Chapter 8E -- Inverse Sine Function · Chapter 8E -- Inverse Sine Function The graph of y=sinx is...

Chapter 8E -- Inverse Sine Function

The graph of y = sin x is beautiful, but clearly not 1-1.

But similar to other cases we studied earlier, the domain could be restricted, so that y = sin x is 1-1 on the restricted domain.

Consider

y = sin x for - π2≤ x ≤ π

2.

Is y = sin x

a 1-1 function on

this restricted domain?

_____________________

Reflect the curve drawn above

through the line y=x.

This new curve is the graph of

_________________________

Domain Range

y = sin x

y = arcsin x = sin−1x

©Fry TAMU Math 150 Chapter 8 E-H Fall 2013

1

y = arcsin x = sin−1x ! ! iff

Note for −1≤ x ≤1 and - π2≤ y ≤ π

2 sin−1(sin y) = and sin(sin−1 x) =

1. sin−1 32

⎛

⎝⎜

⎞

⎠⎟ ! ! ! ! ! ! 2. sin−1 − 2

2

⎛

⎝⎜

⎞

⎠⎟

3. sin−1 sinπ6

⎛⎝⎜

⎞⎠⎟ ! ! ! ! ! ! ! 4. sin sin−1 2

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

5 sin−1 sin 2π3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟! ! ! ! ! ! 6. cos sin−1 2

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟


2

Chapter 8F - Inverse Trigonometric Functions

Like y = sin x , y = cos x is not a 1-1 function unless the domain is restricted.

For y = cos x , we will

restrict the domain to

0 ≤ x ≤ π

Consider the plot

y = cos x for 0 ≤ x ≤ π .

Reflect y = cos x

through the line y = x

This reflected curve is

the curve of

___________________

Domain Range

y = cos x

y = arccos x = cos−1x

! ! y = arccos x =cos−1x iff


3

Note for −1≤ x ≤1 and 0 ≤ y ≤ π cos−1(cos y) = ! and cos(cos−1x) =

1. Evaluate

a) cos−1(−1) =b) cos−1 2

2

⎛

⎝⎜

⎞

⎠⎟ =

c) cos−1 − 12

⎛⎝⎜

⎞⎠⎟ = d) cos−1 3

2

⎛

⎝⎜

⎞

⎠⎟ =

e) cos cos−1 − 1

2⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= f) cos−1 cos 5π

4⎛⎝⎜

⎞⎠⎟ =

Inverse Tangent Function

Recall the graph of y = tan x

To define an inverse function

we limit the domain to

-π2< x < π

2


4

Examine the plot of

y = tan x for

-π2< x < π

2

Is y = tan x a 1-1

function on this

restricted domain?

_____________

!

Reflect y = tan x

through the line

y = x .

This new curve is the graph of ___________________________________

Domain Range

y = tan x

y = arctan x =tan−1x

! ! y = arctan x =tan−1x ! ! iff

Note for −∞ < x < ∞ and -π2< y <π

2 tan−1(tan y) = and tan(tan−1 x) =


5

2. Evaluate

a) arctan(−1) ! ! ! ! ! ! b) arctan 33

⎛

⎝⎜

⎞

⎠⎟

c) sin arctan 45

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟! ! ! ! ! ! d) cos(arctan2)

e) cot sin−1 −1213

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

! ! ! ! ! f) tan(arccos(−3))

g) cos(arctan(−3))


6

Chapter 8G - Law of Sines and Law of Cosines

Given a general triangle, labeled as below

Two interesting truths exist:

A. The Law of Sines! ! ! sinAa

= sinBb

= sinCc

B. The Law of Cosines:! ! ! c2 = a2 + b2 − 2abcosC

Notice that the Pythagorean Theorem is a special case of the Law of Cosines!

In a right triangle , C = so cosC = and we have


7

Understanding the possibilities!

In geometry, we learned that if 2 triangles have corresponding sides of the same length, then they are congruent. It is usually referred to as SSS Congruence Theorem.

So if we are given a set of 3 numbers that represent the lengths of the sides of a triangle, then there are two possibilities: either there is no triangle with those side lengths or there is exactly 1 triangle with those side lengths.

1. If a = 3, b = 4, and c = 8,

we know that no triangle exists

because 3 + 4 < 8

2. If a = 3, b = 4, and c = 5, then we could use the Law of Cosines to determine the measures of the 3 angles.


8

There was also the AAS Congruence Theorem. It told us that if 2 triangles had 2 pairs of corresponding angles congruent and one pair of sides congruent (the side not between the angles), then the triangles were congruent. This could be interpreted another way. If we are given 2 angle measures whose sum is less than 180° and the length of one of the sides of the triangle (that side not being between the given angles), then we can determine the angle measure of the third angle and the length of the other 2 sides of the triangle using the Law of Sines.

3. If A = 75° , C= 65° , and a=3, solve the triangle.


9

The third theorem to consider is the ASA Congruence Theorem. It told us that if 2 triangles had 2 corresponding congruent angles and their included side also congruent, then the triangles themselves were congruent.

4. If If A = 75° , C= 65° , and b = 2, solve the triangle.

Here we can be confident that exactly one solution exists.


10

Finally we get to the congruence theorem that does not exist: SSA. If you have 2 triangles, and you know that 2 sides on one triangle are congruent to 2 sides on the other triangle and one of the angles from one triangle is congruent to an angle on the other triangle, but it’s not the angle between the congruent sides, then you cannot be sure that the triangles are congruent.

For us this means that if we are given 2 numbers that represent the lengths of the sides of a triangle and an angle measure that is not between the given sides, then there are three possibilities for that set of numbers.

1. The given conditions might be such that no triangle exists.

2. The given conditions might be such that exactly 1 triangle exists.

3. The given conditions might be such that 2 different triangle exist.

Example: Triangle ABC does not necessarily exist. If it does, determine the missing parts of the triangle from the given information.

5. A = 40° , a = 20, b = 15


11

6. A = 75° , a = 51, b = 71

7. A = 37° , a = 12, b = 16.1


12

8. From (http://cims.nyu.edu/~kiryl/Precalculus/Section_6.5-The%20Law%20of%20Sines/The%20Law%20of%20Sines.pdf)

a)


13

http://cims.nyu.edu/~kiryl/Precalculus/Section_6.5-The%20Law%20of%20Sines/The%20Law%20of%20Sines.pdf




b)


14

c)


15

9. (From http://www.math.usu.edu/~rheal/online1060/chapter7/7-2.pdf)


16

http://www.math.usu.edu/~rheal/online1060/chapter7/7-2.pdf


10. (From http://www.math.usu.edu/~rheal/online1060/chapter7/7-3.pdf )


17



Chapter 8H - Solving Trigonometric Equations

For what values of θ does sin2θ + cos2θ =1?

sin2θ + cos2θ =1 is true for _________________

because sin2θ + cos2θ =1 is a trigonometric ______________________.

Solving Trigonometric Equations:

1. Solve sinθ = 32


18

2. Solve sin2 θ( ) = 34


19

3. a) Solve cos 3θ( ) = −1

b) Solve cos 3θ( ) = −1 , 0 ≤ θ ≤ 2π


20

4. a) Solve sin x( ) tan x( ) = sin x

b) Which of these solutions is on −π, π[ ]


21

5. Consider the identity sin2x = 2sin xcos x .

It could be rewritten as 12

⎛⎝⎜

⎞⎠⎟ sin2x = sin xcos x .

a) What is the amplitude of the function f (x) = 12

⎛⎝⎜

⎞⎠⎟ sin2x ? ______________________

b) What is the period? _________________! ! c) Plot f (x) = 12

⎛⎝⎜

⎞⎠⎟ sin2x


22

d) For what values of θ does (sin2θ)(cos2θ) =1 ?

e) For what values of θ on −π, π[ ] does 4 (sin2θ)(cos2θ) =1 ?


23

6. Solve cos x = sin2x on −π, π[ ] .


24

7. Solve 2 sin x2

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟2

+ 3sin x2

⎛⎝⎜

⎞⎠⎟ +1= 0 on [−4π , 4π ] .


25

Chapter 8E -- Inverse Sine Function · Chapter 8E -- Inverse Sine Function The graph of y=sinx is...

Documents

Transcript of Chapter 8E -- Inverse Sine Function · Chapter 8E -- Inverse Sine Function The graph of y=sinx is...