Chapter+8

** Chapter 8: Trigonometric Functions!!! **

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= Section 8.1: Angles and their Measure = ** General Vocabulary ** : 1. __Ray__-also know as half line, is a portion of a line that starts a pt V (vertex) on the line and extends indefinitely in one direction. 2.__Angle-__ 2 rays drawn with a common vertex. One side is the initial side, and one side is the terminal side. 3.__Standard Position__- when the vertex is at the origin and its initial side lies on the x-axis. Here is a picture that ties all three together:



4. Quadrantal Angles are when the terminal side lies on an axis as well.

There are two ways to measure angles: 1. Degrees 2. Radians __-Degrees__ 360 degrees = 1 revolution, so one degree equals 1/360 of a revolution. The quadrantal angles have measures of the negative and positive multiples of 90°.

- Another, more complicated way, of measuring angles is Radians.

__-Radians__ Radians come from the length of the radius. The radian is the angle, and no matter what your radius is, the angle is always the same. A whole revolution of the circle is 2 pie. (This is how we get the equation for the circumference C=pie r squared) ** __Finding the Length of an Arc__ ** The equation you use to do this is ** s=rθ ** where s is the arc length, r is the radius of the circle, and θ is the central angle. You must always use radians for θ. So…. S=2×¼So S= ½ meters If they give you the measure of the central angle in degrees or revolutions (more typically the will give it to you in degrees), you will have to convert!So how do we do that? Simple! You know that all the way around a circle is 360° and 2 pie radians, so we come up with the conversion formulas: 1 degree = pie over 180 radians & 1 radian = 180 over pie degrees! **
 * Let’s say you want to find the arc length of a circle that has a radius of 2 meters and a central angle of ¼ Radians *

Lets try an example! Convert (a) 60 degrees to radians and (b) 3pi ⁄ 2 radians to degrees. a) 60 × (pie ⁄180)= pie⁄3 b) 3pie⁄2 × (180⁄pie)= 270° __**So the next step is to find the Area of a Sector**__ The yellow area is called a sector. The part of the circle which is on the edge of the sector is called an arc. The equation for finding the area of a sector is ** A=½r∧2θ **. The area is A. R is the radius formed by the central angle of θ. Angle θ must always be in radians!!! (This means you will have to convert if they give you something else. Example:*Find the area of the sector of a circle of radius 4 inches formed by and angle of 45°* __Step 1__: Convert 45° to radians 45° × (pie⁄180) = pie⁄4 radians __Step 2:__ Plug in information to formula A=½(4) ∧2 (pie⁄4) A=2pie square in.!!!! Circular Motion There are two equations you will need to know for this: The linear speed equation and the angular speed equation. Both are very simple and almost the same. They are both units measurement over a unit of time. But the linear speed’s unit is one that measures in a straight line (such as miles, meters, feet, etc.) and angular speed’s unit measure angle of circles (radians, revolutions, and degrees) this is what they formally look like: Linear speed ** V=s⁄t ** *S stands for distance and t stands for time*  Angular Speed ** ω=θ⁄t ** *θ stands for the angle but it MUST BE IN RADIANS! (You are still converting!) and t is the time* There is also a relationship between V and ω (S=rθ). So take the linear speed (V=s⁄t) and replace S with rθ and you get V=rθ⁄t and we know that θ⁄t=ω so… ** V=rω!! ** Here is an example:An object is moving around a circle with a radius of 2 meters. If in 20 seconds a central angle of ⅓ radian is swept out, what is the angular speed of the object? What is its linear speed?Solution:R= 5cm; t=20 seconds; θ=⅓ radians ω=θ⁄t= (⅓)⁄20= ⅓×1⁄20= 1/60 radian/sec v=s/t = rθ⁄t= (5×⅓)⁄20= 5/3 × 1/20= 1/12 cm/sec = 8.2: Right angle Trigonometry =

Trigonometry is based off of triangles. One angle is labeled theta, the hypotenuse is labeled C, the leg next to theta is A, and the leg not connected to theta is B. There are six different trigonometric functions; the Sine of theta (sinθ), the Cosine of theta (cosθ), the Tangent of theta (tanθ), the Cotangent of theta (cotθ), Secant of theta (secθ), and the Cosecant of theta (cscθ). the sine of theta is equal to B divided C, Cosine is A over C, Tangent equals B divided by A, Cotangent is A over B. Secant is C divided by A, and Cosecant is C over B.

Fundamental Identities:

Because Sine of theta equals the reciprocal of Cosecant, Cosecant equals 1 over the sine of theta. the same applies with Cosine and Secant, and Tangent with Cotangent. Tangent also equal to Sine of theta over Cosine of theta. Because Sine equals B over C and Cosine equals A over C, then Sine over cosine equals B over C times C over A. The C's cancel and you get B over A: which equals the tangent of theta. The same applies with cotangent that equals Cosine over Sine. The final identities are the pythagorean identities. This is developed from the pythagorean theorem (A squared + B squared= C squared). When you divide both sides by C squared you end up with (A/C) squared + (B/C) squared= 1; which equals (sinθ) squared + (Cosθ)= 1. the other two are developed by dividing the last equation by (sinθ) squared. this equals Cotθ +1= Cscθ. the same thing happens with tangent and Cosecant replaced by their cofunctions when you divide by Cosine rather than sine. Heres a table in case you forgot: Cofunctions theorem:

Because there is a right angle, the remaining two angles must add up to 90 degrees. If theta switched angles then A and B would switch, so therefore: Sinθ= Cos(90-θ). this applies with other cofunctions. Heres another table to remember these:



= 8.3- Computing the Values of Trigonometric Functions of Given Angles =

First we must create a 45° triangle using the Pythagorean Theorem · c²= a²+ b²
 * How to find the exact value of the trigonometric functions of **** 45° **

Since we know that the other angle in the triangle has to be 45°

We know the triangle is isosceles. Therefore side //a// and //b// are the same length. If we plug in 1 as //a// and //b// then we can solve for //c//

From this info we can find sin, cos, tan, csc, sec, and cot __45°__ sin=√2/2 cos=√2/2 tan= 1 csc=√2 sec= √2 cot=1

__Example problem: find the exact value of each expression__ 1) cot45°+ tan45° · cot45°=1  · tan45°=1  · 1+1 = **//2//**

2) (csc π/4)(tan π/4) • csc π /4 = √2 •tan π/4= 1 •√2 x 1= **//√2//**

First we must create a 30°-60°-90° triangle using the Pythagorean Theorem. · a²+b²=c²
 * How to find the exact value of the trigonometric functions of 30° and 60° **

By using this rule we conclude that the sides are: a=1 b= √3 c=2

Using these three sides we can then, like before, find sin,cos, tan, cos, sin, and cot for the angle 30°, and 60°. __30°__ sin=1/2 cos= √3/2 tan= √3/3 csc= 2 sec= 2 √3/3 cot= √3

__60°__ sin= 1/2 cos= √3/2 cot= √3/3 csc= 2 √3/3 sec= 2 tan=√3

__Example problem: find the exact value of each expression__ 1) csc30°+tan45° · csc30°=2  · tan45°=1  · 2+1= 3 2) (Sin30°)(sec60°) · Sin30°=1/2 · Sec60°=2 · 1/2 x 2= 1

=** Section 8.4: Trigonometric Functions of General Angles **=

Key Section Vocab: · Coterminal Angles: Two angles in standard position that have the same terminal side · Reference Angle: The angle formed by the terminal side and either the positive or negative x-axis when using non-acute angles. · Trigonometric Functions of θ: the ratios of sin θ, cos θ, tan θ, csc θ, sec θ, & cot θ with a point on the terminal side and the distance from the point to the origin.

Section Goals: · Find the exact value of the trigonometric functions for general angles (so far it's only been acute). · Use coterminal angles to find the exact value of a trigonometric function. · Determine the sign of the trigonometric functions of an angle in a given quadrant (positive or negative). · Find the reference angle of a general angle. · Use the Pythagorean Theorem on reference angles. · Find the exact value of trigonometric functions of an angle given one of them and the quadrant of the angle. In this section, we learned how to find the six **trigonometric functions of general angles** (angles other than acute angles).

The initial side is still along the positive x axis. The terminal side is called r. We use a point on r called (a,b) (a,b cannot be the origin). You can use the Pythagorean Theorem to find r. If you draw a line from the point (a,b) to the x axis (never the y axis) you have formed a right triangle.

Once you have a, b, and r, you can use them to find the six trigonometric functions of θ with these ratios: These ratios will be the same for any point on the line r as long as r is the same is r.
 * Sin θ= b/r || Cos θ= a/r || Tan θ= b/a ||
 * Csc θ= r/b || Sec θ =r/a || Cot θ= a/b ||

Here’s an example of how to use a point on the terminal side to find the six trigonometric functions:

The point is: (2, -3)



First we use the Pythagorean Theorem to find the hypotenuse of the triangle formed (shown above):

a 2 +b 2 = r 2 2 2 + (-3) 2 =r 2 13=r 2 r=√13 a=2 b=-3

Now we can use the table to find the six trigonometric functions: It is a little different to find the value of each trigonometric functions with quadrantal angles. Quadrantal angles are 0, π/2, π, and 3π/2.
 * Sin θ= -3/√13 → (-3√13)/13 || Cos θ= 2/√13→(2√13)/2 → √13 || Tan θ= -3/2 ||
 * Csc θ= √13/-3 || Sec θ= √13/2 || Cot θ= 2/-3 ||

Here’s a chart that sums up the values of the six trigonometric functions for the quadrantal angles:


 * ** θ(radians) ** || ** θ(degrees) ** || ** Sin θ ** || ** Cos θ ** || ** Tan θ ** || ** Csc θ ** || ** Sec θ ** || ** Cot θ ** ||
 * 0 || 0 || 0 || 1 || 0 || undefined || 1 || undefined ||
 * π/2 || 90 || 1 || 0 || undefined || 1 || undefined || 0 ||
 * π || 180 || 0 || -1 || 0 || undefined || -1 || undefined ||
 * 3π/2 || 270 || -1 || 0 || undefined || -1 || undefined || 0 ||

You don’t have to memorize the table; you can just draw the angles and figure out the values from there.

You can also use the unit circle which you will learn about in the next section.

Since coterminal angles have the same terminal side, the values of trigonometric functions are the same for all angles that are coterminal.
 * Coterminal Angles: **

If an angle is more than revolution around, and it is measured in degrees, you can take out 360 degrees until the angle is smaller than 360 degrees and then you can measure it and the values of the trigometric functions are equal.  Here’s an example:



In both pictures, angle a and b are coterminal because they both have the same terminal side. The signs of the Trigonometric Functions: ** All angles (except quadrantal angles) lie in one of the four quadrants. The sign of the value will be positive or negative depending on which quadrant it is in.

This shows what the sign of each of the six functions would be: You can remember which pair is positive in each quadrant by remembering: **A **ll **S **tudents **T **ake **C **lasses (**A **ll, **S **in, **T **an, **C **os) going counter-clockwise.

Once you know the sign of two of these, you can figure out what quadrant it is in.

Example: Sin θ ˂ 0, Tan θ ˂ 0. What quadrant is θ in?

Answer: Θ is in quadrant IV because sinθ and tanθ are both negative so that means cosθ is positive and it’s in quadrant IV.

Reference angles help us turn a nonacute angle into an acute angle so it’s easier to find values for. It is easiest to find the reference angle by drawing the angle and seeing how much is between the terminal side and the x-axis.
 * Reference Angles: **

Like this: The reference angle here is 45° because the amount left-over to get to the x-axis after 135° is 45°. We already know how to find the values for a 45° angle.

We know have everything we need to know to find the six trigonometric values given, for example, one of them and what quadrant it’s in.
 * Finding the exact value of trigonometric functions: **

Here’s an example: Sinθ= 5/13 & 90° <span style="font-family: 'Arial','sans-serif'; font-size: 14pt; line-height: 115%;">˂ θ <span style="font-family: 'Arial','sans-serif'; font-size: 14pt; line-height: 115%;">˂ 180°. What are the remaining trigonometric functions?

Answer: This angle is in the second quadrant because 90° <span style="font-family: 'Arial','sans-serif'; font-size: 14pt; line-height: 115%;">˂ θ <span style="font-family: 'Arial','sans-serif'; font-size: 14pt; line-height: 115%;">˂ 180°. This tells us the sinθ and cscθ are positive, but everything else is negative. (Remember **A**ll **S**tudents **T**ake **C**lasses)

Sinθ= 5/13, and sinθ is opposite/hypotenuse so that means the side opposite of θ is 5 and the hypotenuse is 13. Then use the Pythagorean Theorem to find the adjacent. We can call the adjacent x for now.

5 2 +x 2 =132 5 2 +x 2 =169 x 2 =144 x=12 adjacent=12 opposite=5 hypotenuse=13

Now that we know the adjacent is 12, we can find all six trigonometric functions.

Watch this video for more clarification:
 * Sinθ=5/13 || Cosθ=-12/13 || Tanθ=-5/12 ||
 * Cscθ=13/5 || Secθ=-13/12 || Cotθ=-12/5 ||

<span style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: lime; font-size: 16pt; line-height: 115%;">media type="youtube" key="jUEkz0mGvg4" height="385" width="480"

=<span style="background-color: #12ff00; font-family: 'Arial','sans-serif'; font-size: 10pt;">Section 8.5: Properties of Trigonometric Functions =

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Because my section was created on pages (a mac program), which has no apparent equation symbols, I have been forced to improvise.

My Symbols: · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">@=theta · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">&=pi · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">^=degrees · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">{x}=square root of x

Key Section Vocab: · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Unit Circle: A circle based on the angle theta on the x,y coordinate plane which has a radius from the origin (it's center) of 1. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Period: The value of x that it takes for a graph to begin to repeat itself.

Section Goals: · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Use the unit circle to find the values of the six trig. functions. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Know the domain and ranges of the six trig. functions. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Know the periods of the six trig. functions and how to use them. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Know the even/odd properties of the six trig. functions and how to use them.

1. THE UNIT CIRCLE

If you have a right triangle with a point on the unit circle the x value is equal to cosine and the y value is equal to sine following this rational: · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Theta is the angle formed by the hypotenuse and x axis. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">The x value in the point on the unit circle is the adjacent side. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">The y value is the opposite side. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">The hypotenuse must be one because of the unit circle. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">To find sine of theta you take the opposite side over the hypotenuse. In this situation this is y over 1. As a result, y is equal to sine of theta. · <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">The same works for cosine. Cosine is equal to the adjacent over hypotenuse, which is x over 1.

You can also find that point on the unit circle by computing cosine for the x value and sine for the y value.

Example Problem: Find the values of the six trig. functions if you have the point (-1/2, {3}/2)

1. We know that cos@=a, and sin@=b. 1. So cos@=-1/2, and sin@= {3}/2. 2. tan@=sin@/cos@ > ({3}/2)/-1/2 > -{3}. 3. The other three functions are just reciprocals. 4. Therefore: csc@=-2, sec@=2/{3} > 2{3}/3, cot@=-1/{3}

Click on the link to see a visual illustration. <span style="background-color: #ffffff; color: blue; font-family: 'Arial','sans-serif'; font-size: 10pt;">[|Unit Circle Graphic.tiff]

<span style="background-color: #ffffff; font-family: 'Arial','sans-serif'; font-size: 10pt;">You can also watch this video for more clarification. [|Unit Circle Video]

You try some now: Find the values of the trig. functions from the the given point on the unit circle. Remember to simplify! 1. (-{2}/2, -{2}/2) 2. ({5}/3, 2/3) 3. (-{5}/5, 2{5}/5)

Solutions (Highlight to show): <span style="background-color: #000000; font-family: 'Arial','sans-serif'; font-size: 10pt;">1. sin=-{2}/2, cos= -{2}/2, tan=1, csc={2}, sec= -{2}, cot=1 2. sin=2/3, cos={5}/3, tan=2{5}/5, csc=3/2, sec=2{5}/5, cot={5}/2 3. sin=2{5}/5, cos=-{5}/5, tan={5}, csc={5}/2, sec={5}, cot={5}/5 <span style="background-color: #ffffff; font-family: 'Arial','sans-serif'; font-size: 10pt;">

2. DOMAIN AND RANGE OF TRIG. FUNCTIONS

These are key in graphing the trig. functions.

Domain - x can be angle angle theta that is a real number, so the domain of sine is all real numbers. Range - Because theta forms the unit circle, there is no possible way for the range to go outside of 1, so the range of sine is [-1,1].
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Sine: ** <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

Domain - Exact same as sine. All real numbers. Range - Cosine is the y value of the unit circle, but once again it will never stray out of a radius of one. The range of cosine is again [-1,1].
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cosine: **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

Domain - All real numbers except odd multiples of 90^. This is because Tan=Sin/Cos and if Cosine is 0 it is impossible. The values that cosine are zero are the multiples of 90^. Range - All real numbers. Tan=Sin/Cos so there is nowhere to limit it.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Tangent: **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

The domain and range of the other three functions are derived from the 1st three and so their reasoning is similar.


 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Domain/Range of Reciprocal Functions **

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;"> 4. PERIODS OF TRIG. FUNCTIONS
 * **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Function ** || **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Domain ** || **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Range ** ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cosecant || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">All real numbers except multiples of 180^. || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">(-Infinity, -1] [1, Infinity) ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Secant || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">All real numbers except odd multiples of 90^. || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">(-Infinity, -1] [1, Infinity) ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cotangent || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">All real numbers except multiples of 180^. || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">All real numbers. ||

The periods are also helpful in graphing, telling you when the graph will just repeat again.

360^ or 2&. sin(@ + 2& //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">k //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">)=sin@. //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">k //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">is an integer. Really this is breaking it up into how many 2& there are, and the leftover value for @ is what your looking for.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Sine: **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

Exact same thing as sine.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cosine: **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

180^ or &. tan(@ + & //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">k //<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">)=tan@. Once again, its really saying break it up into how many & there are, and the leftover value for @ is what your looking for. The reciprocal functions are the same as their reciprocals. (the period for sine is the same as the period for cosecant).
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Tangent: **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

Example Problem: Use the periodic properties to find the value of sin420^. 1. The period for the sine function is 360^ or 2&. 2. How many times does 360 go into 420? 1, with a remainder of 60. 3. This means that sin420^ is the same as sin60^. 4. sin60^={3}/2.

You try some now: Use the period properties to find the value of the expression. 1. sin405^ 2. tan(21&)^ 3. cot390^

Solutions (Highlight to show): <span style="background-color: #000000; font-family: 'Arial','sans-serif'; font-size: 10pt;">1. {2}/2 2. 0 3. {3 } <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">

4. EVEN/ODD PROPERTIES OF TRIG. FUNCTIONS

This is symmetry. If the function is odd, f(-x)=-f(x), then it has y-axis symmetry. If it is even, f(-x)=--f(x), it has x-axis symmetry.

 If you forgot all about symmetry, watch this video. [|Even/Odd Video] <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">
 * **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Function ** || **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Even/Odd ** || **<span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Why ** ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Sine || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Odd || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">sin@=b, sin(-@)=-b > -sin@ ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cosine || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Even || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">cos@=a, cos(-@)=a > cos@ ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Tangent || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Odd || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">tan(-@)= sin(-@)/cos(-@) > -sin@/cos@=-tan@ ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cosecant || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Even || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">cot(-@)=1/tan(-@) > 1/-tan@= -cot@ ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Secant || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Even || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">sec(-@)=1/cos@=sec@ ||
 * <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Cotangent || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">Odd || <span style="font-family: 'Arial','sans-serif'; font-size: 10pt;">csc(-@)=1/-sin@=-csc@ ||

Example Problem: Use the even/odd properties to find the value of sin(-45^). 1. The sine function is odd so sin(-45^) becomes -sin(45^) 2. The sin45^={2}/2. 3. Now make that negative, so it becomes -{2}/2.

You try some now: Use the even/odd properties to find the value of the expression. 1. sin(-60^) 2. tan(-30^) 3. cos(-&/4)

Solutions (Highlight to show): <span style="background-color: #000000; font-family: 'Arial','sans-serif'; font-size: 10pt;">1. -{3}/2 2. -{3}/3 3. {2}/2

=** SECTION 8.6: GRAPHS OF THE SINE AND COSINE FUNCTIONS **=

Key Section Vocab: · Sinusoidal graphs: sine and cosine graphs are sinusoidal because they are similar. · Amplitude: The distance stretched from the center to the max or min amount. For a function: y= A sin x, the amplitude is |A|. · Cycle: one period of the graph of sine or cosine. The amount for the graph to get back where it started is shown.

Section Goals: · Graph transformations of the sine function · Graph transformations of the cosine function · Determine the amplitude and period of sinusoidal functions · Graph sinusoidal functions: y=Asin(ωx) · Find an equation for a sinusoidal graph

In this section we learned what the sine and cosine functions look like and how to use them.

This is how we write the trigonometric functions when graphing:
 * y=f(x)=sin x ||  || y=f(x)=cos x ||   || y=f(x)=tan x ||


 * y=f(x)=cscx ||  || y=f(x)=sec x ||   || y=f(x)=cot x ||

The independent variable, x, is an angle (measured in radians).

The period of the sine function has a period of 2π. <span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">This is what one period of the graph of y= sin x looks like:
 * Graphing the sine function: **



Here are some points for this period of the graph of sine:
 * x value (in radians) || y value ||
 * 0 || 0 ||
 * π/6 || 1/2 ||
 * π/2 || 1 ||
 * 5π/6 || 1/2 ||
 * π || 0 ||
 * 7π/6 || -1/2 ||
 * 3π/2 || -1 ||
 * 11π/6 || -1/2 ||
 * 2π || 0 ||

These x values can be paired with their corresponding y value to form a point. For example, (π,0).

The graph continues like this in both directions.

__Properties of the sine function:__ · Domain: all real numbers · Range: [-1,1] · The sine function is odd. It has origin symmetry. · It has a period of 2π · x-intercepts: -2π, -π, 0, π, 2π, etc… · y-intercept : 0

Here’s an example of how to graph transformations of the sine function: Graph: y = -2 sin x + 1 Start with the graph y = sin x



The period is 2π. The amplitude is |-2|=2. Now we can graph y = 2 sin x



<span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Now the graph has and amplitude. Next, we can flip the graph over the x-axis because that’s what the “-2” tells us to do.

The last thing we need to do is shift the graph up one unit because that’s what the “+ 1” tells us to do.

This is the graph of y = -2 sin x + 1

The period of the cosine function has a period of 2π. This is what one period of the graph of y= cos x looks like:
 * Graphing the cosine function: **

<span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Here are some points for this period of the graph of cosine:


 * x value (in radians) || y value ||
 * 0 || 1 ||
 * π/3 || 1/2 ||
 * π/2 || 0 ||
 * 2π/3 || -1/2 ||
 * π || -1 ||
 * 4π/3 || -1/2 ||
 * 3π/2 || 0 ||
 * 5π/3 || 1/2 ||
 * 2π || 1 ||

These x values can be paired with their corresponding y value to form a point. For example, (π,-1). The graph continues like this in both directions.

__Properties of the sine function:__ · Domain: all real numbers · Range: [-1,1] · The sine function is even. It has y-axis symmetry. · It has a period of 2π. · x-intercepts: -3π/2, -π/2, π/2, 3π/2, 5π/2, etc… · y-intercept : 1

Here’s an example of how to graph a transformation of the sine function: <span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Graph: y = 2 cos(3x)

Start with the graph of y = cos x

<span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Now we can graph y = 2cos(x) by giving the previous graph an amplitude of 2.



The last thing to do is to find the period. 2π/ω = 2π/3 This is the graph of y=2cos(3x)

<span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Here is what sinθ and cosθ look like when they are graphed together.

From the graph, we can see that sin x= cos(x-(π/2)). Because they are similar in this way, sine and cosine are sinusoidal graphs.

Let’s review how to find the amplitude and period of a sinusoidal function.
 * Amplitude: |A| ||  || Period: 2π/ω ||

Example: What is the amplitude and period of y=-4cos(2x)?

Answer:
 * Amplitude: |-4|= 4 ||  || Period: 2π/2=π ||

Now you have everything you need to know to graph a sinusoidal function. Once you know how to find the amplitude and period from an equation, try finding the equation given the amplitude and the period.

Example: <span style="font-family: 'Calibri','sans-serif'; font-size: 14pt; line-height: 115%;">Find the equation of the sine function with these characteristics:
 * Amplitude: 3 ||  || Period: 2 ||

Answer: Y= ± 3sin(πx) We know that it has to be + or - 3 because the amplitude is the absolute value of A so A can be positive or negative. ω=2 because to get the period of a sine function, you take 2π/ω=2. That means for this function, 2π/ω must equal 2. If 2π/ω=2, ω=π.

You can also figure out how to write an equation given a graph.

Example:

Answer: The highest the line goes is ¾ and the lowest it goes is -¾. This means that the amplitude is ¾.

The period is 1 because after that is when the graph starts repeating itself.

We know that this is a sine function because the y-intercept is one.

A is positive because it goes up and down like the normal sine function does. ω=2π because the period is 1. That means 2π/ω=1 so ω must be 2π.

Now we have all the information we need to write the equation.


 * y=3/4sin(2πx)**

= 8.7 Graphs of Tangent, Cotangent, Cosecant & Secant Functions =

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 130%;">**Objectives:** 1. Graph Transformation of the Tangent Function and Cotangent Function 2. Graph Transformations of the Cosecant Function and Secant Function __Tangent__ -The Domain of Tan > all real numbers except odd multiples of Pi/2 -The range of tan> all real numbers - Tangent has a period of pi

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 130%;">**I**f you graph COS and SIN on the same axis, there is an asymptote at wherever COS is 0. Where SIN is 0, TAN is 0.

How we get the graph: tanθ= cot (pi/2 - θ) cot (-θ+ pi/2) cot − (θ- pi/2) -Domain: All real numbers except multiples of pi -Range: All real numbers - Period: Pi **
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 130%; font-weight: normal;">__Cotangent__
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 130%; font-weight: normal;">Notice when tan is increasing, cot is decreasing

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 130%; font-weight: normal;">__Cosecant__ We derive the graph for cosecant from the graph for sine. Here’s what sine looks like: So, since csc=1/sin and has a domain of multiples of pi, we can graph the domain as asymptotes and flip the crests of the sine graph. Now you have a bunch of parabolas that make up your cosecant graph! <span style="font-family: Arial,Helvetica,sans-serif; font-size: 120%;"> __Secant__ To find the graph of sec, you do the same thing you did with the sine graph, but this time use the cos graph. __Transformations__ All the rules that you use for transforming regular graphs apply to these too!! <span style="background-color: #ffffff; font-family: Arial,Helvetica,sans-serif; font-size: 120%;">
 * COS**
 * SEC**

And that's all you have to do!! You will get into further detail in other sections.

= Section 8.8: Phase Shift; Sinusoidal Curve Fitting =

Objectives: 1. Determine the Phase Shift of Sinusoidal Function 2. Graph Sinusoidal Functions: y= Asin (wx-phase shift)

1. How to determine the Phase Shift of a sinusoidal function. As taught from previous sections, we know that the amplitude is the number in front of the sin, cos, or tan. The period of sin and cos are 2pie and the period of tan is pie. Let’s say you were given the equation f(x) = 2sin(3x - p /2) Let’s try to determine the phase shift. We know that 2 is the amplitude, so now let’s try to find the period. To find this, we find the original period, 2 pie, because it is sin and divide it by the number in front of x, which is three. So the period is 2Pie/3. Now for the phase shift, we know that it is moved to the right because of the minus in the parentheses. So now we factor out the three in front of the x so it’s by itself and you get - (- p /2) / 3 = p /6. You always factor out the number in front of the x and that’s how you get the phase shift! Pretty simple! So now that we know the phase shift let’s graph it! 2. How to Graph a Sinusoidal Function with phase shift. Let’s graph the same equation f(x) = 2sin(3x - p /2). We already figured out the Amplitude=3 Period=2pie/3 Phase shift=pie/6 This is the original graph of cosine without any shifts or anything. But now we have all of these changes! So amplitude makes it higher and lower and the period makes it shorter or longer and the phase shift moves it left or right. So now you try to graph it! After you do look and see if you can get the graph below