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Math 1303 - Part III -Identities:
cos(A + B ) = _______________________________
cos (A – B ) = _____________________________________
sin (A + B ) = _______________________________
sin( A – B ) = ______________________________________
examples:
Find cos 255o . ___________________
find sin ( - 165 o ) = ____________________
Write each of the following as a single trig. function of a single angle.
cos 3 cos 2 - sin 3 cos 2 = __________________________
sin  cos /2 - cos  sin /2 = ____________________________
sin 2 sin  + cos 2 cos  = _____________________________
Other identities:
Double angle Identities
Look at sin 2A. We can write sin 2A = sin ( A + A ) . Now what ?
What about cos 2A ?
By definition of the tangent function we can write similar identities for
tan(A + B ) and the tan 2A .
tan ( A + B ) = ____________________________
tan 2A = _______________
End of Day 17
Exam on Day 18
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Day 19
June 21, 2002
Half-angle Identities –
Use the double angle identity cos 2A = ______________________ and the
Pythagorean identity  ________________
To create two new identities:
sin A =
cos A =
Both of these can also be written as
sin A/2 =
cos A /2 =
2) graph each of the following
b) y = 4cos2 x/4 - 4 sin2 x/2
a) y = 4 sin 3x cos 3x
c) y = 4sinx cos3x - 4cosx sin 3x
3) Let sin A = -3/5 with A not in quadrant III and sin B = 12/13 with B not in quadrant I.
Find
a) sin 2A = ____________
b) cos 2B = __________________
c) sin (A + B ) = ______________________
d) cos ( A + B ) = ______________________
e) What quadrant is A + B in ? ___________________ Why ?
f) sin A / 2 = ___________________
g) cos A / 2 = ____________________
h) tan 2A = ____________________
4) Find sin A if sin (A + B ) = 3/5 with A + B not in Quadrant II and cos B = -3/ 5, B not in
quadrant II.
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HW: page 264: 29, 33, 35, 36, 43, 47, 51,
HW: 37, 39, 41, 43, 45, 47, 51, 57
page 272: 1, 5, 9, 17, 21, 23, 25, 29, 31, 33,
page 278: 1, 5, 9, 15, 17, 19, 23, 33,
Equations
1) Find ALL POSSIBLE solutions of the equations tan x = 0
2) Solve for  if 0 <  < 360o , sin2  - sin  - 1 = 0
3) Solve for x if 0 < x < 2 ,
4)
sin 2 - cos  = 0
cos 2 = -1 , find all  if 0 <  < 2
_
5) \/3 sec = 2
6) If sin  = 2/3 and  is not in QI find
a) sec  = __________
b) cos 2 = ___________
b) sin 2 = _______________
sin 2 = _____________
7) Solve.
csc 3 = 2
8) cos  - sin  = 1
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Another type of equation – parametric equations
A parametric equation is an equation in which both x and y depend on a third variable ( a parameter ).
Sometimes these are useful and sometimes they are necessary.
y = t2
ex. x = 2t
ex. x = sin t
, y = cos t
Relations - Functions – inverse relations – inverse functions
Recall basic ideas about functions and relations
Inverse relations:
Let f(x) = 2x + 4 be given – rewrite in the form y = 2x + 4.
What happens when we interchange the x and the y variables.
Solve for y and label this new relation ( function ? ) g(x) = ____________
Look at the relationship between f(x) and g(x). If ( p, q) is a point on f, give me a point on g.
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Notation: Let f(x) = 2x – 3
We can define
and g(x) = 4 – x
f + g:
f–g:
fg:
f/g:
Begin with a number: say 4 double it and add 3 --after you finish, take the resulting value and square it
In terms of functions:
Let f(x) = _______________
and g(x) = __________
(double your number and add 3 )
(square the value)
Is there one function that does both functions at the “same” time – in the same equation ?
Notation: We write g o f (x) and call it a composition function.
g o f (x) = g ( f (x) ).
ex. from above.
ex. let f(x) = 3x – 2
Find g o f:
find g o f ( 2 ) = __________
g(x) = x + 5
Find f o g :
f o g ( - 1 ) = __________
Certain functions provide special results:
ex. f(x) = 2x – 3
Find f o g ( 1 ) = _______
g(x) = ( x + 3 ) / 2
f o g ( - 2 ) = ____________
This is the idea of inverse relations. Look at the example at the previous page at the beginning of our
discussion of inverse relations. This is one way of creating inverse relations.
Look at trig. functions and how they relate to their inverse relations.
We have solved equations of the form sin  = 2/3 and found the values of  that make this statement
true.
Change the notation of the statement above.
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Function
Domain
Range
f(x) = sin x
f(x) = cos x
f(x) = tan x
f(x) = csc x
f(x) = sec x
f(x) = cot x
all real numbers
except /2, - /2 and
coterminal angles
all real numbers
except 0,  and all
coterminal angles
all real numbers
except /2, 3/2 and
all coterminal angles
all real numbers
except 0,  and all
coterminal angles
Define the inverse relation of following trig. functions. What is the domain and range of each.
y = sin x  ________________
D:
R:
y = cos x  _______________
D:
R:
y = tan x  _______________
D:
R:
Change the range of the three relations so that they represent functions. The domain is still the same.
y = arcsin x = sin-1 x
D:
R:
y= arccos x = cos-1 x
D:
R:
y = arctan x = tan-1 x
D
R:
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Graphs:
y = arcsin x
y=arccos x
y=arctan x
Problems: Simplify each of the following.
1) sin –1 ½ = x  _____________
2) sin-1 ( - 1 ) = x  __________
3) arctan 3  ______________
4) arc cos  = x  _____________
Since All of the above represent angles , we could go back and talk about the topics discussed earlier in terms of these inverse
trig. functions.
5)
sin ( A ) = ?
If A = sin-1 4/7  ____________________
6) cos ( arc cos 7/9 ) = _______________________
7) sin ( tan-1 3/8 ) = ________________________
We can work with identities.
8) sin ( sin-1 3/5 + cos-1 12/13 ) = ________________________
9) sin ( 2 arcsin 5/6 ) = _____________________
10) cos ( 2 sin-1 (- 5/8 ) ) = _____________________________
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We can also solve equations.
11)
12)
Oblique Triangles
Law of Sines
sin A/ a = sin B / b = sin C / c
Can be used if one pair is known ( A, a; B, b; or C, c ) and either another angle or another side.
ex.
ex.
Law of Cosines
c2 = a2 + b2 - 2abcos C
Can be used if all three sides are known or two sides and the included angle.
ex.
ex.
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Examples of these type of problems.
ex. 1
ex. 2
ex.3
The area of a triangle –
several ways to find area - see section 7.4 on page 353.
1) (Two sides and the included angle ) ½ ab sin C
or ½ ac sin B or ½ bc sin A see ex. 1 p. 354
2) ( Two angles and one opposite side ) S = a 2 sin B sin C / 2 sin A see ex. 2 p. 355
3) (Three sides ) see ex. 3 page 357
Heron’s Formula:
____________________
S = \/ s(s – a ) ( s – b) ( s – c ) , where s = ½ ( a + b + c )
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Vectors : magnitude and direction.
Def. Vectors are said to be equal provided __________________________________________
Def. mV is a vector having the same magnitude as V but it is m times as large as vector V.
if m > 0 , then it also has the same direction.
if m < 0, then it has an opposite ( 180o ) direction
Sum and Difference of two vectors.
Def. A vector of length 1 is called a unit vector.
Special vectors of length one: unit vectors in the direction of the positive x-axis and the pos. y-axis.
i: ___________
j : ______________
Ai and Bj are vectors that are parallel to i and j but of length A and B respectively.
Any vector can be written in terms of a horizontal and a vertical component.
ex. Let V = 3i + 4j. Draw it and its vertical and horizontal components.
From this drawing we can define the length of a vector by using the Pythagorean Identity.
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Ch. 8 Complex Numbers
Question Posed by Jerome Cardan –
“ If someone says to you, divide 10 into two parts, one of which multiplied into the other shall produce
40, it is evident that this case or question is impossible.”
With the use of complex numbers we can find the answer to the question posed above.
___
Def. We define i = \/ - 1 . with the condition that i2 = - 1
Def. A complex number is a number of the form a + bi, where a and b are any real numbers and i is
defined as above.
Def. In the notation a + bi
1) a is called the real part of a + bi
2) b is called the imaginary part of a + bi
3) if b = 0, then a + bi is a copy of the real number a.
4) if b  0 , then a + bi is called an imaginary number.
5) The length of a +bi is defined as we did with complex numbers
The modulus of
________
a + bi = | a + bi | = \/ a2 + b2
6) The conjugate of a + bi is the number a - bi
ex. Use - 3 - 4i to find
1) _______
2)_________
5)__________
6)_________
5)__________
6)_________
5)__________
6)_________
ex. Use 4i to find parts 1, 2, 5, and 6
1) _______
2)_________
ex. Use 4 + 0i to find 1, 2, 5, 6
1) _______
2)_________
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Ambiguous Case. side-side-angle
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