Download Chapter 7: Analytic Trigonometry

Lecture Notes
Algebra II Honors
Chapter 7: Analytic Trigonometry
7.1 Trigonometric Identities
Below are the basic trig identities discussed in previous chapters.
Reciprocal
csc(x) =
1
sin(x)
sec(x) =
1
cos(x)
sin(x)
cos(x)
cot(x) =
cos(x)
sin(x)
cot(x) =
1
tan(x)
Quotient
tan(x) =
Pythagorean
sin 2 (x) + cos 2 (x) = 1 1 + tan 2 (x) = sec2 (x) 1 + cot 2 (x) = csc2 (x)
Even-Odd
sin(–x) = –sin(x)
cos(–x) = cos(x)
tan(–x) = –tan(x)
Cofunction (the function of angle u equals the cofunction of the complementary angle)
⎛π
⎞
sin ⎜ − u ⎟ = cos ( u )
⎝2
⎠
⎛π
⎞
cos ⎜ − u ⎟ = sin ( u )
⎝2
⎠
⎛π
⎞
tan ⎜ − u ⎟ = cot ( u )
⎝2
⎠
⎛π
⎞
cot ⎜ − u ⎟ = tan ( u )
⎝2
⎠
⎛π
⎞
sec ⎜ − u ⎟ = csc ( u )
⎝2
⎠
⎛π
⎞
csc ⎜ − u ⎟ = sec ( u )
⎝2
⎠
These may be used to establish proofs of other identities, to solve trig equations, or to simplify
problems in calculus.
In order to prove the identity below, we work to turn the LHS into the RHS.
We agree to work on one side only of the identity.
Often it is helpful to put everything in terms of sin and cos and find common denominators.
1 + csc(x)
= sec(x)
cos(x) + cot(x)
1
sin(x) + 1
1+
sin(x)
sin(x)
→
cos(x) cos(x)
cos(x) ⋅ sin(x) + cos(x)
+
1
sin(x)
sin(x)
sin(x) + 1
sin(x)
1
⋅
→
→ sec(x)
sin(x) cos(x) [sin(x) + 1]
cos(x)
Example : Pr ove
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Lecture Notes
Algebra II Honors
1 − cos(x)
sin(x)
. Work with the RHS and multiply by conjugate.
=
sin(x)
1 + cos(x)
sin(x) 1 − cos(x) sin(x) ⋅ [1 − cos(x) ] sin(x) ⋅ [1 − cos(x) ]
⋅
=
=
1 + cos(x) 1 − cos(x)
1 − cos 2 (x)
sin 2 (x)
1 − cos(x)
=
sin(x)
Example: Prove
Example: Substitute x = 3 · sin(θ) into the following expression and simplify. Take 0 ≤ θ < π/2
9 − x 2 = 9 − ( 3 ⋅ sin(θ) ) = 9 − 9sin 2 (θ)
2
9 ⋅ ⎡⎣1 − sin 2 (θ) ⎤⎦ = 9 ⋅ cos 2 (θ) = 3 ⋅ cos(θ)
Note: you must memorize the basic identities. You need not write the reasons for each step, but
show every step. Remember you are only allowed to work on one side of an identity. You are not
solving an equation; do not move quantities from one side to another, etc.
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Lecture Notes
Algebra II Honors
7.2 Addition and Subtraction Formulas
sin ( x + y ) = sin ( x ) ⋅ cos ( y ) + cos ( x ) ⋅ sin ( y )
sin ( x − y ) = sin ( x ) ⋅ cos ( y ) − cos ( x ) ⋅ sin ( y )
cos ( x + y ) = cos ( x ) ⋅ cos ( y ) − sin ( x ) ⋅ sin ( y )
cos ( x − y ) = cos ( x ) ⋅ cos ( y ) + sin ( x ) ⋅ sin ( y )
tan ( x + y ) =
tan ( x − y ) =
tan ( x ) + tan ( y )
1 − tan ( x ) ⋅ tan ( y )
tan ( x ) − tan ( y )
1 + tan ( x ) ⋅ tan ( y )
These are proved in the text or left for exercises.
•Example: Find exactly sin(15°) = sin(π/12)
sin (15° ) = sin ( 45° − 30° ) = sin ( 45° ) ⋅ cos ( 30° ) − cos ( 45° ) ⋅ sin ( 30° )
2 3
2 1
6− 2
⋅
−
⋅ =
2 2
2 2
4
⎛π
⎞
− u ⎟ = cos ( u )
⎝2
⎠
⎛π
⎞
⎛π⎞
⎛π⎞
sin ⎜ − u ⎟ = sin ⎜ ⎟ cos ( u ) − cos ⎜ ⎟ sin ( u ) =
⎝2
⎠
⎝2⎠
⎝2⎠
•Example: Prove sin ⎜
1 ⋅ cos ( u ) − 0 ⋅ sin ( u ) = cos ( u )
Another helpful identity converts a sum of sin and cos into a single sin function.
For A and B real numbers, let k = A 2 + B2 and let φ satisfy
cos ( φ ) =
A
and sin ( φ ) =
B
A 2 + B2
A 2 + B2
•Then A ⋅ sin ( x ) + B ⋅ cos ( x ) = k ⋅ sin ( x + φ )
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Lecture Notes
Algebra II Honors
7.3 Double Angle, Half Angle, and Product ↔ Sum Formulas
Double Angle
sin ( 2x ) = 2 ⋅ sin ( x ) ⋅ cos ( x )
cos ( 2x ) = cos 2 ( x ) − sin 2 ( x ) = 1 − 2 ⋅ sin 2 ( x ) = 2 ⋅ cos 2 ( x ) − 1
tan ( 2x ) =
Lowering Powers
sin 2 ( x ) =
2 ⋅ tan ( x )
1 − tan 2 ( x )
1 − cos ( 2x )
2
cos 2 ( x ) =
1 + cos ( 2x )
2
tan 2 ( x ) =
1 − cos ( 2x )
1 + cos ( 2x )
Half Angle: Choose the correct sign for the quadrant of x/2
1 − cos ( x )
1 + cos ( x )
⎛x⎞
⎛x⎞
sin ⎜ ⎟ = ±
cos ⎜ ⎟ = ±
2
2
⎝2⎠
⎝2⎠
sin ( x )
⎛ x ⎞ 1 − cos ( x )
tan ⎜ ⎟ =
=
sin ( x )
1 + cos ( x )
⎝2⎠
Product to Sum
1
⋅ ⎡sin ( x + y ) + sin ( x − y ) ⎤⎦
2 ⎣
1
cos ( x ) ⋅ sin ( y ) = ⋅ ⎡⎣sin ( x + y ) − sin ( x − y ) ⎤⎦
2
1
cos ( x ) ⋅ cos ( y ) = ⋅ ⎡⎣cos ( x + y ) + cos ( x − y ) ⎤⎦
2
1
sin ( x ) ⋅ sin ( y ) = ⋅ ⎡⎣cos ( x − y ) − cos ( x + y ) ⎤⎦
2
sin ( x ) ⋅ cos ( y ) =
Sum to Product
⎛x+y⎞
⎛x−y⎞
sin ( x ) + sin ( y ) = 2 ⋅ sin ⎜
⎟ ⋅ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛x+y⎞
⎛x−y⎞
sin ( x ) − sin ( y ) = 2 ⋅ cos ⎜
⎟ ⋅ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛x+y⎞
⎛x−y⎞
cos ( x ) + cos ( y ) = 2 ⋅ cos ⎜
⎟
⎟ ⋅ cos ⎜
⎝ 2 ⎠
⎝ 2 ⎠
⎛x+y⎞
⎛x−y⎞
cos ( x ) − cos ( y ) = −2 ⋅ sin ⎜
⎟ ⋅ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
Proofs may be found in the text or are left as exercises for the reader.
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Lecture Notes
Algebra II Honors
•Example: Prove tan ( 3x ) =
tan ( 3x ) = tan ( 2x + x ) =
3 ⋅ tan ( x ) − tan 3 ( x )
1 − 3 ⋅ tan 2 ( x )
tan ( 2x ) + tan ( x )
1 − tan ( 2x ) ⋅ tan ( x )
=
2 tan ( x )
tan ( x )
3 ⋅ tan ( x ) − tan 3 ( x )
+
1 − tan 2 ( x )
1
1 − tan 2 ( x )
=
1 − 3 ⋅ tan 2 ( x )
1 2 tan ( x ) tan ( x )
−
⋅
1 1 − tan 2 ( x )
1
1 − tan 2 ( x )
3 ⋅ tan ( x ) − tan 3 ( x )
1 − tan 2 ( x )
⋅
1 − tan 2 ( x )
1 − 3 ⋅ tan 2 ( x )
=
3 ⋅ tan ( x ) − tan 3 ( x )
1 − 3 ⋅ tan 2 ( x )
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Lecture Notes
Algebra II Honors
7.4 Inverse Trigonometric Functions
You are given 1/2 = sin(x) and asked to find x. You might try a graphical approach, looking for
the intersection(s) of y = 1/2 and y = sin(x).
1.0
0.5
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-5
5
10
- 0.5
- 1.0
There seem to be many possible solutions. This is similar to knowing 1/4 = x2 and finding
x = ±1/2. Such situations are very confusing to calculators, which desperately want to return an
answer but can only return a single value. We solved this problem by introducing a principal
square root function (√ ) as the inverse of y = x2 that only returns the positive root. Essentially we
restricted the domain of y = x2 to only 0 ≤ x < ∞ so that the range of y = √x would be the same.
Thus we made a one-to-one function of the restricted square function that had as an inverse the
principal square root function.
Let's try something similar. We can restrict the domain of y = sin(x) so that the resulting function
(called the restricted sine function y = Sin(x) - note the capital S, it's quite sensitive about that) is
one-to-one. There are many ways to do this. Do it my way, and we'll all get along famously.
Choose the domain of y = Sin(x) to be [–π/2, π/2]. Its range is [–1, 1], and as you can see below,
it is one-to-one. For this function, define an inverse function with domain [–1, 1] and range
[–π/2, π/2]. Recall the x and y values swap between function and inverse, their graphs being
reflections in the line y = x.
What to call the inverse of the restricted sine function? Back in the day, when everyone was
taking trig functions of angles, the notation was y = Arcsin(x), meaning y was the angle whose
Sine was x. The math went all loopy and decided to take trig functions of real numbers, and the
(bad) notation that evolved has us today calling the inverse function y = Sin–1(x). The –1 in the
exponent is not a real exponent - it just signifies inverse. In particular, Sin–1(x) ≠ 1 / Sin(x).
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Lecture Notes
Algebra II Honors
y = Sin@xD
1.0
0.5
-1.5
- 1.0
- 0.5
0.5
1.0
1.5
- 0.5
- 1.0
y = ArcSin@xD
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1.0
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-1.0
-0.5
0.5
1.0
- 0.5
- 1.0
- 1.5
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Lecture Notes
Algebra II Honors
Now when you ask your calculator to find sin–1(1/2) it reports 30° or π/6 radians.
It always returns an angle in I if your x value is positive, and an angle in IV if your x value is
negative. (Try it: sin–1(–1/2) = –30° or – π/6 radians.)
Note: Sin(Sin–1(x)) = x, for –1 ≤ x ≤ 1 and Sin–1(Sin(x)) = x, for –π/2 ≤ x ≤ π/2
Be careful of the domains!
•Example: Sin–1(sin(210°)) = ? Not 210° - that's not in the domain of Sin–1. Proceed as follows:
Sin–1(–1/2) = –30°. These inverse functions must always return a value from –90° to +90°.
What about inverses for all the other trig functions? It is all very similar to what we have done for
Sin–1, but some of the domain restrictions may at first glance seem a bit odd. Let's just assume
there is some sort of "useful in calculus" reason and make the best of it.
y = Cos–1(x) or y = Arccos[x] has domain [–1, 1] and range [0, π]
Note this function return values in I (for +x) or II (for –x)
y = Tan–1(x) or y = Arctan(x) has domain [–∞, ∞] and range (–π/2, π/2)
y = Csc–1(x) or y = Arccsc(x) has domain (–∞, –1] ∪ [1, ∞) and range (0, π/2] ∪ [π, 3π/2)
y = Sec–1(x) or y = Arcsec(x) has domain (–∞, –1] ∪ [1, ∞) and range [0, π/2) ∪ (π, 3π/2]
y = Cot–1(x) or y = Arccot(x) has domain [–∞, ∞] and range (0, π)
Graphs of these functions may be found in the text. Note there are no keys on the TI-86
corresponding to the last three functions above. To find Sec–1(2), think sec(θ) = 2 = 1 / cos(θ)
Thus you wish to find the angle θ such that cos(θ) = 1/2 or Cos–1(1/2) = 60°.
•Example: Find Cos–1(–1/2). This should return an angle in II and it does: 120° or 2π/3 radians
•Example: Find cos(Sin–1(3/5)) exactly. We want cos(θ), where θ = Sin–1(3/5) or sin(θ) = 3/5.
2
4
⎛3⎞
We could use an identity: cos ( θ ) = ± 1 − sin ( θ ) = ± 1 − ⎜ ⎟ = (choose + as Sin–1(+) in I)
5
⎝5⎠
2
However, a sketch of a right triangle with side opposite θ = 3, hypotenuse 5, and hence adjacent
to θ having side of 4 let's us see the cos(θ) = 4/5 easily.
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Lecture Notes
Algebra II Honors
7.5 Trigonometric Equations
Equations with trig functions are to be solved for the numbers or angles that make them true.
•Example: Solve 2 · cos(θ) – 1 = 0.
Isolate a single trig function as you would in solving a regular equation: cos(θ) = 1/2
There are infinitely many solutions; between 0° and 360°, both 60° and 300° have cos = 1/2.
We write: θ = 60° + 360° · n or 300° + 360° · n, where n is any integer.
Alternatively, if solving for real numbers or angles in radians: π/3 + 2πn or 5 π/3 + 2πn.
•Example: Solve sin2(x) = 4 – 5 · cos2(x) for x on the interval [0, 2π).
Recall sin2(x) = 1 – cos2(x). The equation becomes: 1 – cos2(x) = 4 – 5 · cos2(x) →
4 · cos2(x) = 3 → cos(x) = ±√3/2 → x = π/6, 5π/6, 7π/6, and 11π/6.
•Example: Solve
3 ⋅ sin ( 2x ) = cos ( 2x )
Be careful when dividing both sides of an equation by a variable. Check that if the variable is 0,
you are not losing solutions. Dividing both sides here by sin(2x) is safe; no values that makes
sin(2x) = 0 also make cos(2x) = 0, so we are losing no solutions.
3 = cot ( 2x ) or tan ( 2x ) =
3
→ 2x = 30° + 180° · n or 2x = 210° + 180° · n
3
(Recall the periods of tan and cot are 180°.) Thus x = 15° + 90° · n or 105° + 90° · n
This collapses to x = 15° + 90° · n, where n is any integer.
•Example: Solve sin(2x) = x graphically. Graph both, and use ISECT on the TI-86.
1.5
1.0
0.5
-1.5
- 1.0
- 0.5
0.5
1.0
1.5
- 0.5
- 1.0
- 1.5
We find x = –0.9477, 0, and 0.9477 satisfy the given equation.
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