chapter 8e -- inverse sine function · chapter 8e -- inverse sine function the graph of y=sinx is...
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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
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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⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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) =
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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))
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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.
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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.
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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.
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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6. A = 75° , a = 51, b = 71
7. A = 37° , a = 12, b = 16.1
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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8. From (http://cims.nyu.edu/~kiryl/Precalculus/Section_6.5-The%20Law%20of%20Sines/The%20Law%20of%20Sines.pdf)
a)
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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9. (From http://www.math.usu.edu/~rheal/online1060/chapter7/7-2.pdf)
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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10. (From http://www.math.usu.edu/~rheal/online1060/chapter7/7-3.pdf )
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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3. a) Solve cos 3θ( ) = −1
b) Solve cos 3θ( ) = −1 , 0 ≤ θ ≤ 2π
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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4. a) Solve sin x( ) tan x( ) = sin x
b) Which of these solutions is on −π, π[ ]
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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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
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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d) For what values of θ does (sin2θ)(cos2θ) =1 ?
e) For what values of θ on −π, π[ ] does 4 (sin2θ)(cos2θ) =1 ?
©Fry TAMU Math 150 Chapter 8 E-H Fall 2013
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