17-1 Trigonometric Functions in Triangles Chapter 17 17-1 Trigonometric Functions in Triangles

Special Right Triangles CONCEPT SUMMARY 2x x 30-60-90 Triangle 45-45-90 Triangle The hypotenuse is 2 times the short leg! The hypotenuse is the times the leg! The long leg is the times the short leg Special Right Triangles

2x x

2x x

10 is the short leg since it is opposite the 30° Angle Hypotenuse = 2(Short Leg) y = 2(10) y = 20

Definitions

Exact Trig Values CONCEPT SUMMARY 2x x Trig Values

Find the value of each variable Find x first Find y next Exact Value Approximation Example 4-1a

Find the value of each variable Find x first Find y next Exact Value Approximation Example 4-1b

Find the value of each variable x = 12(cos 63) = 5.4479 y = 12(sin 63) = 10.6921 Example 4-1c

SHORT-RESPONSE TEST ITEM A wheelchair ramp is 3 meters long and inclines at Find the height of the ramp to the nearest tenth centimeter. Example 5-2a

Answer: The height of the ramp is about 0.314 meters, Y W Multiply each side by 3. Simplify. Answer: The height of the ramp is about 0.314 meters, Example 5-2b

Method 2 The horizontal line from the top of the platform to which the wheelchair ramp extends and the segment from the ground to the platform are perpendicular. So, and are complementary angles. Therefore, Y W Example 5-2c

Answer: The height of the ramp is about 0.314 meters, Multiply each side by 3. Simplify. Answer: The height of the ramp is about 0.314 meters, Example 5-2d

Answer: The roller coaster car was about 285 feet above the ground. SHORT-RESPONSE TEST ITEM A roller coaster car is at one of its highest points. It drops at a angle for 320 feet. How high was the roller coaster car to the nearest foot before it began its fall? Answer: The roller coaster car was about 285 feet above the ground. Example 5-2e

Evaluate the six trigonometric functions of the angle  shown in the right triangle. 13 SOLUTION The sides opposite and adjacent to the angle are given. To find the length of the hypotenuse, use the Pythagorean Theorem.

If , find the other 5 trig functions. 17 8 15 SOLUTION Draw a triangle such that one angle has the given cosine value use the Pythagorean theorem

HW #17-1 Pg 732 1-23

17-2 More fun with Trigonometric Functions Chapter 17 17-2 More fun with Trigonometric Functions

ANGLES IN STANDARD POSITION

GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS (x, y) r Let be an angle in standard position and (x, y) be any point (except the origin) on the terminal side of . The six trigonometric functions of are defined as follows.

GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS sin = y r csc = , y  0 r y y y r r

GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS cos = x r sec = , x  0 r x x r x r

tan = , x  0 y x cot = , y  0 x y y y x x GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS tan = , x  0 y x cot = , y  0 x y y y x x

r r = x 2 +y 2. (x, y) Pythagorean theorem gives GENERAL DEFINITION OF TRIGNONOMETRIC FUNCTIONS r (x, y) Pythagorean theorem gives r = x 2 +y 2.

Let  be an angle in standard position and let P(x, y) be a point on the terminal side of  . Using the Pythagorean Theorem, the distance r from the origin to P is given by . The trigonometric functions of an angle in standard position may be defined as follows.

terminal side of an angle in standard position. Evaluate the Evaluating Trigonometric Functions Given a Point Let (3, – 4) be a point on the terminal side of an angle in standard position. Evaluate the six trigonometric functions of . r (3, – 4) SOLUTION Use the Pythagorean theorem to find the value of r. r = x 2 + y 2 = 3 2 + (– 4 ) 2 = 25 = 5

you can write the following: Evaluating Trigonometric Functions Given a Point r Using x = 3, y = – 4, and r = 5, you can write the following: (3, – 4) 4 5 sin = = – y r csc = = – r y 5 4 cos = = x r 3 5 sec = = r x 5 3 tan = = – y x 4 3 cot = = – x y 3 4

' The values of trigonometric functions of angles greater than 90° (or less than 0°) can be found using corresponding acute angles called reference angles. Let be an angle in standard position. Its reference angle is the acute angle (read theta prime) formed by the terminal side of and the x-axis. '

' ' '  90 < < 180; < < 2 = 180 – – Degrees: 90 < < 180;  2 < < ' = 180 Degrees: ' – Radians: =  – '

' ' ' 3   180 < < 270; < < 2 = – 180 – Degrees: 180 < < 270;  3 2 < < ' – 180 = Degrees: ' Radians: = –  '

' ' ' 3 2 270 < < 360; < < 2 = 360 – 2 – Degrees: 270 < < 360; 2 3 2 < < ' – 360 = Degrees: ' Radians: = – 2 '

' ' Use these steps to evaluate a trigonometric function of Evaluating Trigonometric Functions Given a Point CONCEPT SUMMARY EVALUATING TRIGONOMETRIC FUNCTIONS Use these steps to evaluate a trigonometric function of any angle . Find the reference angle . ' 1 2 Evaluate the trigonometric function for angle . ' 3 Use the quadrant in which lies to determine the sign of the trigonometric function value of .

Signs of Function Values Evaluating Trigonometric Functions Given a Point CONCEPT SUMMARY EVALUATING TRIGONOMETRIC FUNCTIONS Signs of Function Values Quadrant I Quadrant II Quadrant III Quadrant IV sin , csc : + sin , csc : + cos , sec : – cos , sec : + tan , cot : – tan , cot : + sin , csc : – sin , csc : – cos , sec : – cos , sec : + tan , cot : + tan , cot : –

' ' Evaluate tan (– 210). SOLUTION Using Reference Angles to Evaluate Trigonometric Functions Evaluate tan (– 210). ' = 30 SOLUTION = – 210 The angle – 210 is coterminal with 150°. The reference angle is = 180 – 150 = 30. ' The tangent function is negative in Quadrant II, so you can write: tan (– 210) = – tan 30 = – 3

HW #17-2 Pg 739-740 1-61 Odd, 62-63

17-3 Radians, Cofunctions, and Problem solving Chapter 17 17-3 Radians, Cofunctions, and Problem solving

Theorem 17-1 The radian measure  of a rotation is the ratio of the distance s traveled by a point at a radius r from the center of rotation to the length of the radius.

One Radian is the measure of an angle in standard position whose terminal side intercepts an arc of length r. The arc length s of a sector with radius r and central angle  is given by the formula: s = r 

A B

Linear Speed Angular Speed Linear Speed

A child is spinning a rock at the end of a 2-foot rope at the rate of 180 revolutions per minute (rpm). Find the linear speed of the rock when it is released.

The rock is moving around a circle of radius r = 2 feet. The angular speed of the rock in radians is:

The linear speed of the rock is: The linear speed of the rock when it is released is 2262 ft/min  25.7 mi/hr.

Linear Speed on Earth Earth rotates on an axis through its poles Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 40° north latitude is about 3033.5 miles. Therefore, a location on Earth at 40° north latitude is spinning on a circle of radius 3033.5 miles. Compute the linear speed on the surface of Earth at 40° north latitude.

Co-Function Identities The sine of an angle is the cosine of the complement and the cosine of an angle is the sine of the complement The same is true of each trig function and its co-function

HW #17-3 Pg 746-747 1-47 Odd, 48-49

17-4 Finding Function Values Chapter 17 17-4 Finding Function Values

Let's label the unit circle with values of the tangent Let's label the unit circle with values of the tangent. (Remember this is just y/x)

 = - 360° + 45°  = - 315°  = 45°  = 360° + 45° = 405° What is the measure of this angle? You could measure in the positive direction and go around another rotation which would be another 360°  = - 360° + 45°  = - 315°  = 45° You could measure in the positive direction  = 360° + 45° = 405° You could measure in the negative direction There are many ways to express the given angle. Whichever way you express it, it is still a Quadrant I angle since the terminal side is in Quadrant I.

If the angle is not exactly to the next degree it can be expressed as a decimal (most common in math) or in degrees, minutes and seconds (common in surveying and some navigation). 1 degree = 60 minutes 1 minute = 60 seconds  = 25°48'30" degrees seconds minutes To convert to decimal form use conversion fractions. These are fractions where the numerator = denominator but two different units. Put unit on top you want to convert to and put unit on bottom you want to get rid of. Let's convert the seconds to minutes 30" = 0.5'

 = 25°48'30" = 25°48.5' = 25.808° 48.5' = .808° 1 degree = 60 minutes 1 minute = 60 seconds  = 25°48'30" = 25°48.5' = 25.808° Now let's use another conversion fraction to get rid of minutes. 48.5' = .808°

What is the length of this segment?

HW #17-4 Pg 753 1-73 Odd, 74-78

17-6 Trig Functions and Relationships Chapter 17 17-6 Trig Functions and Relationships

Let’s consider the length of this segment?

HW #17-6 Pg 767-768 1-33 Odd, 34-37

17-8 Algebraic Manipulations Chapter 17 17-8 Algebraic Manipulations

Row 2, 4, 6 Row 1. 3. 5

Row 2, 4, 6 Row 1. 3. 5

Row 1, 2, 3. 4, 5, 6

HW #17-8 Pg 772-773 1-47 Odd, 48-51 

Test Review

Determine the quadrant in which the terminal side of the angle lies

Find one positive angle and one negative angle coterminal with the given angle.

Rewrite each degree measure in radians and each radian measure in degrees.

Find the arc length of a sector with the given radius r and central angle 

Evaluate the trigonometric function without using a calculator.

Evaluate the trigonometric function without using a calculator.

Find the values of the other five trigonometric functions of .

Verify the identity.

Verify the identity.

HW #R-17 Pg 776-778 1-46