]> Trigonometry Notes

Trigonometry

From the Greek trigonometria, meaning "three angle measure" (the word first appears in print in the late 16^th century). Perhaps you can guess that this will have quite a bit to do with triangles…

Right Triangles

We'll consider right triangles, since they make things easier, but the following comments about similar triangles and ratios of side lengths are true for any kind of similar triangles.

Consider the following set of nested similar triangles:

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Since $Δ ABC ~ Δ ADE$ , $\frac{BC}{CA} = \frac{DE}{EA}$ . Similar statements can be made about sides of the other similar triangles in the diagram, as long as they correspond.

The Basic Trigonometric Ratios

This ratio of side lengths is a function of the angle θ; thus, function names were given to the various ratios of sides that can be constructed. To make the definitions easier, we first draw a standard right triangle:

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Now we can say

Sine: $sin θ = \frac{opposite}{hypotenuse}$ , Cosine: $cos θ = \frac{adjacent}{hypotenuse}$ , and Tangent: $tan θ = \frac{opposite}{adjacent}$ . Note that the symbol $sin$ is a function name, just like $\log$ or $\ln$ .

Also note that we've only defined three of the possible six ratios!

Reciprocal Trigonometric Ratios

The other possible ratios are just the reciprocals of the first three; hence, they are often called the reciprocal functions.

Cosecant: $csc θ = \frac{hypotenuse}{opposite}$

Secant: $sec θ = \frac{hypotenuse}{adjacent}$

Cotangent: $cot θ = \frac{adjacent}{opposite}$

Special Right Triangles

There are two very special right triangles, for which you must memorize these ratios. They are often referred to by the names "45-45-90" and "30-60-90" right triangles. Here are the patterns for these:

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Examples

[1.] Find all six trigonometric ratios of the angle θ.

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Note that the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.

$sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$ , $tan θ = \frac{5}{12}$ , $sec θ = \frac{13}{12}$ , $csc θ = \frac{13}{5}$ , $cot θ = \frac{12}{5}$ .

[2.] Find all six trigonometric ratios of the angle γ.

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Note that you'll first need to use the Pythagorean Theorem to find the length of the third side!

$4^{2} + x^{2} = 5^{2} ⇒ 16 + x^{2} = 25 ⇒ x^{2} = 9 ⇒ x = 3$ ; thus

$sin γ = \frac{3}{5}$ , $cos γ = \frac{4}{5}$ , $tan γ = \frac{3}{4}$ , $sec γ = \frac{5}{4}$ , $csc γ = \frac{5}{3}$ , $cot γ = \frac{4}{3}$

[3.] Find all six trigonometric ratios of the angle α.

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This time, you'll have to recognize the type of triangle in order to find the lengths of the missing sides. In particular, this is a 30-60-90, so the side adjacent to α is $7 \sqrt{3}$ and the hypotenuse is 14.

$sin α = \frac{7}{14} = \frac{1}{2}$ , $cos α = \frac{7 \sqrt{3}}{14} = \frac{\sqrt{3}}{2}$ , $tan α = \frac{7}{7 \sqrt{3}} = \frac{1}{\sqrt{3}}$ , $sec α = \frac{2}{\sqrt{3}}$ , $csc α = 2$ , $cot α = \sqrt{3}$ .

Points in the Plane

Pick a point in the plane…drop a perpendicular to the x-axis, then connect both ends to the origin. You've now made a triangle out of a point. Every point makes a triangle; thus every point makes an angle to which sine, cosine, et. al., may be applied!

The Rule

Most points make something that you would call a triangle.

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Note that the angle is always measured from the positive x-axis. Angles going counter-clockwise are positive (and referred to as being in standard position), and angles going clockwise are negative (yes, negative!). It is perfectly fine to talk about angles that are greater than 180° (or even greater than 360°!). More on the various peculiarities of angles later…

For a point $(x, y)$ that is r units from the origin, we have

$sin θ = \frac{y}{r}$ ; $cos θ = \frac{x}{r}$ ; $tan θ = \frac{y}{x}$ ; $sec θ = \frac{r}{x}$ ; $csc θ = \frac{r}{y}$ ; $cot θ = \frac{x}{y}$ .

The Exceptions

Okay…there are a few points that don't make something that you would call a triangle. Nonetheless, the definitions given above still hold! For those points that cause bad things (e.g., division by zero), the ratio is not defined.

Examples

[4.] Find the value of all six trigonometric functions for the standard position angle β that passes through the point $(- 15, 8)$ .

First of all, the distance to the origin is $\sqrt{{(- 15)}^{2} + 8^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17$ . That makes our ratios $sin β = \frac{8}{17}$ , $cos β = - \frac{15}{17}$ , $tan β = - \frac{8}{15}$ , $sec β = - \frac{17}{15}$ , $csc β = \frac{17}{8}$ , $cot β = - \frac{15}{8}$ .

[5.] Find the value of all six trigonometric functions for the standard position angle φ that passes through the point $(- 20, - 21)$ .

Once again, let's find r first: $\sqrt{{(- 20)}^{2} + {(- 21)}^{2}} = \sqrt{400 + 441} = \sqrt{841} = 29$ . This makes $sin φ = - \frac{21}{29}$ , $cos φ = - \frac{20}{29}$ , $tan φ = \frac{21}{20}$ , $sec φ = - \frac{29}{21}$ , $csc φ = - \frac{29}{20}$ , $cot φ = \frac{20}{21}$ .

The Unit Circle

The Idea

Every point in the plane makes a triangle; every point makes an angle to which sine, cosine, et. al., may be applied…however, there are a lot of points that make the same angle (and a lot of angles that pass through the same point)! Thus, we reduce and simplify: let's just consider the points that are one unit away from the origin. This creates the unit circle, and simplifies the trigonometric ratio definitions given above:

$sin θ = y$ ; $cos θ = x$ ; $tan θ = \frac{y}{x}$ ; $sec θ = \frac{1}{x}$ ; $csc θ = \frac{1}{y}$ ; $cot θ = \frac{x}{y}$

That takes care of the multiple points with the same angle…however, that leaves the multiple angle issue. Once an angle measure exceeds 360°, it points in the same direction as another angle: specifically, the angles with measures ${(360 + x)}^{°}$ and $x^{°}$ point in the same direction. More generally, if the difference between two angles is an integer multiple of 360°, then those angles point in the same direction.

A Better Angle Measure

This is the time to fix something…it turns out (for various, difficult-to-explain reasons) that degrees are not really the best way to measure angles. An alternate way to measure angles is with arc length along the unit circle.

One radian is the angle which subtends an arc of length one unit on the unit circle. Here's a picture:

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Radians are dimensionless; thus, there is no symbol to indicate radians. If you are writing about degrees, you need to write the degree symbol!

An angle of 180° is the same as π radians. Use this fact if you want/need to convert between the two methods.

Radians make things easier…for example, arc length: if you use degrees, then finding the length of an intercepted (subtended) arc requires that you write out a proportion. In radians, the length of an arc is just $r θ$ . There are more reasons…especially when you get to calculus. For now, just memorize a few special (hint, hint) radian measures…you'll see these again in PreCalculus.

The Ideal Points

There are a few nice points on the unit circle that we all have to memorize…happily, they mostly come from special right triangles!

Angle (Degrees)	Angle (Radians)	Coordinates
0°	0	$(1, 0)$
30°	$\frac{π}{6}$	$(\frac{\sqrt{3}}{2}, \frac{1}{2})$
45°	$\frac{π}{4}$	$(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
60°	$\frac{π}{3}$	$(\frac{1}{2}, \frac{\sqrt{3}}{2})$
90°	$\frac{π}{2}$	$(0, 1)$

I've only listed the ones in the first quadrant…from those, you can use the symmetry of a circle to find the rest. Here's a picture (measured in radians):

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(Note that this diagram has the denominators rationalized)

Examples

[6.] Find $sin 60^{°}$ .

Memorization: $\frac{\sqrt{3}}{2}$ .

[7.] Find $cos \frac{3 π}{4}$ .

Check the diagram, or use symmetry: $- \frac{1}{\sqrt{2}}$ .

[8.] Find $tan 390^{°}$ .

$390^{°} {(360 + 30)}^{°} =$ , so $tan 390^{°} = tan 30^{°} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$ .

Trigonometric Graphs

The Idea

I mentioned (hinted?) earlier that sine and cosine are functions. Functions can be graphed…and shifted…

Sine

Measure the angle horizontally—in radians!—and measure the sine of the angle vertically. Connect the dots to create the graph of $y = sin (x)$ .

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That graph never ends…since drawing forever is hard, we usually just draw one period of the graph (the length it takes for the graph to start repeating itself). The period of sine is $2 π$ .

The amplitude of a sine graph is half the distance from the highest to lowest y-values. The amplitude of sine is 1.

Once you've got the basic graph, you're ready to tackle $y = a \cdot sin (b (x - c)) + d$ . Guess what each of those letters do?

Incidentally, the value c is called the phase shift.

Cosine

Measure the angle horizontally—in radians!—and measure the cosine of the angle vertically. Connect the dots to create the graph of $y = cos (x)$ .

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What are the amplitude and period of this graph?

Given this graph, could you graph $y = a \cdot cos (b (x - c)) + d$ ?

Tangent

Measure the angle horizontally—in radians!—and measure the tangent of the angle vertically. Connect the dots to create the graph of $y = tan (x)$ .

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(those red lines are vertical asymptotes)

I wonder what $y = a \cdot tan (b (x - c)) + d$ looks like…

Examples

[9.] Sketch $y = 2 sin (x - \frac{π}{4})$ .

Sketch a normal sine graph; multiply the y-values by 2; shift right $\frac{π}{4}$ . I've left the original sine graph on for comparison:

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[10.] Sketch $y = cos (4 x) + 1$ .

Sketch a normal cosine graph; divide the x-values by 4 (reduce the period by a factor of 4); shift up one. Once again, I've left the original graph for comparison.

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Page last validated 2010-08-15