos\theta$
Divide each side by r.
$r=10cos\theta$

Polar Form of Complex Numbers

A complex number is written in the form $z=a+bi$ . The polar form is written as $x=r(cos\theta+isin\theta)$ where $r=\sqrt{a^{2}+b^{2}}$ and $\theta=tan^{-1}\frac{b}{a}$ when a > 0 and $\theta^{-1}\frac{b}{a}+\pi$ when a < 0.

Example: Express the complex number 4 + 4i in polar form.

First identify that a = 4 and b = 4 from the form of a complex number. Find r using
$r=\sqrt{a^{2}+b^{2}}$
$r=\sqrt{4^{2}+4^{2}}$
$r=\sqrt{16+16}$
$r=\sqrt{32}$
$r=4\sqrt{2}$
Find $\theta$ using $\theta=tan^{-1}\frac{b}{a}$
$\theta=tan^{-1}\frac{4}{4}$
$\theta=tan^{-1}1$
$\theta=\frac{\pi}{4} Plugging these values we found the polar form of a complex number gives$ 4\sqrt{2}(cos\frac{\pi}{ $}+isin\frac{\pi}{4})$ .

De Moivre’s Theory

$z^{n}=[r(cos\theta+isin\theta]^{n}=r^{n}(cos\ n\theta+isin\ n\theta)$

Example: Find $(4+4i)^5$ .

The first step is to express the $4+4i$ in polar form. This has already been done in the previous example.
$[4\sqrt{2}(cos\frac{\pi}{4}+isin\frac{\pi}{4})]^6$
Using De Moivre’s Theory we can redistribute the exponent to each term.
$4\sqrt{2}^{6}(cos6(\frac{\pi}{4})+isin6(\frac{\pi}{4}))$
$32768(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})$
$32768(0-1i)$
$-32768i$

References

Blitzer, Robert. Precalculus. 4th ed. Upper Saddle River: Pearson, 2010. Print.

Carter, John and Gilbert, Cuevas. Precalculus. Columbus: The McGraw-Hill Companies, 2011. Print.

Hostetler, Robert and Larson, Ron. Precalculus. Boston: Houghton Mifflin Company, 2007. Print.

Picture Credits:

Law of Sines
http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php

Law of Cosines
http://www.msdgeometry.com/joomla/index.php?option=com_content&view=article&id=164:la
w-of-sines-and-law-of-cosines&catid=51:chapter-8&Itemid=68

Unit Circle
http://www.mathpeer.com/index.php?act=trigonometry&trig=radian_measure_and_the_circular_
functions

“All Students Take Calculus”
http://en.wikipedia.org/wiki/All_Students_Take_Calculus