4. Solving Trigonometric Equations
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Chapter 7
4.
Solving Trigonometric Equations
This lesson focuses on teaching how to solve trigonometric equations using various techniques. It emphasizes the role of inverse functions and trigonometric identities in finding solutions. Understanding the domain and range of trigonometric functions is also highlighted as crucial for effective problem-solving. Real-world examples, such as calculating angles in engineering projects or determining time intervals in physics, are used to demonstrate the practical applications of these equations. The lesson aims to equip learners with the skills needed to tackle trigonometric equations in both academic and real-world settings.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Trigonometric Equations
Slide of 11
This lesson will focus on analyzing and solving different trigonometric equations. For that purpose, inverse trigonometric functions and trigonometric identities will be used.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
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Riding a Ferris Wheel
On Sunday, Magdalena and her younger sister Paulina went to the amusement park Adventurally with their father. They all had a lot of fun going on numerous rides, including a Ferris Wheel. When they first saw it close up, the girls were so amazed by its size that they asked one of the workers for more details about it.
The worker replied that it has a diameter of 46 meters and it turns at a rate of 1.5 revolutions per minute. When Magdalena's and Paulina's father heard this, he said that the height of their seat h above the ground in meters after t minutes can be modeled by the following function. h=23-22cos 3π t Then he asked his daughters two questions.
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b What height will the girls be at after 3.5 minutes of riding the Ferris wheel? Round the answer to one decimal if needed.
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Presenting Inverse Trigonometric Functions
There are functions that undo the trigonometric functions, so to speak. These functions are called inverse trigonometric functions.
Concept
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Inverse Trigonometric Functions
The inverse trigonometric functions are the inverse functions of the trigonometric functions. For example, the inverse sine is the inverse function of the sine function. The main inverse trigonometric functions are shown in the table below.
| Trigonometric Function | Inverse Trigonometric Function |
|---|---|
| f(x)=sinx | f^(-1)(x)=sin^(-1)x |
| f(x)=cosx | f^(-1)(x)=cos^(-1)x |
| f(x)=tanx | f^(-1)(x)=tan^(-1)x |
| Inverse Trigonometric Function | Domain | Range |
|---|---|---|
| y=sin^(-1)x | [-1,1] | -π/2 ≤ x ≤ π/2 |
| y=cos^(-1)x | [-1,1] | 0≤ x ≤ π |
| y=tan^(-1)x | All real numbers | -π/2 ≤ x ≤ π/2 |
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Trigonometric Equations and Their Solutions
Contrary to trigonometric identities — which are true for all values of the variable for which both sides are defined — some equations involving trigonometric functions are true only for certain values of the variable. Now such equations will be presented.
Concept
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Trigonometric Equation
A trigonometric equation is an equation that includes one or more trigonometric functions. Consider the following example trigonometric equations. I. sin θ = 34 & II. 3cosθ = - 2 [0.2cm] III. 3tan^2θ &-1 = sec^2θ If a trigonometric equation consists only trigonometric functions and constants, the solution is found by looking for the values of the argument that make the equation true. Solving trigonometric equations is similar to solving algebraic equations. sin θ &= 34 &⇓ θ &= ? If possible, it is useful to rewrite a trigonometric equation such that they have the same trigonometric function on its both sides. sin θ=sin α, where α is an angle or an expression However, in cases where it is not possible, the inverse function to the trigonometric function that appears in the equation can be used. sin θ = 34 ⇓ arcsin(sinθ)= arcsin(3/4) Since the functions on the left-hand side are inverses, they cancel each other out and can be simplified to θ. The value of the right-hand side, on the other hand, can be found by using a calculator. θ=0.848062... rad or θ=48.59^(∘) Note that because trigonometric functions are periodic, they can have numerous angles corresponding to the same trigonometric value. This point can be illustrated by the following graph that shows at least four different solutions of the given equation.
To sum up, the following facts can be used to solve trigonometric equations. Note that n and m written below are integers.
| Equation | Solutions |
|---|---|
| sinθ=sinα | θ=nπ &+(- 1)^nα &⇓ ifn is odd, &θ=(2m+1)π-α ifn is even,& θ=2mπ+α |
| cosθ=cosα | θ=2nπ±α |
| tanθ=tanα | θ=nπ+α |
Example
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Decoding the Message with the Solutions to Trigonometric Equations
Today Magdalena’s math class started learning how to solve trigonometric equations. To encourage students to practice solving some more equations at home, Magdalena's teacher gave the students a fun task. A secret message has been encoded. The only way to read the message is to solve two equations to decode it.
Input the answers to the first equation first. The smaller number in each pair should be written first. Write the numbers one after the other without any commas or spaces. Students who decode the message will get a bonus point on the upcoming test. The teacher said that 0^(∘) ≤ θ≤ 360^(∘).
a 4tanθ=3+tanθ
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b cos 2θ=8-15sinθ
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Hint
a Gather all the trigonometric functions and constants on different sides of the equation and simplify.
b Use the Double-Angle Identity for cosine to write cos 2θ in terms of sinθ. Then gather all the terms on one side of the equation and rewrite them as a product of two expressions.
Solution
a To solve the given equation, start by rewriting it so that all the trigonometric function appear on one side of the equation and the constants on the other side.
b Start by analyzing the given equation.
cos 2θ=8-15sinθ
cos(2θ)=1- 2sin^2(θ)
1-2sin^2 θ=8-15sinθ
AddEqn
LHS+2sin^2 θ=RHS+2sin^2 θ
1=2sin^2 θ+8-15sinθ
SubEqn
LHS-1=RHS-1
0=2sin^2 θ+7-15sinθ
RearrangeEqn
Rearrange equation
2sin^2 θ+7-15sinθ=0
CommutativePropAdd
Commutative Property of Addition
2sin^2 θ-15sinθ+7=0
2sin^2 θ- 15sinθ+7=0
Rewrite
Rewrite 15sinθ as 14sinθ+sinθ
2sin^2 θ-( 14sinθ+sinθ)+7=0
Distr
Distribute - 1
2sin^2 θ-14sinθ-sinθ+7=0
(2sinθ-1)(sinθ-7)=0
2sinθ-1=0
AddEqn
LHS+1=RHS+1
2sinθ=1
DivEqn
.LHS /2.=.RHS /2.
sinθ=1/2
sin^(-1)(LHS) = sin^(-1)(RHS)
sin^(- 1)(sinθ)=sin^(- 1)(1/2)
f^(-1)(f(x)) = x
θ=sin^(- 1)(1/2)
Example
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Math Challenge With a Gift
Magdalena really liked the math joke that her teacher encoded with the solutions to the exercise. She and her friend Davontay decided to give each other two more equations to solve. If they solve them correctly, they will gift each other an envelope with another math joke they came up with or heard somewhere.
Magdalena received the following two trigonometric equations from Davontay. He told her to write all the possible solutions in the form of an equation where n is an integer number.
a cos2θ+4cosθ=- 3
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b 3tan^2 θ-1=sec^2 θ
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Hint
a Use the Double-Angle Identity for cosine to rewrite cos2θ in terms of cosθ.
b Apply one of the Pythagorean Identities and simplify the equation.
Solution
a In the first equation, the same trigonometric function appears two times but it has different arguments. Therefore, use the Double-Angle Identity for cosine to rewrite the equation.
cos2θ+4cosθ=- 3
cos(2θ)=2cos^2(θ)-1
2cos^2 θ-1+4cosθ=- 3
AddEqn
LHS+3=RHS+3
2cos^2 θ+2+4cosθ=0
CommutativePropAdd
Commutative Property of Addition
2cos^2 θ+4cosθ+2=0
DivEqn
.LHS /2.=.RHS /2.
cos^2 θ+2cosθ+1=0
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
(cosθ+1)^2=0
SqrtEqn
sqrt(LHS)=sqrt(RHS)
cosθ+1=0
SubEqn
LHS-1=RHS-1
cosθ=- 1
b Start by analyzing the given equation.
3tan^2 θ-1=sec^2 θ
Substitute
sec^2 θ= 1+tan^2 θ
3tan^2 θ-1= 1+tan^2 θ
SubEqn
LHS-tan^2 θ=RHS-tan^2 θ
2tan^2 θ-1=1
AddEqn
LHS+1=RHS+1
2tan^2 θ=2
DivEqn
.LHS /2.=.RHS /2.
tan^2 θ=1
SqrtEqn
sqrt(LHS)=sqrt(RHS)
tanθ=± 1
Example
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Making a Bet on Equations
The teacher gave the class a couple of exercises to solve for homework. She also warned that one of the equations has extraneous solutions and that the students should identify them. To make things interesting, Magdalena and Davontay decided to make a bet about which equation has extraneous solutions.
After each chose an equation, they started solving them to see who guessed correctly. The winner will get the last piece of cake left in the fridge. To solve the equations, they must write all the solutions in radians such that 0≤ θ≤ 2π.
a 2cos^2 θ+3sinθ=3
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b sinθ-cosθ=0
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Hint
a Use the Pythagorean Identity to rewrite cosθ in terms of sinθ. Then factor the equation and apply the Zero Product Property.
b Move cosθ to the right side of the equation and then square both sides. Check the solutions by substituting them into the original equation and seeing if it remains true.
Solution
a Start by examining the first equation.
2cos^2 θ+3sinθ=3
Substitute
cos^2 θ= 1-sin^2 θ
2( 1-sin^2 θ)+3sinθ=3
Distr
Distribute 2
2-2sin^2 θ+3sinθ=3
CommutativePropAdd
Commutative Property of Addition
- 2sin^2 θ+3sinθ+2=3
SubEqn
LHS-3=RHS-3
- 2sin^2 θ+3sinθ-1=0
MultEqn
LHS * (- 1)=RHS* (- 1)
2sin^2 θ-3sinθ+1=0
2sin^2 θ-3sinθ+1=0
Rewrite
Rewrite 3sinθ as 2sinθ+sinθ
2sin^2 θ-(2sinθ+sinθ)+1=0
Distr
Distribute - 1
2sin^2 θ-2sinθ-sinθ+1=0
FactorOut
Factor out 2sinθ
2sinθ(sinθ-1)-sinθ+1=0
FactorOut
Factor out - 1
2sinθ(sinθ-1)-(sinθ-1)=0
FactorOut
Factor out sinθ-1
(2sinθ-1)(sinθ-1)=0
lc2sinθ-1=0 & (I) sinθ-1=0 & (II)
(I), (II): LHS+1=RHS+1
l2sinθ=1 sinθ=1
DivEqn
(I): .LHS /2.=.RHS /2.
lsinθ= 12 sinθ=1
2cos^2 θ+3sinθ=3
Substitute
θ= π/6
2cos^2 ( π/6)+3sin( π/6)? =3
▼
Simplify left-hand side
\ifnumequal{30}{0}{\cos\left(0\right)=1}{}\ifnumequal{30}{30}{\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(\dfrac{\pi}{2}\right)=0}{}\ifnumequal{30}{120}{\cos\left(\dfrac{2\pi}{3}\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(\dfrac{3\pi}{4}\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(\dfrac{5\pi}{6}\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(\pi\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(\dfrac{7\pi}6\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(\dfrac{3\pi}{2}\right)=0}{}\ifnumequal{30}{300}{\cos\left(\dfrac{5\pi}3\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(\dfrac{7\pi}4\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(\dfrac{11\pi}6\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(2\pi\right)=1}{}
2(sqrt(3)/2)^2+3sin(π/6)? =3
\ifnumequal{30}{0}{\sin\left(0\right)=0}{}\ifnumequal{30}{30}{\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{30}{45}{\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{90}{\sin\left(\dfrac{\pi}{2}\right)=1}{}\ifnumequal{30}{120}{\sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{135}{\sin\left(\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{30}{180}{\sin\left(\pi\right)=0}{}\ifnumequal{30}{210}{\sin\left(\dfrac{7\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{225}{\sin\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\sin\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{30}{270}{\sin\left(\dfrac{3\pi}{2}\right)=\text{-} 1}{}\ifnumequal{30}{300}{\sin\left(\dfrac{5\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{30}{315}{\sin\left(\dfrac{7\pi}4\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\sin\left(\dfrac{11\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{360}{\sin\left(2\pi\right)=0}{}
2(sqrt(3)/2)^2+3(1/2)? =3
PowQuot
(a/b)^m=a^m/b^m
23/4+3(1/2)? =3
MoveLeftFacToNum
a*b/c= a* b/c
6/4+3/2? =3
ReduceFrac
a/b=.a /2./.b /2.
3/2+3/2? =3
AddFrac
Add fractions
3=3 ✓
| Solution | Substitute | Evaluate | True or False |
|---|---|---|---|
| θ= π/6 | 2cos^2 ( π/6)+3sin( π/6)? =3 | 2(sqrt(3)/2)^2+3(1/2)? =3 | 3=3 ✓ |
| θ= 5π/6 | 2cos^2 ( 5π/6)+3sin( 5π/6)? =3 | 2(- sqrt(3)/2)^2+3(1/2)? =3 | 3=3 ✓ |
| θ= π/2 | 2cos^2 ( π/2)+3sin( π/2)? =3 | 2(0)^2+3(1)? =3 | 3=3 ✓ |
Therefore, there are three solutions to the equation and none of them are extraneous.
b Again, begin by analyzing the given equation.
sin^2 θ=cos^2 θ
Substitute
cos^2 θ= 1-sin^2 θ
sin^2 θ= 1-sin^2 θ
AddEqn
LHS+sin^2 θ=RHS+sin^2 θ
2sin^2 θ=1
DivEqn
.LHS /2.=.RHS /2.
sin^2 θ=1/2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sinθ=± sqrt(1/2)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
sinθ=± 1/sqrt(2)
ExpandFrac
a/b=a * sqrt(2)/b * sqrt(2)
sinθ=± sqrt(2)/2
| Solution | Substitute | Evaluate | True or False |
|---|---|---|---|
| θ_1= π/4 | sin( π/4)-cos( π/4)? =0 | sqrt(2)/2-sqrt(2)/2? =0 | 0=0 ✓ |
| θ_2= 3π/4 | sin( 3π/4)-cos( 3π/4)? =0 | sqrt(2)/2-(- sqrt(2)/2)? =0 | sqrt(2)≠ 0 * |
| θ_3= 5π/4 | sin( 5π/4)-cos( 5π/4)? =0 | - sqrt(2)/2-(- sqrt(2)/2)? =0 | 0=0 ✓ |
| θ_4= 7π/4 | sin( 7π/4)-cos( 7π/4)? =0 | - sqrt(2)/2-sqrt(2)/2? =0 | - sqrt(2)≠ 0 * |
It can be concluded that θ= 3π4 and θ= 7π4 are extraneous solutions. Therefore, only θ= π4 and θ= 5π4 are solutions to the equation. This means that Davontay bet on the right equation and he will get the last piece of the cake!
Example
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Determining the Cardinal Direction
Magdalena's math teacher designed a labyrinth in the school athletics field for her students. To determine which direction to go at each crossroad, she made signs with certain clues. At one of the crossroads, the clue said to follow the direction that is not a solution to either of the two given trigonometric equations.
The clue also advised to graph the solutions on a unit circle. Write all the possible solutions in the form of general equations where n is an integer number.
a sinθcotθ-cos^2 θ=0
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b tan^2 θ-sec^2 θ=cos(- θ)
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Hint
a Start by applying the Cotangent Identity.
b Use the Negative-Angle Identity and one of the Pythagorean Identities involving secant.
Solution
a In order to solve the given equation, first use the Cotangent Identity.
sinθcotθ-cos^2 θ=0
cot(θ) = cos(θ)/sin(θ)
sinθ (cosθ/sinθ) -cos^2 θ=0
MoveLeftFacToNum
a*b/c= a* b/c
sinθcosθ/sinθ-cos^2 θ=0
ReduceFrac
a/b=.a /sinθ./.b /sinθ.
cosθ-cos^2 θ=0
FactorOut
Factor out cosθ
cosθ(1-cosθ)=0
These solutions together can be described by the following general equation. θ_1=π/2+π n Similarly, the table of trigonometric values can be used to find that Equation (II) has one solution, 0, which also repeats every period of 2π. II. cosθ=1 ⇒ lθ_2=0+2π n or θ_2=2π n However, recall that θ=π n cannot be a solution to the equation because cotangent is undefined for these values. This means that the equation itself is undefined for these values. Therefore, only θ_1= π2+π n are solutions to the initial equation. Finally, graph these solutions on a unit circle to see which directions are indicated.
As shown on the unit circle, the solutions are located at two out of the four cardinal directions, north and south. Therefore, these are the directions the students should not choose.
b First, start by recalling the Negative-Angle Identity for cosine.
tan^2 θ-sec^2 θ=cosθ
Substitute
sec^2 θ= 1+tan^2 θ
tan^2 θ-( 1+tan^2 θ)=cosθ
Distr
Distribute - 1
tan^2 θ-1-tan^2 θ=cosθ
SubTerm
Subtract term
- 1=cosθ
RearrangeEqn
Rearrange equation
cosθ=- 1
The solution is located in the western direction, which means that students should not choose it. Considering the solutions to the first equation, the only direction left is east, so the students should turn east at the crossroad.
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Which Equation Has More Solutions?
In the next exercise, the teacher showed the class two trigonometric equations and asked them to guess which equation has more solutions. The class was evenly split; roughly half the class voted for the first equation, while the other half voted for the second equation. 
Therefore, the teacher said to solve both equations and see who was right. How many solutions does each equation have if 0^(∘)≤ θ≤ 360^(∘)?
a sin 2θ+sqrt(3)/2=sqrt(3)sinθ+cosθ
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b cos(π/2-θ)=cscθ
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Hint
a First, use the Double-Angle Identity for sine. Then gather all the terms on one side of the equation and factor them into two parentheses.
b Apply the Cofunction Identity for cosine and the Reciprocal Identity for cosecant.
Solution
a In order to simplify the equation, start by recalling the Double-Angle Identity for sine.
sin 2θ+sqrt(3)/2=sqrt(3)sinθ+cosθ
sin(2θ)=2sin(θ)cos(θ)
2sin θcosθ+sqrt(3)/2=sqrt(3)sinθ+cosθ
SubEqn
LHS-(sqrt(3)sinθ+cosθ)=RHS-(sqrt(3)sinθ+cosθ)
2sin θcosθ+sqrt(3)/2-(sqrt(3)sinθ+cosθ)=0
Distr
Distribute - 1
2sin θcosθ+sqrt(3)/2-sqrt(3)sinθ-cosθ=0
CommutativePropAdd
Commutative Property of Addition
2sin θcosθ-cosθ-sqrt(3)sinθ+sqrt(3)/2=0
(2sin θ-1)(cosθ-sqrt(3)/2)=0
lc2sin θ-1=0 & (I) cosθ- sqrt(3)2=0 & (II)
AddEqn
(I): LHS+1=RHS+1
l2sin θ=1 cosθ- sqrt(3)2=0
DivEqn
(I): .LHS /2.=.RHS /2.
lsin θ=1/2 cosθ- sqrt(3)2=0
AddEqn
(II): LHS+sqrt(3)/2=RHS+sqrt(3)/2
lsin θ= 12 cosθ= sqrt(3)2
b The second equation can be simplified by using the Cofunction Identity for cosine and the Reciprocal Identity for cosecant.
cos(π/2-θ)=cscθ
cos(π/2-θ)=sin(θ)
sinθ=cscθ
csc(θ) = 1/sin(θ)
sinθ=1/sinθ
MultEqn
LHS * sinθ=RHS* sinθ
sin^2 θ=1
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sinθ=± 1
Example
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The Height of the Tide
It was finally the weekend and Madgalena and her family went on a boat ride on the local river. A man who worked there said that there was a very high tide recently.
Answer
10:40PM and 1:20AM
Solution
In order to find at what times of the day the tide was 4 feet above the mean level of water, substitute 4 for h into the given equation and solve it for t. Note that to do this, the inverse function of cosine will have to be used.
It can be concluded that the tide was 4 feet above the mean level of water about 1.33 hours before and after midnight. Note that 0.33 is roughly 13, and one third of an hour is 20 minutes. Therefore, these times are 10:40PM and 1:20AM, respectively.
h=5cos 2π/13t
Substitute
h= 4
4=5cos 2π/13t
DivEqn
.LHS /5.=.RHS /5.
4/5=cos 2π/13t
cos^(-1)(LHS) = cos^(-1)(RHS)
cos^(-1)(4/5)=cos^(-1)(cos 2π/13t)
f^(-1)(f(x)) = x
cos^(-1)(4/5)=2π/13t
RearrangeEqn
Rearrange equation
2π/13t=cos^(-1)(4/5)
MultEqn
LHS * 13/2π=RHS* 13/2π
t=13/2πcos^(-1)(4/5)
UseCalc
Use a calculator
lt_1=1.331412... t_2=- 1.331412...
RoundDec
Round to 2 decimal place(s)
lt_1≈ 1.33 t_2≈ - 1.33
Closure
Share
Report error
Answering Questions About the Ferris Wheel
It was previously stated that when Magdalena and Paulina were at the amusement park Adventurally, they were so amazed by the size of the Ferris wheel that they asked a worker about how large it is.
The worker replied that it has a diameter of 44 meters and it turns at a rate of 1.5 revolutions per minute. When Magdalena's father heard this, he said that in that case the height of their seat h above the ground in meters after t minutes can be modeled by the following function. h=23-22cos 3π t Then, their father asked them two questions.
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b At what height will they be after 3.5 minutes of riding?
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Solution
a The park worker told the girls that the height of their seat on the Ferris wheel above the ground, represented by h, is modeled by the following function.
h=23-22cos 3π t
Substitute
h= 34
34=23-22cos 3π t
SubEqn
LHS-23=RHS-23
11=- 22cos 3π t
DivEqn
.LHS /(- 22).=.RHS /(- 22).
11/- 22=cos 3π t
MoveNegDenomToFrac
Put minus sign in front of fraction
- 11/22=cos 3π t
ReduceFrac
a/b=.a /11./.b /11.
- 1/2=cos 3π t
RearrangeEqn
Rearrange equation
cos 3π t=- 1/2
2/9 min* 60sec/1min
▼
Simplify
MoveRightFacToNum
a/c* b = a* b/c
2min/9 * 60sec/1min
MultFrac
Multiply fractions
2min* 60sec/9* 1min
ReduceFrac
a/b=.a /3min./.b /3min.
2* 20sec/3* 1
Multiply
Multiply
40sec/3
CalcQuot
Calculate quotient
13.333333... sec
RoundDec
Round to 1 decimal place(s)
≈ 13.3 sec
b This time the girls want the height of the seat after riding for 3.5 minutes. Substitute 3.5 for t into the given equation and solve for h.
h=23-22cos 3π t
Substitute
t= 3.5
h=23-22cos 3π ( 3.5)
Multiply
Multiply
h=23-22cos 10.5π
UseCalc
Use a calculator
h=23-0
SubTerm
Subtract term
h=23
Solving Trigonometric Equations
Exercise 2.1
The number of hours of daylight d in Mackinaw City, Michigan, can be approximated by a trigonometric equation. d = 4 sin (2π/365t) + 12 In this equation, t is the number of days after March 21^(st).
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We know that the number d hours of daylight in Mackinaw City, MI, can be approximated by the following equation, where t is the number of days after March 21^(st). d = 4sin 2π365t+12 We want to find the days when Mackinaw City has exactly 10 hours of sunlight. Let's write an equation for the values of t that correspond to d = 10. d = 4sin 2π365t+12 ⇓ 10= 4sin 2π365t+12 To solve the given equation, we will first isolate sin 2π365t in the equation. Then we will use the unit circle to find the exact values of t that satisfy the equation. Let's solve the equation for sin 2π365t.
The value of sin 2π365t that corresponds to d = 10 is - 12. The sine of an angle in standard position is the y-coordinate of the point of intersection P of its terminal side and the unit circle. P(x,y)=(cosθ,sinθ) To solve the equation sin 2π365t =- 12, we need to consider the points on the unit circle that have a y-coordinate of - 12. Recall that sine is negative in Quadrants III and IV.
We can now draw two congruent right triangles, each with a leg on the x-axis. Since the radius of the unit circle is 1, the hypotenuse of each triangle is also 1. Furthermore, since the y-coordinate of both points is - 12, the length of the leg which is not on the x-axis of each triangle is 12.
For both triangles one of the legs is half the hypotenuse, so they are 30^(∘)-60^(∘)-90^(∘) triangles. In this type of triangle, the smaller angle measures 30^(∘) or π6 radians. Knowing that a half turn is π radians and a full turn measures 2π radians, we can calculate the desired angle measures. We will add π6 to π, and subtract π6 from 2π.
We found two solutions for the equation sin 2π365t =- 12. 2π365t= 7π6 and 2π365t= 11π6 This gives us two equations for t. Let's solve them one by one, starting with 2π365t= 7π6.
One of the days when Mackinaw City will have exactly 10 hours of daylight is 213 days from March 21^(st). To find the second day, let's solve the second equation!
The two days when Mackinaw City will have exactly 10 hours of daylight are 213 and 335 days from March 21^(st).
We want to use the information from Part A to find which days of the year have less than 10 hours of daylight. We know that the city gets exactly 10 hours of daylight on the 213^(rd) and 335^(th) day after March 21^(st). Furthermore, the number of hours of daylight each day is modeled by the equation d = 4sin 2π365t+12. Let's graph this equation.
From the graph we can say that the days between the 213^(th) and 335^(th) days get no more than 10 hours of daylight. Note that the shortest day in a year occurs around December 21^(st). That is, the number of hours of daylight must fall from 213^(rd) day to December 21^(st) and rise from December 21^(st) to the 335^(th) day.
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Tadeo uses an air conditioner that cools his house when the outside temperature is above 24^(∘)C. Tadeo is able to model the outside temperature in degrees Celcius using the function f(t), where t is the number of hours past midnight.
f(t) = 26-6cos π12t
Find the time interval in which the air conditioner cools Tadeo's house.
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We will suppose that the outside temperature is modeled by the given function. f(t) = 26-6cos( π12t) We are told that an air conditioner cools Tadeo's house when the outside temperature is above 24 degrees Celcius. Let's set f(t) as 24 and solve it for t. By doing this, we will find times when it is 24^(∘)C outside.
The air conditioner is turned on about 4.7 hours after midnight. Note that 0.7 hours is 42 minutes. In other words, the air conditioner starts to work at 4:42AM. Let's now consider the graph of the function.
We can see that the graph is symmetric about the line x=12. Since the difference between 12 and 4.7 is 7.3, we can determine the time when the air condition is turned off by adding 7.3 to 12. This is equal to 19.3.
About 19.3 hours after midnight the air conditioner is turned off. Since 0.3 hours is 18 minutes, the air conditioner stops at 19:18, or 7:18PM. As a result, the air conditioner runs from 4:42AM to 7:18PM.
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Find all of the solutions in degrees of the trigonometric equation.
sin^2θ - sinθ/cosθ = 0
Write the answer without the degrees symbol.
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Let's first rewrite the given trigonometric equation.
We can now factor the equation and use the Zero Product Property to solve for sinθ and tan θ. Then we will use the unit circle to find the exact values of θ that satisfy the equation.
Solving the Equation
Let's start by factoring the expression on the left-hand side. Then, let's use the Zero Product Property to solve the equation for sin θ and tan θ.
Now, let's solve the obtained equations for θ.
tan θ = 0
To solve the equation tan θ = 0, recall the graph of the tangent function. Let's also draw the line y = 0 to see for which values of θ we have tan θ = 0.
We see that the first non-negative solution to the equation is θ = 0^(∘). We also see that solutions repeat every 180^(∘). With this information, we can write the general form for all the solutions to the equation tan θ = 0. 0^(∘) + k * 180^(∘) Here, k is any integer number.
sin θ = 1
To solve the equation sin θ = 1, recall the graph of the sine function. Let's also draw the line y = 1 to see for which values of θ we have sin θ = 1.
We see that the first positive solution to the equation is θ = 90^(∘). We also see that solutions repeat every 360^(∘). With this information, we can write the general form for all the solutions to the equation sin θ = 1. 90^(∘) + k * 360^(∘) Here, again, k is any integer number.
Listing All Solutions
While solving equations tan θ =0 and sin θ = 1, we have found the following solutions.
| Equations | Solutions |
|---|---|
| tanθ =0 | 0^(∘)+k * 180^(∘) |
| sinθ = 1 | 90^(∘) + k * 360^(∘) |
These are also all the values that could be the solutions to our original equation. However, we need to check whether there are no extraneous solutions.
Extraneous Solutions
Since we only have two trigonometric functions in the original equation, and out of them only cos θ makes the equation undefined for some values of θ. The cosine function is the only possible source of extraneous solutions. Let's recall its graph and plot the solutions of the form 90^(∘)+k* 360^(∘).
We see that for every solution of the form 90^(∘)+k* 360^(∘), the cosine function is zero, and therefore makes the original equation undefined. Therefore, these solutions are extraneous solutions to our original equation. This means that all solutions to the original equation are multiples of 180^(∘). 0^(∘)+k * 180^(∘)
Alternatively, we can start by multiplying the equation by cos θ. After solving the resulting equation for θ, we exclude the values of θ that make cos θ equal to zero. This is because cos θ is the denominator of the original equation. Let's find the solutions using this strategy.
We can now use the Zero Product Property to solve for sin θ.
Now, let's solve the obtained equations for θ. To solve the equation sin θ = 0, recall the graph of the sine function. Let's also draw the line y = 0 to see for which values of θ we have sin θ = 0.
We see that the first non-negative solution to the equation is θ = 0^(∘). We also see that solutions repeat every 180^(∘). 0^(∘) + k * 180^(∘), where k ∈ Z Earlier, we found the general form for all the solutions to the equation sin θ = 1 to be the following. 90^(∘) + k * 360^(∘), where k ∈ Z We found the following solutions.
| Equations | Solutions |
|---|---|
| sinθ =0 | 0^(∘)+k * 180^(∘) |
| sinθ = 1 | 90^(∘) + k * 360^(∘) |
These are also all the values that could be the solutions to our original equation. Recall that cos θ cannot be zero. We have shown that for every solution of the form 90^(∘)+k* 360^(∘), the cosine function is zero, and therefore they are extraneous solutions to our original equation. We found the same set of solutions for θ. 0^(∘)+k * 180^(∘)
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Solve the equation for θ if θ is measured in radians.
2sqrt(2)sin^2θ+ (2-sqrt(2))sin θ = 1
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We are given the following equation. 2sqrt(2)sin^2θ+ (2-sqrt(2))sin θ = 1 We will first solve the above equation for sin θ. Then, we will find the exact values of θ which solve our equation. Let's do these things one at a time.
Solving the Equation for sin θ
We will start by writing all the terms on the left-hand side of the equation. Then, we will solve the obtained equation by factoring and using the Zero Product Property.
Let's solve these equations for θ.
Solving sin θ = 12 for θ
To solve the equation sin θ = 12, let's recall the graph of the sine function. Let's also draw the line y = 12 to see for which values of θ we have sin θ = 12.
We see that the first two positive solutions to the equation are θ = π6 and θ= 5π6. We also see that the solutions repeat every 2π radians. With this information, we can write the general form for all the solutions to the equation sin θ = 12. π/6 + 2π k and 5π/6 + 2π k Here, k is any integer number.
Solving sin θ = - sqrt(2)2 for θ
To solve the equation sin θ = - sqrt(2)2, recall the graph of the sine function again. This time we will draw the line y = - sqrt(2)2 to see for which values of θ we have sin θ = - sqrt(2)2.
We see that sin θ = - sqrt(2)2 at θ = 5π4 and at θ = 7π4. We also see that the solutions appear every 2π radians. With this information, we can write the general form for the solutions to our equation. 5π/4 + 2 π k and 7π/4 + 2 π k Here, again, k is any integer number.
Finding All Solutions
While solving equations sin θ =- sqrt(2)2 and sin θ = 12, we have found the following solutions.
| 2sqrt(2)sin^2θ+ (2-sqrt(2))sin θ = 1 | |
|---|---|
| sinθ = 1/2 | sinθ = - sqrt(2)/2 |
| π/6 + 2 π k and 5π/6 + 2 π k | 5π/4 + 2 π k and 7π/4 + 2 π k |
These are also all the solutions to our original equation.
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Solving Trigonometric Equations
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