Objectives:• Convert between degree and radian measure• Pythagorean identities• Find the trigonometric values of any angle measure• Unit circle

• Graph trigonometric functions

• Solve for lengths and angle measures of any type of triangle.

• Application of trigonometry• Verify identities

Uses of trigonometry:

Used to describe the motion of any object that behaves in a circular,oscillating or periodic manner.

Angle: Consists of two rays or half lines that originate at a common pointcalled the vertex.

These two rays have names:

• Terminal side: The ray that rotates to generate the angle.

• Initial side: The ray that does not move.

Angles are commonly denoted using lower case Greek letters.

,,,alpha, beta, theta, gamma (respectively)

To better describe the formation of angles we superimpose an xy plane on theangle with the vertex at the origin.

y

x

B

A0

Angles are generated by the terminalside rotating counterclockwise orclockwise.

If angles are generated by theterminal side moving counterclockwise the angle is positive.

If angles are generated by theterminal side moving clockwisethe angle is negative.

The direction of the arrow inside the angle will tell you if the terminalside is generating a positive or negative angle.

y

x

B

A0

If angles are generated by theterminal side moving clockwisethe angle is negative.

If we have an angle of 390°, this is one revolution of the unit circle plus 30°.

So, we say that 390° is coterminal with 30°

Coterminal angles differ by multiples of 360°

Ex 1: Determine if the following angles are coterminal.

a.) 90° and 1170 ° b.) 123 ° and 844 °

Solution:

3360

901170 If you get a whole number they are coterminal.If the result is not a whole number then they are not coterminal.

002777778.2360

123844

YES NO

To define the measure of an angle, we first add the unit circle centered at theorigin to the coordinate system.

y

x1

1

-1

-1

This is called a unit circle because the length of the radiusis 1.

The equation of the unit circleis: 122 yxThe domain is: [-1, 1]

To generate angles we must consider the terminal and initial sides.

The initial side is aligned with the x-axis and the terminal side starts at the x-axis and rotates to generate the angle.

y

x

t

This point is P(t). P is the function and t is the input value.

For every P(t) on the unit circle we can define its measure by using degreesor radians.

Radian Measure: For any real number t, the angle generated by rotatingcounterclockwise from the positive x-axis to the pointP(t) on the unit circle is said to have radian measure t.

x

t

y

(0,1)

(1,0)

(0,-1)

(-1,0)

2

1

2

3

-1

-2

-3

…and so on…

We can see that an angle that measures 90° is the same measure as 2

We can also see that 180° is

We will use this fact to convert between degrees and radian measure.

To convert degree measure to radian measure you multiply the degree measureby:

180

Ex: Convert the following degree measure to radian measure.

a.) 150° b.) 225° c.) -72°

6

5

180

150

180150.)

a

3

5

36

45

180

225

180225.)

b

5

2

180

72

18072.)

c

To convert radian measure to degree measure you multiply the radian measureby:

180

Ex 2: Convert the following radian measures to degree measure.

611

.)5

12.)

67

.) cba

210

61260180

67

.)

a

432

52160180

512

.)

b

330

61980180

611

.)

c

3.2 The Sine and Cosine Functions

Recall: The terminal side rotates to generate the angles.

There are infinitely many points on the unit circle that the terminalside “could” generate.

We will only “memorize” a few of them.

We will memorize all angles on the unit circle that are in increments of 30°and 45°.

y

x

45°135°

225° 315°

0°180°

90°

270°

60°

30°

120°

150°

210°

240° 300°

330°

Unit Circle

An ordered pair has the form: (x, y)

The Sine and Cosine Functions:

Suppose that the coordinates of the point P(t) on the unit circle are (x(t), y(t)). Then the sine of t, written sin t, and the cosine of t, written cos t,are defined by

sint t = y(t) and cos t = x(t)

Our new ordered pairs are of the form: P(t) = (cos t, sin t)

Finding the cosine and sine values of the common angles on the unit circle.y

x

45°135°

225° 315°

0°180°

90°

270°

60°

30°

120°

150°

210°

240° 300°

330°

We create righttriangles by drawinglines perpendicularto the x or y axis.It does not matterwhich axis.

To find the cosine and sine of 30° we must use the triangle we created.

30°

60°

1st: We know the third angle is 60° by the triangle sum theorem.

2nd: We know the length of the hypotenuse since this is a unit circle: r = 1

1

3rd: Using the properties of a 30-60-90 right triangle we can find the other two sides.

Since the base of this triangle is on the x-axis this side would represent thecos t. The height would represent the sin t.

Recall: * The side opposite the 30° angle is half of the hypotenuse. * The side opposite the 60° angle is the product of the short leg and the square root of 3.

If the hypotenuse has length 1, then the side opposite 30° is ½ .

The side opposite the 60° angle is

23

321

So, P(t) =

21

,23

6

PThis is how you will solve for all sine and cosine on the unit circle dealing with 30 – 60 – 90 right triangles.

Finding the coordinates of a point using a 45 – 45 – 90 right triangle.

45°

45°

1

We know: • The third angle is 45°

• The hypotenuse has length 1.

• In a 45 – 45 – 90 right triangle the legs are the same length. We can call both legs the same variable since they are equal.

122 aa Now solve for a.

22

42

22

21

21

21

21

21

12

2

2

a

a

a

a

a

We rationalized the denominator in this stepbecause we do not leave radicals in thedenominator.

Since the legs have the same length, the cosine and sine values are the same.

22

,22

4

PtP

Pythagorean Identity:

For all real numbers t,

(sin t)2 + (cos t)2 = 1

Because of the Pythagorean Identity, sine and cosine have bounds:

For all real numbers t,

1cos11sin1 tandt

The cosine function is even

For all real numbers t,

cos(-t) = cos t

The sine function is odd

For all real numbers t,

sin(-t) = - sin t

A Periodic Function is a function that repeats the same thing over and over again.

Trigonometric Functions are periodic because they repeat.

Sine and Cosine functions have a period of 2From your unit circle you can see that the ordered pairs on your unit circledo not begin to repeat until after one complete revolution of the unit circle360°

ttandtt cos)2cos(sin)2sin(

Reference Number

For any real number t, the reference number r associated with t is theshortest distance along the unit circle from t to the x-axis. The referencenumber r is always in the interval

2,0

Ex 2: What is the reference number of 6

11 Since the answer is always inthe above interval, it is

6

Ex 3: Determine the values of t in that satisfy]2,0[ 0cos3)(sin2 2 tt

This is simply a quadratic function. Use what you know.

You can view this quadratic as… 032 2 xy

We cannot work with it this way; we can only have one unknown. We needto turn both variables into y (sine) or x (cosine).

Pythagorean Identity: (sin t)2 + (cos t)2 = 1

Manipulate this identity so that it is something that you can use.

22

22

)(cos1)(sin

1)(cos)(sin

tt

tt

We can replace (sin t)2 with (1 – (cos t)2)

0cos3))(cos1(2 2 tt Now distribute

02cos3)(cos2

0cos3)(cos222

2

tt

tt Now factor and set each factor equalto zero.

To make things easy, and we don’t like our leading coefficient being negative,multiply both sides by -1.

02cos3)(cos2 2 tt Now factor

0232 2 xx It may help you to view it like this to factor.

(2x + 1)(x – 2) = 0 Set each factor equal to zero and solve.

2x + 1 = 0 and x – 2 = 0

221 xandx Replace x with cos t since x

represents cos t.

2cos21

cos tandt Where on the unit circle is this true?

34

32

:21

cos tandtatt cos t = 2 will never happen; it is

outside the bounds of cosine.

Lesson 3.3: Graphs of the Sine and Cosine Functions

Sine Curve

We will use the values that we memorized from the unit circle to graph oneperiod of the sine function.

Sin t

t

The more points you plot the more precise your graph will be. I will plotonly the 45° increments of the angles.

4

2

2

4

7

2

3

4

5

4

3

1

-1

4

Cosine Curve

cos t

t

4

2

2

4

7

2

3

4

5

4

3

1

-1

4

Ex 1: Use the graphs of y = sin x and y = cos x to sketch the graphs of

2sin

2sin

xyandxy

Trigonometric functions are no different than any other function when “shifts” are involved… minus to the right …plus to the left.

Again, we must be sure the leading coefficient is 1 before we try and seethe shifts.

Sin t

t

4

2

2

4

7

2

3

4

5

4

3

1

-1

4

graph of y = sin x

Note: Your shift will help guide you as to what you should count by on the x-axis.

x

sin x

Q: What is the last value going to be that we will write on the x-axis?

A: 2

5

2

52

2

3

2

1

-1

NOW YOU TRY THE SECOND ONE!!!

2sin

xy

x

sin x

2

52

2

3

2

1

-1

2

2sin

xy

You try…

2cos

2cos

xyandxy

We can see from the previous 4 graphs that when we shift one graph to theright or left that we obtain the graph of a different equation, such as:

The graph of: xx cos2

sin

xx sin2

cos

xx cos2

sin

xx sin2

cos

These are all true onlybecause we are shifting by

2

When you shift by otherincrements differentequalities occur.

When shifting by increments of pi…

)sin()sin( xxy

)cos()cos( xxy

Ex 1: Sketch the graph of

2cos2)(

xxf

All trig functions are of the form:

y = A sin(Bx + C) The name of the trig function changes only.

|A| is the amplitude

Period is B2

Shift is B

C

Don’t forget!!! The leading coefficientmust be 1 before you can see the shift!!!

2cos2)(

xxf

Amplitude is: 2; so the height will go to +2 and -2 on y-axis.

B = 1, so the period has not changed.

Shift: right units2

First, I will graph y = 2cosx, then shift this graph to the right units.2

Since the period did not change and the shift is units, we will count by on the x-axis. 2

4

x

cos x2

-2

2

2

3

2

Notice that the zeros of this function donot change. Amplitude is just a verticalelongation or compression of the graph.

Now, we shift this graph!

y = 2cos x

x

cos x2

-2

22

3

2

2

5

2cos2)(

xxf

Ex 2: Sketch the graph of

23cos2)(

xxf

B = 3, not 1 so we must find the length of the new period.

Recall: period is found by:B2

3

2 Always reduce if necessary.

Instead of this graph being graphed from 0 to 2π, the entire cosine curvewill be graphed between 0 and

3

2

Recall: To see a horizontal shift the leading coefficient must be 1 and it is not.

We must factor out the 3 from the quantity.

63cos2)(

23cos2)(

xxf

xxf

We will shift this graph to the right units. 6

x

cos x

2

-2

2

6

11

3

5

2

3

3

4

6

7

6

5

3

2

236

The main points of the cosinecurve (y = cos x) are:

)1,2(

0,2

3

)1,(

0,2

)1,0(

Divide all of theseorginal x-coordinatesby 3 and multiply they-coordinates by 2,these will be the locations of your newpoints.

New points:

2,

3

2,0,

2,2,

3,0,

6),2,0(

withshift

2,

6

5,0,

3

2,2,

2,0,

3,2,

6

Ex 3: Sketch the graph of )2cos(3 xy

x

cos x

2

-2

4

32442

3.4 Other Trigonometric Functions

The Tangent, Cotangent, Secant, and Cosecant Functions:

The tangent, cotangent, secant and cosecant functions, written respectively astan x, cot x, sec x, and csc x are defined by the quotients

xx

xcossin

tan xx

xsincos

cot

xx

cos1

sec x

xsin

1csc

Note: Tangent and secant are only defined when cos x ≠ 0 cotangent is cosecant are only defined when sin x ≠ 0

Trig. function Sign in quadrant…

I II III IV

Sin x

Cos x

Tan x

Csc x

Sec x

Cot x

+ + - -

+ - - +

+ - + -

+ + - -

+ - - +

+ - + -

Ex: .sin 232 xthatandxSuppose

Determine the values of the other trigonometric functions.

Solution: Since we know that csc x is the reciprocal of sin x, write the reciprocal of sin x.

23

sin1

csc x

x

Next we must find cos x because the remaining trig. functions contain it.

What do you know that involves both sine and cosine that willhelp you find cos x? Pythagorean Identity.

3

5cos

9

5cos

9

5)(cos

9

4

9

9)(cos

9

41)(cos

1)(cos9

4

1)(cos3

2

1)(cos)(sin

2

2

2

2

22

22

x

x

x

x

x

x

x

xx The interval for sin x is given. Quadrant II

Cosine is negative in QII, so…

3

5cos x

5

52

5

5

5

2

53

6

5

3

3

2

cos

sintan

35

32

x

xx

cot x = 2

5 or you could have started

at the beginning to find thissolution.

sec x is the reciprocal of cos x. Write it and simplify:

5

53

5

5

5

3

5

3sec x

Note: You are now ready to write your answers. Make sure they have the correct sign for quadrant II.

2

5cot

5

53sec

2

3csc

5

52tan

3

5cos

3

2sin:

x

x

x

x

x

xGiven Ex 2: 2

3

2

3cos

tthatandtthatGiven

Find the values of the other trigonometric functions.

Solution: We can see that we are in QIII. The given isthe cosine value for

6

7

2

1

6

7sinsin

x

3

3

3

3

3

1

3

2

2

1tan

23

21

x

2csc t

3

32

3

2sec x

33cot x

The graph of the tangent function.

The tangent function is zero when the sine function is zero because sine isin the numerator of the tangent function.

The tangent function is undefined when the cosine function is zero becausethe cosine function is in the denominator of the tangent function.

The tangent function will be zero at: 2,,0

The tangent function is undefined at:2

3,

2

t

tan t

6

5

3

2

236632

y = tan x

1

-1

Sine and cosine have periods of , therefore, tangent will also repeat itselfon that same period.

2

Ex 3: Sketch the following graphs:

13

2tan.

32tan.

2tan.

xyc

xyb

xya

Solution: a.) B = 2; set up your inequality:

x

x

0

220

The length of your new period is .

Since we divided the period by 2 we will also have to divide therestrictions by 2:

Original restrictions:

2

3

2

and

New Restrictions:

4

3

4

and

a.) y = tan 2x

b.)

32tan

xy

c.)

32tan

xy This graph is the graph from part b

reflected over the x-axis.

c. contd.)1

32tan

xy

Now we shift it up one unit.

Note: The zeros from the graph are not obvious. To find them we would set the function equal to zero.

13

2tan0

x

Which implies…

32tan1

x If we let x =

32

x

We have… tan x = 1; where does this occur, when x = ?

We will find this later this chapter.

Now that we have the graphs of sine, cosine and tangent we cangraph the remaining trig. Functions using the reciprocal technique.

Do you remember those properties of graphing reciprocals?

… as f(x) increases, its reciprocal… decreases!

Graph: y = sin x.

y= csc x is undefined at: Where sin x = 0. There, we havevertical asymptotes.

The Cosecant function:

Now, use the fact that as y = sin x increases, y = csc x decreases andvice versa.

Ex: Graph y = sec x

Ex: Graph y = cot x

Be careful with y = cot x.

The restrictions for y = tan x are different from y = cot x. They havedifferent vertical asymptotes.

Ex: Sketch the graph of

23csc

21

xy

Note: There are several strategies for graphing:

1. You could start with the parent function and build on it one change at a time.

2. You can find the most important pieces of information of this new graph and make the changes to the pieces, and then plot your new points and new vertical asymptotes.

This time, lets make the changes to the parent function:

y = csc x

y = csc x has vertical asymptotes at: 2,,0

y = csc 3x has vertical asymptotes at:3

5,

34

,,3

2,

3,0

Now, the amplitude changes to ½

Now, we shift the graph to the left units.6

The vertical asymptotes go with it!!! They get shifted too!

3.5 Trigonometric Identities

Using the first Pythagorean Identity we can manipulate it and obtain two more.

Pythagorean Identities:

22 )(sec1)(tan xx

22 )(csc1)(cot xx

Sum and Difference Formulas for Sine and Cosine

For every pair of real numbers and , we have1x 2x

212121 sincoscossin)sin( xxxxxx

212121 sinsincoscos)cos( xxxxxx and

Ex 1: Determine the following:

12

sin.)

a

12

7cos.)

b

Using the values from the unit circle that we have committed to memorywe will find a combination of two that equal the function we are trying toevaluate.

What equals ?

12

4312

We now label these two 21 xandx

Using the difference formula (we subtracted above and for sine you dothe operation of the combination that you used to equal your function), wewill evaluate

12

sin

4sin

3cos

4cos

3sin

43sin

4sin

3cos

4cos

3sin

43sin

2

2

2

1

2

2

2

3

43sin

4

2

4

23

43sin

Both terms have a common factor:4

2

Factor it out!

134

2

43sin

You try part b. Hint: For cosine you use the opposite formula compared to thesign you used to obtain your function.

If you subtracted to get then you will use the sum formula, and visa versa.

12

7cos

b.)

43cos

12

7cos

4sin

3sin

4cos

3cos

43cos

2

2

2

3

2

2

2

1

43cos

314

2

43cos

We can also find the tangent of functions on the unit circle…

)cos(

)sin()tan(

21

2121 xx

xxxx

2121

2121

sinsincoscos

sincoscossin

xxxx

xxxx

We can find tangent of t by using this formula or we can continueto manipulate the formula to obtain a smaller one.

)tan( 21 xx2121

2121

sinsincoscos

sincoscossin

xxxx

xxxx

We maintain equality as long as we do the same thingto both the numerator and the denominator.

Lets divide the numerator and denominator by

)tan( 21 xx

21

21

21

21

21

21

21

21

coscos

sinsin

coscos

coscoscoscos

sincos

coscos

cossin

xx

xx

xx

xxxx

xx

xx

xx

Now simplify

)tan( 21 xx21

21

tantan1

tantan

xx

xx

21 coscos xx

Ex 2: Determine

12

tan

4312

4tan

3tan1

4tan

3tan

43tan

To use this formula it is a must that you can easily findthe tangent values of the measures you have committedto memory.

131

13

43tan

13

13

43tan

YOU pick which is easier for you!!!

Double angle formulas for sine and cosine

sin 2x = sin(x + x) = sin x cos x + sin x cos x = 2sin x cos x

22222 )(sin21)(sin)(sin1)(sin)(cos2cos xxxxxx

Ex 3: Determine all the values of x in [0, 2π] that satisfy each equation.

a.) cos x = sin 2x b.) sin x = cos 2x c.) 1 = sin x + cos x

Solution: a.) We need to get everything into sin x and cos x because that is what we know on the unit circle.

cos x = sin 2x = 2sin x cos x

cos x = 2sin x cos x

0 = 2sin x cos x – cos x = cos x(2sin x – 1)

0 = cos x(2sin x – 1)

cos x = 0 and 2sin x – 1 = 0

sin x = ½

2

3

20cos

andxwhenx

6

5

62

1sin

andxwhenx

b.) sin x = cos 2x2)(sin21sin xx

01sin)(sin2 2 xx

0)1)(sin1sin2( xx

6

5

62

1sin

orxwhenx

2

31sin

xwhenx

c.) 1 = sin x + cos x Square both sides.

2)cos(sin1 xx

22 )(coscossin2)(sin1 xxxx Re-order expression

xxxx cossin2)(cos)(sin1 22

xx cossin211

xx cossin20

x2sin0 In order for sine to ever be equal to 0, 2x must be a multiple of pi.

x could be any of the following to make this statement true:

2,

2

3,,

2,0 xxxxx CHECK!!!

2,

2

3,,

2,0 xxxxx

NOTE: We squared the equation earlier, so this could introduce extraneous solutions. This must have happened because the original expression, 1 = sin x + cos x is only satisfied if the following is true:

2,

2,0 xxx

When x = the statement does not equal 1. 2

3

Half Angle Formulas

Half angle formulas come from double angle formulas involving “2x.”

Ex: Double Angle formula half angle formula

x2cos

2cos

x

Instead of doubling the angle it is cut in half.

Half Angle Formula:

For any real number x we have

2

cos1

2cos

2

cos1

2sin

22xx

andxx

If we replace x with 2x we have

2

2cos1)(cos

2

2cos1sin

xxand

xx

Ex 4: Determine .8

7tan,

8

7cos,

8

7sin

and

Solution:

8

7sin

2

8

72cos1

8

7sin

28

72cos1

8

7sin

Substitution for x

Reduce fraction

24

7cos1

8

7sin

Now take cosine

22

21

8

7sin

Now use algebra and simplify expression

22

21

8

7sin

22

2

2

2

8

7sin

22

22

8

7sin

2

1

2

22

8

7sin

4

22

8

7sin

2

22

8

7sin

Now we must determine the sign.

This function is in quadrant II-sine is positive here.

2

22

8

7sin

You try the next one!!!

8

7cos

2

22

4

22

22

22

22

2

2

2

22

21

24

7cos1

Since is in the second quadrant, cosine is negative.8

7

2

22

8

7cos

8

7cos

8

7sin

8

7tan

22

22

22

22

2

222

22

Verifying Identities

Using all of the trigonometry we’ve learned so far we can use this informationto prove identities. (make one side of the equation look like the other)

Ex 5: verify the identity222 )(tan))(sec)(cos1( xxx

It does not matter which side you start with to make it look like the other side.

Many times we only have to manipulate one side to look like the other.

Sometimes we work both sides together and as we go along they will be thesame.

Solution: I will start with the left side. I notice that I can use the PythagoreanTheorem.

222 )(tan)(sec)(sin xxx Remember, your goal is to turn it into tangent.

22

2 )(tan)(cos

1)(sin x

xx 2

2

2

)(tan)(cos

)(sinx

x

x

22 )(tan)(tan xx

Now you try one!!!

Ex 6: Verify: cot x + tan x = sec x csc x

xxx

x

x

xcscsec

cos

sin

sin

cos

xxx

x

x

xcscsec

cos

)(sin

sin

)(cos 22

xxxx

xxcscsec

cossin

)(sin)(cos 22

xxxx

cscseccossin

1

xxxx

cscseccos

1

sin

1

xxxx cscseccscsec

3.6 Right Triangle Trigonometry

θ

c b

a

Trigonometric Functions of an Angle in a Right Triangle

For the angle θ in the right triangle shown, we have

b

a

a

c

b

ca

b

c

a

c

b

cot,sec,csc

,tan,cos,sin

Recall: SohCahToa

opp

adj

adj

hyp

opp

hyp

adj

opp

hyp

adj

hyp

opp

cot,sec,csc

,tan,cos,sin

Ex 1: Suppose that an acute angle θ is known to satisfy . Determine the other trigonometric functions of this angle. 5

3sin

Lets use a right triangle and label what we know.Solution:

θ

35

a

We must find a beforewe can find the others, except for csc θ

The Pythagorean Theorem will help us find the value of a.

4

16

259

53

2

2

222

222

a

a

a

a

cba

= 4

Now that we know all of the values for a, b, andc, we can write all of the solutions:

3

4cot,

4

5sec,

3

5csc

,4

3tan,

5

4cos,

5

3sin

Note: The last example said that the angle was acute which means it lies in the first quadrant. If the angle was in the second, third, or fourth, we would have solved the problem like we just did, and then change the signs as necessary.

Ex 2: Find the value of the six trigonometric functions if the 5

3cos

Using the Pythagorean theorem we find that the opposite side equals 4.

Solution:

4

3cot,

3

5sec,

4

5csc

,3

4tan,

5

3cos,

5

4sin

Ex 3: A climber who wants to measure the height of a cliff is standing 35 feet from the base of the cliff. An angle of approximately 60° is formed by the lines joining the climber’s feet with the top and bottom of the cliff, as shown. Use this information of approximate the height of the cliff.

cliff x

35 ft

60°

Solution: When you have a right triangle situation and at least one angle and oneside is known, follow these steps to find the missing piece.

1. Ask yourself: What side do I have and what side do I want?

We have the adjacent side and we want the opposite side.

2. Which trig function involves the two answers from question #1?

The tangent function will be used.

3560tan

x

adj

opp

3560tan

x Solve for x.

35732050808.1

x

x ≈ 60.6 ft.

Ex 4: Two balls are against the rail at opposite ends of a 10 foot billiard table. The player must hit the ball on the left with the cue ball on the right without touching any of the other balls on the table. This is done by banking the cue ball off the bottom cushion, as shown. Where should the cue ball hit the bottom cushion, and what is the angle that its path makes with the bottom cushion.

2

3

θ θ

x 10 - x

tan2 x x

10

3

xx

10

322(10 – x) = 3x

20 – 2x = 3x x = 4 Since ;2

tanx

2

1

4

2tan

Unfortunately, we do not know a value on our unit circle where this is true, but thisis the best we can do for now, even though we need the angle measure, not thetangent of that angle.

We can answer where the cue ball should hit the bottom cushion: The cue ball should hit the bottom cushion 2 feet from the bottom left or 8 feet from the bottomright.

Ex 5: An engineer is designing a drainage canal that has a trapezoidal cross section, as shown. The bottom and sides of the canal are each L feet long, and the side makes an angle θ with the horizontal.

a.) Find an expression for the cross-sectional area of the canal in terms of the angle θ with the horizontal.

b.) If the canal is S feet long, approximate the angle θ that will maximize the capacity of the canal.

L

L L

a.) Area of a trapezoid:

hbbA )(2

121

We know the length of one base (L) and can find the height by making a right triangle.

We have angle θ and the hypotenuse; we want theopposite side:

Lsin θ

Lcos θ Lcos θ

hbbA )(2

121

sincos222

1LLLA

sin)cos( LLLA

sin)cos1( LLA

sin)cos1(2 LA

b.) Capacity means volume, so a canal that is S feet long has the capacity…

SAV

32 sin)cos1( feetSLV

To find the exact value of θ will have to wait until the next section!

3.7 Inverse Trigonometric Functions

Recall properties of inverse functions:

Properties of Inverse Functions

Suppose that f is a one-to-one function. Then the inverse f-1 isunique, and …

1. The domain of f-1 is the range of f.2. The range of f-1 is the domain of f.3. If x is in the domain of f-1 and y is in the domain of f, then

f-1(x) = y if and only if f(y) = x.4. f(f-1(x)) = x when x is in the domain of f-1.5. f-1(f(x)) = x when x is in the domain of f.6. The graph of y = f-1(x) is the graph of y = f(x) about the line y = x.

Sine

The sine function is not one to one because it is periodic. If fails the horizontalline test for every horizontal line between -1 and 1 on the y-axis.

However, we can restrict the domain of the sine function so that it is one-to-one.

The sine function is one to one on the interval:

2,

2

On this interval, the sine function does have an inverse and is denoted:Arcsine function or simply: y = arcsin x

Note: csc x = (sin x)-1

The Arcsine Function

The arcsine function, denoted arcsin, has domain (-1,1) and rangeand is defined by

arcsin x = y if and only if sin y = x

2,

2

Arcsine Properties

sin(arcsin x) = x when x is in [-1, 1] and

arcsin(sin x) = x when x is in

2,

2

The graph of y = arcsin x is the reflection of the restricted function of y = sin xreflected about the line y = x.

Notice how steep the ends ofthe curve are on the inverse.This corresponds to the flatness on the sine curve.

Ex 1: Find

4

3sinarcsin.)

3sinarcsin.)

2

1arcsin.)

cba

62

1arcsin.)

a

Solution: Remember that we have a restricted interval now for sine. So we ask ourselves: Where inthe restricted interval does sin = ½ ?

33sinarcsin.)

b Recall the arcsine properties: arcsin(sin x) = x.

4

3sinarcsin.)

c Here we have to be careful – our answer must

lie in the restricted interval. is not in the restricted interval. 4

3

Where in the restricted interval is the the same value?4

3sin

...4

sin4

3sin So

44

3sinarcsin.)

c

All of the other trigonometric functions are defined by making domain restrictions, too.

Cosine

The Arccosine Function

The arccosine function, denoted arccos, has domain [-1, 1] and range [0, π].and is defined by

arccos x = y if and only if cos y = x

Arccosine Properties

cos(arccos x) = x when x is in [-1, 1] and arccos(cos x) = x when x isin [0, π].

Ex 2: Find

4cosarccos.)

3cosarccos.)

2

1arccossin.)

2

1arccoscos.)

d

c

b

aSolution:

2

1

2

1arccoscos.)

a

2

3

3

2sin

2

1arccossin.)

b

33cosarccos.)

c

.44

cosarccos.) domainrestrictedtheinnotisthisexceptd

Where in the restricted domain does ? 2

2

4cos

At 4

Tangent

The Arctangent Function

The arctangent function, denoted arctan, has domain (-∞, ∞) and range , and is defined by

arctan x = y if and only if tan y = x

2,

2

Arctangent Properties

tan(arctan x) = x for x in (-∞, ∞) and arctan(tan x) = x for x in .

2,

2

The graph of y = arctan x is the graph of y = tan x reflected over the line y = x.

Graph together with students on board

Ex 3: Find

5

3arcsin

5

12arctancos

Solution: We can see that this is of the form: )cos( 21 xx

We will use the difference formula for cosine.

5

3arcsinsin

5

12arctansin

5

3arcsincos

5

12arctancos

5

3arcsin

5

12arctancos

5

3

5

12arctansin

5

3arcsincos

5

12arctancos

5

3arcsin

5

12arctancos

We know the value of the very last identity.

The remaining identities must be found using a right triangle:

θ

12

5

5

12tan But this is not one of our identities…

5

12arctan)arctan(tan

To get θ by itself we take the inverse of both sides.

5

12arctan

22 12513

Now, using this triangle find values of identities involving5

12arctan

5

3

5

12arctansin

5

3arcsincos

5

12arctancos

5

3arcsin

5

12arctancos

5

3

5

12arctansin

5

3arcsincos

5

12arctancos

5

3arcsin

5

12arctancos

These values are: 13

5

13

12

5

3

13

12

5

3arcsincos

13

5

5

3arcsin

5

12arctancos

All we have to do now is draw another triangle to help us find

5

3arcsincos

θ

5

3sin

5

3arcsin

35

4

Now, find the value of your identity:5

4

5

3arcsincos

5

3

13

12

5

4

13

5

5

3arcsin

5

12arctancos

Now evaluate

65

16

5

3arcsin

5

12arctancos

Arcsecant

The arcsecant function has the same restricted interval as the cosine function.

Recall that y = sec x has a vertical asymptote at 2

The Arcsecant Function

The arcsecant function, denoted arcsec, has domain (-∞, -1] U [ 1, ∞) and

range and is defined by

arcsec x = y if and only if sec y = x

,22

,0

Arcsecant Properties

Sec(arcsec x) = x when x is in (-∞, -1] U [1, ∞) and arcsec(sec x) = x when

x is in

,22

,0

Ex: Graph y = arcsec x

Graph with students on board.

3.8 Applications of Trigonometric Functions

A Cessna Citation III business jet flying at 520 miles per hour isdirectly over Logan, Utah, and heading due south towardPhoenix. Fifteen minutes later an F-15 Fighting Eagle passes overLogan traveling westward at 1535 miles per hour. We would like toDetermine a function that describes the distance between the planesin terms of the time after the F-15 passes over Logan until it reaches the California border 20 minutes later.

Distance = rate x time

d = 520(t + 0.25) = 520t + 130

d = 1535t

d(t)

22 )130520()1535()( tttd

Recall: the distance formula here involvestime in hours: 20 minutes is 1/3 of an hour

t = 1/3

22 )130520()1535()( tttd

22 )1303

1520()

3

11535(

3

1

d

milesd 5953

1

It is more likely that the planes will not be traveling in paths that areperpendicular. Lets change the problem a little.

A Cessna Citation III business jet flying at 520 miles per hour isdirectly over Logan, Utah, and heading due south towardPhoenix. Fifteen minutes later an F-15 Fighting Eagle passes overLogan traveling 24° west of south at 1535 miles per hour towardNellis Airforce base 395 miles away.

24°

d(t)

1535t

520t + 130

Since this situation does not includea right angle, we can not use thePythagorean Theorem…directly.

Law of Cosines

Suppose that a triangle has sides of length a, b, and c and corresponding opposite angles α, β, and γ as shown. Then

cos2222 bccba

B

A C b

c a

α γ

β

We also get…

cos2222 accab

cos2222 abbac

Ex 1: A triangle has sides of length 6 and 8, and the angle between these sides is 60°. What is the length of the third side?

60°

6 x

8

Solution: Using the Law of Cosines…

60cos)8)(6(286 222x

21

961002x

522 x 132x

Ex 2: If two sides of a non right triangle are of lengths 15 and 25 and the included angle measures 35°, find the missing side and one of the other angles.

15 x

25

35°

35cos)25)(15(22515 22x

35.15x

cos)25)(35.15(22535.1515 222 ------------------------------------------------------------------

cos5.7676256225.235225 cos5.7676225.860225

cos5.7676225.635

cos8281726384.0 0085.34

34

γ

We can now go back and solve the aircraft problem…

24cos)130520)(1535(2)130520()1535())(( 222 tttttd

Lets let t = 0.333

24cos)13016.173)(155.511(2)13016.173()155.511())333.0(( 222d

milesd 2653

1

Ex 3: A picture in an art museum is 5 feet high and hung so that its base is 8 feet above the ground. Find the viewing angle θ(x) of a 6-foot tall viewer who is standing x feet from the wall.

APB and APC are bothright triangles. We can Use the PythagoreanTheorem.

249 xAC

24 xAB

Using the Law of Cosines…

)(cos4942)49()4(25 2222 xxxxx

)(cos494225325 222 xxxx

22

2

4942

25325))(cos(

xx

xx

)49)(4(

14arccos)(

22

2

xx

xx

Using a graphing calculator we can see that the best viewing of thepainting is where the maximum occurs which is when x 3.7 feet.This is the distance the viewer should stand from the wall.

When the angles and one side of a triangle are known we can use the Law of Sines to find the other missing parts.

The Law of Sines

Suppose that a triangle has sides of length a, b and c withcorresponding opposite angles α, β and γ as shown. Then

B

A Cα γ

βc a

b

cba

sinsinsin

Ex 4: Find the missing angles.

60° β

1326

8

α

Solution:

If we knew α we couldfind β.

β = 180 – 60 – α

We could use the Law of Cosines to find α…

but it is easier to use the Law of Sines.

cos)132)(6(2)132(68 222

132

60sin

8

sin 132

60sin8sin

132

60sin8arcsin

13

32arcsin

13

392arcsin 1.46;9.73

Ex 5: The aircraft carrier Carl Vinson leaves the Pearl Harbor naval shipyard and heads due west at 28 knots. A helicopter is 175 nautical miles from the carrier 35° south of west.

a.) On what course should the helicopter travel at its cruising speed of 130 knots to intercept the aircraft carrier? b.) How long will it take.

Solution: Draw a picture of the given information.

First, find θ which will give thecourse the helicopter should fly.

tt 130

35sin

28

sin

1235.0130

35sin28sin

t

t

71235.0arcsinThe helicopter should fly 42°north of east.

Since we know θ = 7° we also know the third angle is 138°.

t130

35sin

175

138sin

35sin175138sin130t

138sin130

35sin175t

hourst 154.1

It will take the helicopter about 1 hour and 9 minutes to interceptthe aircraft carrier.

Ex 6: Think Pair Share A campground lies at the west end of an east-west road in a relatively flat, but dense, forest. The starting point for a hike lies 30 kilometers to the northeast of the campground. A hiker begins at the starting point and travels in the general direction of the campground, reaching the road after 25 kilometers. Approximately how far is the campground from the road?

camp

3025

A C C'

45° γ γ'

BLets assume triangle ABC givesthe correct solution. If the hikertraveled along the line BC' wouldmean the hiker was way offcourse – BUT IT COULD HAPPEN! This gives us anisosceles triangle.

First, we need to find γ:25

45sin

30

sin This could be the valueof γ or γ'

25

45sin

30

sin

58 or

Since γ is an obtuse angle it cannot be 58°, this must be the value ofγ'.

γ = 122°

γ is part of a linear pair and the other angle is the same as γ' since theyare the base angles of an isosceles triangle.

The angle at B is: 180 – 45 – 122 = 13°

25

45sin13sin AC

AC = distance of hiker from camp 7.95 km.

If the hiker was lost, but measured the distance from the road correctly, B is 77°, therefore the distance would be…

25

45sin77sin

CA

kmCA 5.3445sin

77sin25

If the hiker was badly off course he is about 34.5 km from camp.

Heron’s Formula

Heron’s formula is used to find the area of a triangle when only the lengths of the sides of the triangle are known.

Heron’s Formula

A triangle with sides of length a, b and c has area given by

)2)(2)(2(4

1cPbPaPP

where P is the perimeter of the triangle, P = a + b + c

Ex 6: Find the area of the triangle.

7 9

12

P = 7 + 9 + 12 = 28

)2428)(1828)(1428(2825.0A

514A