Transcript B - mor media international

Trigonometry—Applying ASTC, Reference, Coterminal Angles
In the old days, when people didn’t have a calculator and only had
a table of reference for the trigonometric function values of angles
from 0° to 90°, it’s crucial to know how to express the function in
terms of a positive acute angle (i.e., its reference angle).
Express as a function of a positive acute angle.
1. sin 123° =
4. sin 456° =
2. cos 234° =
5. cos 12,345° =
3. tan –345° =
6. tan –6,789° =
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S
A
T
C
30°
How to find the sine, cosine, tangent of special angles without a calculator:
A special angle is an angle that are multiples of 30, 45 and 60.
Examples:
1. sin 120 =
2. cos 135 =
3. tan 225 =
60°
45°
4. sin 210 =
5. cos 300 =
6. tan 330 =
45°
Trigonometry—Secant, Cosecant and Cotangent
Page 16
The Other Three Functions
We have learned the 3 basic definitions—sine, cosine and tangent. There are 3
more definitions—secant (sec), cosecant (csc) and cotangent (cot).
Hypotenuse
Opposite leg

Adjacent leg
The basic three:
Opposite leg O
sin  

Hypotenuse H
Adjacent leg A
cos 

Hypotenuse H
Opposite leg O
tan  

Adjacent leg A
The other three:
Hypotenuse H
sec 

Adjacent leg A
Hypotenuse H
csc 

Opposite leg O
Adjacent leg A
cot  

Opposite leg O
It turns out most people don’t remember the definitions of secant, cosecant and
cotangent, instead, they remember them as the ____________ of the basic three.
sec 
1
csc 
1
cot  
1
Q: How can we memorize the other three as the reciprocals the basic
three without mixing them up?
A: Just remember they pair each other with a different initial letter:
tan 
sec 
csc 
Trigonometry—Secant, Cosecant and Cotangent (cont’d)
If  is in Q I, we know:
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sin  = (+), cos  = (+), tan  = (+). What about ...
csc  = ( ), sec  = ( ), cot  = ( )
If  is in Q II, we know: sin  = (+), cos  = (–), tan  = (–). What about ...
csc  = ( ), sec  = ( ), cot  = ( )
If  is in Q III, we know: sin  = (–), cos  = (–), tan  = (+). What about ...
csc  = ( ), sec  = ( ), cot  = ( )
If  is in Q IV, we know: sin  = (–), cos  = (+), tan  = (–). What about ...
csc  = ( ), sec  = ( ), cot  = ( )
How to use calculator to compute cosecant, secant and cotangent:
Unlike sin, cos and tan, most (maybe all) calculators do not have the
built-in functions for csc, sec and cot. In order to evaluate, for
example, csc 41, one must use sin key in the calculator by entering:
1  sin 41 (since csc 41 = 1/sin 41). The only time one should not
use a calculator is when the angle is a special angle, i.e., 30, 45, 60
and multiples of these angles.
Examples:
1. csc 41 =
2. cot 65 =
3. sec 32 =
4. csc 45 =
5. cot 60 =
6. sec 30 =
S
A
T
C
Trigonometry—Introduction to Identities
Page 18
Identities
When two sides of an equation are equal regardless of what the value for the
variable is, it’s called an identity. For example, x + x = 2x is an identity. In
trigonometry, we have many identities, one of which we have seen already:
Reciprocal Identities
1
sec 
cos
1
csc 
sin 
1
cot  
tan 
sin  
sin(90   ) 
cos  
cos(90   ) 
tan  
tan(90   ) 
sec 
sec(90   ) 
csc 
csc(90   ) 
cot  
cot(90   ) 
Other identities includes (concluded from the right triangle above):
Cofunction Identities
Quotient Identities
90°–
r
y

x
Trigonometry—sin 2 + cos 2 = 1
Page 19
One Important Identity: sin 2 + cos 2 = 1
Let see if it’s true when  = 45°, 120°, 36789°.
?
sin 2 45° + cos 2 45° = 1
?
sin 2 120° + cos 2 120° = 1
Why is always true?
Write an equation that connects , y and r:
r

x
y
Write an equation that connects , x and r:
Write an equation that connects y, x and r:
Pythagorean Identities:
sin 2  + cos 2  = 1
tan 2  + 1 = sec 2 
?
sin 2 36789° + cos 2 36789° = 1
1 + cot 2  = csc 2 
Trigonometry—Proving Identities (cont’d)
Page 20
So far, we’ve learned some basic identities: Cofunction Identities and
Reciprocal Identities
1
sec 
cos
1
csc 
sin 
1
cot  
tan 
Quotient Identities
sin 
tan  
cos 
cos 
cot  
sin 
Pythagorean Identities
sin 2  + cos 2  = 1
tan 2  + 1 = sec 2 
1 + cot 2  = csc 2 
We will learn more fundamental identities, but in the mean time, let’s use these
ones to prove some identities that are not so obvious:
1. tan   cot   sec csc
2.
1  sin 
cos

 2sec
cos
1  sin 
Trigonometry—Proving Identities
Page 21
Basic Trigonometric Identities:
Reciprocal Identities
1
sec 
cos
1
csc 
sin 
1
cot  
tan 
Ex. 3:
(sec  tan  ) 2 
1  sin 
1  sin 
Quotient Identities
sin 
tan  
cos 
cos 
cot  
sin 
Pythagorean Identities
sin 2  + cos 2  = 1
tan 2  + 1 = sec 2 
1 + cot 2  = csc 2 
Techniques of proving identities:
1. Use the basic identities wisely.
– Change everything in terms of sine and/or cosine
– If you see (trig. function)2, it might have to do with Pyth. Id.
– Treat an identity in different ways: sin 2  + cos 2  = 1 
sin 2  = 1 – cos 2  = 1  _________________
2. Try to keep one side the same, and only simplify the
complicated side or the side you can do something with.
3. If you need to change both sides, change each side
independently.
4. Basic algebraic manipulation still apply—FOILing,
expanding, factoring, simplifying, rationalizing
denominator, and combining two fractional expressions
into one.
5. The goal is to show left hand side = right hand side.
Trigonometry—Law of Cosines
Page 22
Law of Cosines: c2 = a2 + b2 – 2ab cos C
This side and this angle
are opposite of each other, and
one of them should be the unknown

10
x
To find x (of the triangle shown on the right), we use the Law of
Cosines by setting up the equation as follows:
x2 =

30°
12
The Law of Cosines (LoC) not only can be
used to find an unknown side, it can be also
used to find an unknown angle. For example,
we can use LoC to find  too:
Why is true?
Let’s prove it:
B
c
a
C
A
b
Trigonometry—2nd Formula for the Area of the Triangle
Page 23
Area of a Triangle: A = ½ab sin C (derived from A = ½bh, of course)
B
Proof:
a
c
h
C
b
In words, the area of a triangle is _____________________________________________
Problems:
1.
2.
60°
7
8
30°
10
3.
11
5
6
9
A