MTH 112: Elementary Functions
MTH 112: Elementary Functions
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7.3:Trigonometric equations
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Types of equations
MTH 112: Elementary Functions
An identity is an equation that is true for all values of the variable for
which the expressions are define.
Section 7.3: Trigonometric Equations
Find and use reference angles.
A conditional equation is true for some, but not all, values of the
variable.
Solve trigonometric equations and applications.
Solve inverse trigonometric equations.
A contradiction is an equation that is true for NO values of the variable.
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
√
3
?
Which type of equation is cos(x) + 2 = −
2
7.3:Trigonometric equations
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
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7.3:Trigonometric equations
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Reference angle
Reference angle
For a given angle θ , the reference angle is the positive, acute angle,
θR made by the terminal side and the x-axis.
Boris Iskra
Department of Mathematics. Oregon State University
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
MTH 112: Elementary Functions
7.3:Trigonometric equations
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7.3:Trigonometric equations
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Reference Angle
quadrant
I
II
III
IV
θR (reference angle in degrees)
θR = θ
θR = 180◦ − θ
θR = θ − 180◦
θR = 360◦ − θ
θR (reference angle in radians)
θR = θ
θR = π − θ
θR = θ − π
θR = 2π − θ
The trigonometric function values for θ and the reference angle θR differ by
at most a sign (+ or -).
sin(θ ) = y,
Examples
sin(θR ) = |y|,
Find the reference angle θR for the given angle θ .
3
θ = 240◦ .
5π
θ=
6
θ = −225◦
4
θ = 5 rad
1
2
cos(θ ) = x and tan(θ ) =
y
x
cos(θR ) = |x| and tan(θR ) =
y
.
x
Example 1
On of the previous examples was finding the reference angle θR for θ .
Find sine, cosine and tangent of both angles and compare.
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
7.3:Trigonometric equations
7/18
7.3:Trigonometric equations
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Solving trigonometric equations
Example 2
1
2
in the interval [0, 2π).
1
Find all solutions of sin(x) =
2
Find ALL solution of the equation sin(x) = 12 .
Since sin(π/6) = 12 we know that π/6 is a solution for the
equation and that θ = π/6 is the reference angle for any other
solution.
Note that sin(x) is positive for angles x in quadrant II. The angle
5π
π
θ=
is a solution, since is its reference angle and no other
6
6
angle in [0, 2π) has sine 1/2. The solution set is {π/6, 5π/6}.
Boris Iskra
Department of Mathematics. Oregon State University
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
MTH 112: Elementary Functions
7.3:Trigonometric equations
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2)
7.3:Trigonometric equations
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Solving Trigonometric equations involving multiple angles
Example 3
Recall that the sine function has period 2π , then for any integer k:
Find all solutions of cot(2x) = 1.
1
+ 2πk = sin(π/6) = .
6
2
5π
1
sin
+ 2πk = sin(5π/6) = .
6
2
sin
π
Let θ = 2x. Then
cot(θ ) = 1
Recall that the period of cotangent is π. Search initially for values in [0, π)
Solution set:
π
+ 2πk,
6
5π
+ 2πk, with k integer
6
One solution is given by π4 . Now, contangent function is negative on
quadrant II. Therefore, the only solution to the equation in the interval
[0, π) is π4 .
All solutions of cot(θ ) = 1 are of the form
θ=
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
Boris Iskra
π
+ πk, for k an integer.
4
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
7.3:Trigonometric equations
11/18
7.3:Trigonometric equations
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Example continues
Since θ = 2x:
π π
+ k, for k an integer.
8 2
You can check that these solutions are correct. How?
x=
Boris Iskra
Department of Mathematics. Oregon State University
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
MTH 112: Elementary Functions
7.3:Trigonometric equations
13/18
7.3:Trigonometric equations
Solving a trig equation in factored form
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Example 4
Find all solutions of
3 cos2 (x) = 2 cos(x) + 1 in the interval [0, 2π).
(sin(x) − 2)(csc(x) − 2) = 0 on the interval [0, 2π).
We recognize it as a quadratic equation in cos(x). Strategy:
Solve this quadratic equation in cos(x).
Then, solve it for x.
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
7.3:Trigonometric equations
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7.3:Trigonometric equations
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Examples continues
Example 5
We start by observing that the reference angle is giving by
p
tR = tan−1 ( 5/3)
Solve
2
3 tan (t) − 5 = 0
in the interval [0, 2π).
Note that the tangent is positive whenever the argument (angle) is in
quadrant I or quadrant III.
Solutions:
t = tR and t = π + tR .
Equivalently,
5
tan2 (t) = .
3
Solve
r
tan(t) = ±
In this exercise, MML is asking to approximate tR to the nearest thousandth
and to use the approximation 3.14159 for π.
So
p
tR = tan−1 ( 5/3) ≈ 0.912
5
.
3
So,
r
tan(t) =
5
3
r
or
tan(t) = −
5
3
t = tR + π =≈ 4.054
(approximated to the nearest thousandth)
Boris Iskra
Department of Mathematics. Oregon State University
Boris Iskra
Department of Mathematics. Oregon State University
MTH 112: Elementary Functions
MTH 112: Elementary Functions
7.3:Trigonometric equations
17/18
7.3:Trigonometric equations
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p
Now for tan(t) = − 5/3
Now, we solve the other piece :
p
tan(t) = − 5/3
Observe that tangent is negative whenever its argument (angle) is in quadrant
II and quadrant IV. Using the reference angle tR .
So,
p
p
t = π − tan−1 ( 5/3) and t = 2π − tan−1 ( 5/3)
t3 ≈ 2.230 and t4 ≈ 5.371
are the other solutions for the general equation.
Boris Iskra
Department of Mathematics. Oregon State University
Boris Iskra
Department of Mathematics. Oregon State University