Explore Solving Trigonometric Equations with Inverse Functions and Identities

LHS-tanθ=RHS-tanθ

3tanθ=3

.LHS /3.=.RHS /3.

tanθ=1

Now, try to find the angle θ for which the value of tanθ equals 1. The table of the trigonometric values of notable angles can be useful here. θ=45^(∘) and θ=225^(∘) It can be concluded that this equation has two possible solutions.

b Start by analyzing the given equation.

cos 2θ=8-15sinθ There are two different trigonometric functions involved in the equation, sine and cosine. Therefore, the first step is to rewrite one side of the equation in terms of the other. To do this, the Double-Angle Identity for cosine can be used.

cos 2θ=8-15sinθ

cos(2θ)=1- 2sin^2(θ)

1-2sin^2 θ=8-15sinθ

LHS+2sin^2 θ=RHS+2sin^2 θ

1=2sin^2 θ+8-15sinθ

LHS-1=RHS-1

0=2sin^2 θ+7-15sinθ

Rearrange equation

2sin^2 θ+7-15sinθ=0

Commutative Property of Addition

2sin^2 θ-15sinθ+7=0

Next, to solve this equation, rewrite the left-hand side as a product of two expressions.

2sin^2 θ- 15sinθ+7=0

Rewrite

Rewrite 15sinθ as 14sinθ+sinθ

2sin^2 θ-( 14sinθ+sinθ)+7=0

Distribute - 1

2sin^2 θ-14sinθ-sinθ+7=0

▼

Factor

Factor out 2sinθ

2sinθ(sinθ-7)-sinθ+7=0

Factor out - 1

2sinθ(sinθ-7)-(sinθ-7)=0

Factor out (sinθ-7)

(2sinθ-1)(sinθ-7)=0

By the Zero Product Property, in order for the equation to be true, at least one of the parentheses should be equal to 0. (2sinθ-1)(sinθ-7)=0 ⇓ I. 2sinθ-1=0 or II. sinθ-7=0 Note that adding 7 to both sides of the Equation (II) results in sinθ=7. However, this is not possible, as the values of sine are always between - 1 and 1. Therefore, this equation has no solution. Now consider Equation (I).

2sinθ-1=0

LHS+1=RHS+1

2sinθ=1

.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)

To find the value of sin^(- 1)( 12), or arcsine of 12, think about the angles for which the value of sine is 12. According to the table of the trigonometric ratios of common angles, there are two such angles in the interval between 0^(∘) and 360^(∘). Therefore, the given equation has two possible solutions. θ=30^(∘) and θ=150^(∘) Now that all the solutions have been found, the code mentioned at the beginning can be determined. In Part A, the solutions 45 and 225 were found, while the solutions from Part B were 30 and 150. The code can be found by combining these numbers in the order directed at the start. 4522530150 Try it out to see the hidden message!

Example

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.

The envelope opens and a card is taken from it and increased with the text:'First solve the task' on it

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

b 3tan^2 θ-1=sec^2 θ

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.

cos 2θ=2cos^2 θ-1 Apply this identity and then simplify the equation.

cos2θ+4cosθ=- 3

cos(2θ)=2cos^2(θ)-1

2cos^2 θ-1+4cosθ=- 3

LHS+3=RHS+3

2cos^2 θ+2+4cosθ=0

Commutative Property of Addition

2cos^2 θ+4cosθ+2=0

.LHS /2.=.RHS /2.

cos^2 θ+2cosθ+1=0

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(cosθ+1)^2=0

sqrt(LHS)=sqrt(RHS)

cosθ+1=0

LHS-1=RHS-1

cosθ=- 1

In order to find the values of θ that satisfy this equation, think about the angles for which the cosine equals - 1. According to the table of the trigonometric ratios of common angles, there is at least one such angle. θ=π Since cosine is a periodic function with a period of 2π, for every integer n, the following angles are the solutions to the equation. θ=π+2π n

b Start by analyzing the given equation.

3tan^2 θ-1=sec^2 θ There are two trigonometric functions in the equation, tangent and secant. Recall that one of the Pythagorean Identities relates these two functions. 1+tan^2 θ=sec^2 θ Substitute 1+tan^2 θ for sec^2 θ into the equation and simplify it.

3tan^2 θ-1=sec^2 θ

sec^2 θ= 1+tan^2 θ

3tan^2 θ-1= 1+tan^2 θ

LHS-tan^2 θ=RHS-tan^2 θ

2tan^2 θ-1=1

LHS+1=RHS+1

2tan^2 θ=2

.LHS /2.=.RHS /2.

tan^2 θ=1

sqrt(LHS)=sqrt(RHS)

tanθ=± 1

Next, look for the angles for which tangent is equal to 1 or - 1. θ&=45^(∘) or π4 &θ=135^(∘) or 3π4 θ&=225^(∘) or 5π4 &θ=315^(∘) or 7π4 To find the general equation for the solutions, note that all these angles can be expressed in the following way. π/4 &= π/4+π/2* 0 [0.1cm] 3π/4 &= π/4+π/2* 1 [0.1cm] 5π/4 &= π/4+π/2* 2 [0.1cm] 7π/4 &= π/4+π/2* 3 [0.1cm] Therefore, all the solutions to the equation can be given by the following equation where n is an integer number. θ=π/4+π/2 n After correctly solving both equations, Magdalena can finally read the joke that Davontay found for her.

Example

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

b sinθ-cosθ=0

Who guessed correctly?

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 Since there are two different trigonometric functions, one should be rewritten in terms of the other. To do so, use the Pythagorean Identity, but first rewrite the equation so that cos^2 θ is isolated on one side of the equation. sin^2 θ+cos^2 θ=1 ⇓ cos^2 θ=1-sin^2 θ Substitute 1-sin^2 θ for cos^2 θ and simplify the equation.

2cos^2 θ+3sinθ=3

cos^2 θ= 1-sin^2 θ

2( 1-sin^2 θ)+3sinθ=3

Distribute 2

2-2sin^2 θ+3sinθ=3

Commutative Property of Addition

- 2sin^2 θ+3sinθ+2=3

LHS-3=RHS-3

- 2sin^2 θ+3sinθ-1=0

MultEqn

LHS * (- 1)=RHS* (- 1)

2sin^2 θ-3sinθ+1=0

Next, factor the expression on the left-hand side into two parentheses.

2sin^2 θ-3sinθ+1=0

Rewrite

Rewrite 3sinθ as 2sinθ+sinθ

2sin^2 θ-(2sinθ+sinθ)+1=0

Distribute - 1

2sin^2 θ-2sinθ-sinθ+1=0

Factor out 2sinθ

2sinθ(sinθ-1)-sinθ+1=0

Factor out - 1

2sinθ(sinθ-1)-(sinθ-1)=0

Factor out sinθ-1

(2sinθ-1)(sinθ-1)=0

By the Zero Product Property, at least one of the parentheses must be equal to 0. Two equations can be formed by applying this property.

lc2sinθ-1=0 & (I) sinθ-1=0 & (II)

(I), (II): LHS+1=RHS+1

l2sinθ=1 sinθ=1

(I): .LHS /2.=.RHS /2.

lsinθ= 12 sinθ=1

To solve these two equations, try to find the angles for which the values of sine are 12 and 1. For the first equation, there are two such angles — π6 and 5π6 — while for the second there is only one — π2. θ_1=π/6, θ_2=5π/6, θ_3=π/2 Finally, the solutions should be checked whether there are any extraneous solutions. To do that, substitute each of them into the original equation and see if it is true.

2cos^2 θ+3sinθ=3

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

a/b=.a /2./.b /2.

3/2+3/2? =3

AddFrac

Add fractions

3=3 ✓

The other two solutions can be checked in a similar fashion.

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.

sinθ-cosθ=0 First, move cosθ to the right-hand side of the equation. Then, square each side of the equation. sinθ=cosθ ⇓ sin^2 θ=cos^2 θ Now, to rewrite one trigonometric function in terms of the other, use the Pythagorean Identity. Similarly to Part A, substitute 1-sin^2 θ for cos^2 θ.

sin^2 θ=cos^2 θ

cos^2 θ= 1-sin^2 θ

sin^2 θ= 1-sin^2 θ

LHS+sin^2 θ=RHS+sin^2 θ

2sin^2 θ=1

.LHS /2.=.RHS /2.

sin^2 θ=1/2

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

To find the solutions of this equation, look for the angles for which the sine equals ± sqrt(2)2. The table with the trigonometric ratios of notable angles can be useful here. θ_1=π/4 θ_2=3π/4 [0.8em] θ_3=5π/4 θ_4=7π/4 Finally, substitute each of these solutions into the original equation in order to determine whether any of them is extraneous.

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

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

b tan^2 θ-sec^2 θ=cos(- θ)

Which direction should the students go at that crossroad?

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.

cotθ=cosθ/sinθ When sinθ=0, the cotangent of θ is undefined, as division by zero cannot be calculated. According to the table of trigonometric values of common angles, sinθ equals 0 for θ equal π, 2π, 3π, and so on. All these angles can be described by the expression π n, where n is an integer number. sinθ=0 ⇓ θ=π n Therefore, these values of θ cannot be solutions to the equation. Now, substitute cosθsinθ for cotθ and simplify the equation.

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

a/b=.a /sinθ./.b /sinθ.

cosθ-cos^2 θ=0

Factor out cosθ

cosθ(1-cosθ)=0

By the Zero Product Property, at least one factor in the equation should be equal to 0. From this point, two equations can be formed. ccc I. cosθ=0 & & II. 1-cosθ=0 & & ⇓ & & cosθ=1 The solutions of the equations can be found by identifying the angles for which cosine is 0 and 1. Again, using the table of trigonometric values of common angles, it can be found that there are two solutions to Equation (I), π2 and 3π2. Since cosine is a periodic function, these two solutions repeat every period of 2π. I. cosθ=0 ⇒ lθ= π2+2π n θ= 3π2+2π n The solutions can be visualized on the following diagram.

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.

cos(- θ)=cos θ According to this identity, cosθ can be substituted for cos(- θ) on the right-hand side of the equation. tan^2 θ-sec^2 θ= cos(- θ) ⇓ tan^2 θ-sec^2 θ= cosθ Additionally, one of the Pythagorean Identities that involves secant can be used to simplify the left-hand side of the equation.

tan^2 θ-sec^2 θ=cosθ

sec^2 θ= 1+tan^2 θ

tan^2 θ-( 1+tan^2 θ)=cosθ

Distribute - 1

tan^2 θ-1-tan^2 θ=cosθ

SubTerm

Subtract term

- 1=cosθ

Rearrange equation

cosθ=- 1

To find the solutions to this equation, use the table of trigonometric ratios of notable angles again. According to the table, cosθ=- 1 for θ=π. Furthermore, since cosine is a periodic function, it has the same value for the angles π+2π, π+4π, and so on. Therefore, all the solutions can be described by the following equation. θ=π+2π n Finally, plot the solution on a unit circle.

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.

Example

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.

Students vote for an equation with more solutions by raising a hand

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

b cos(π/2-θ)=cscθ

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=2sinθcosθ Use this identity to rewrite the equation and then simplify it.

sin 2θ+sqrt(3)/2=sqrt(3)sinθ+cosθ

sin(2θ)=2sin(θ)cos(θ)

2sin θcosθ+sqrt(3)/2=sqrt(3)sinθ+cosθ

LHS-(sqrt(3)sinθ+cosθ)=RHS-(sqrt(3)sinθ+cosθ)

2sin θcosθ+sqrt(3)/2-(sqrt(3)sinθ+cosθ)=0

Distribute - 1

2sin θcosθ+sqrt(3)/2-sqrt(3)sinθ-cosθ=0

Commutative Property of Addition

2sin θcosθ-cosθ-sqrt(3)sinθ+sqrt(3)/2=0

▼

Factor

Factor out cosθ

cosθ(2sin θ-1)-sqrt(3)sinθ+sqrt(3)/2=0

Factor out - sqrt(3)/2

cosθ(2sin θ-1)-sqrt(3)/2(2sinθ-1)=0

Factor out (2sin θ-1)

(2sin θ-1)(cosθ-sqrt(3)/2)=0

Two equations can be formed by applying the Zero Product Property.

lc2sin θ-1=0 & (I) cosθ- sqrt(3)2=0 & (II)

(I): LHS+1=RHS+1

l2sin θ=1 cosθ- sqrt(3)2=0

(I): .LHS /2.=.RHS /2.

lsin θ=1/2 cosθ- sqrt(3)2=0

(II): LHS+sqrt(3)/2=RHS+sqrt(3)/2

lsin θ= 12 cosθ= sqrt(3)2

Finally, try to identify angles for which the values of sine is 12 or the values of cosine is sqrt(3)2. θ_1=30^(∘) θ_2=150^(∘) θ_3=30^(∘) θ_4=330^(∘) Here, θ_1 and θ_2 are the solutions to Equation (I), while θ_3 and θ_4 are the solutions to Equation (II). Since 0≤θ≤ 360 and two of the angles are the same, there are three solutions to the equation.

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

sqrt(LHS)=sqrt(RHS)

sinθ=± 1

Now, look for the angles for which sine is equal to 1 or - 1. Here the table of trigonometric ratios of notable angles can be useful. θ_1=90^(∘) θ_2=270^(∘) As can be seen, there are two solutions to the equation. Therefore, the first equation has more solutions.

Example

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.

When Magdalena asked how they measure the height of the tide, the worker said that, in addition to sensors, they also use a formula to determine the height of the tide. h=5cos 2π/13t Here, h is the height of the tide in feet above the mean water level and t is the number of hours past midnight. At what times of the day was the tide 4 feet above the mean level of water? Round the answer to the nearest minute.

Answer

10:40PM and 1:20AM

Hint

Substitute 4 for h and use arccosine to solve the equation for t.

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.

h=5cos 2π/13t

h= 4

4=5cos 2π/13t

.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

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

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.

Closure

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.

a How long will they have to wait until their seat is 34 meters above the ground for the first time? Round the answer to the first decimal.

b At what height will they be after 3.5 minutes of riding?

Hint

a Substitute 34 for h into the equation and solve it for t. After simplifying, apply the arccosine to both sides of the equation.

b Substitute 3.5 for t and calculate the value of h.

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 In order to calculate when the seat would be 34 meters above the ground, substitute 34 for h and solve equation for t.

h=23-22cos 3π t

h= 34

34=23-22cos 3π t

LHS-23=RHS-23

11=- 22cos 3π t

.LHS /(- 22).=.RHS /(- 22).

11/- 22=cos 3π t

MoveNegDenomToFrac

Put minus sign in front of fraction

- 11/22=cos 3π t

a/b=.a /11./.b /11.

- 1/2=cos 3π t

Rearrange equation

cos 3π t=- 1/2

To solve this equation, use the inverse function of cosine — arccosine. Apply it to the both sides of the equation. Since the value of arccos (- 12) is known, the equation can be further simplified until the solution is found. cos^(- 1)(cos 3π t)= cos^(- 1)(- 1/2) ⇓ 3π t=cos^(- 1)(- 1/2) Next, the value of cos^(- 1)(- 12) can be found by looking for an angle for which the value of cosine is - 12. By using the table of the values of notable angle, it can be concluded that there are two such angles, 2π3 and 4π3. 3π t=2π/3 and 3π t=4π/3 What is more, since cosine is a periodic functions, those angles will appear every 2π periods. This fact can be represented by adding 2π k to the found solutions where k is an integer number. 3π t=2π/3+2π k and 3π t=4π/3+2π k Finally, divide both sides of the equations by 3π to find the values of t. cc 3π t/3π=2π3+2π k/3π &3π t/3π=4π3+2π k/3π [0.1cm] ⇓ & ⇓ [0.1cm] t_1=2/9+2k/3 & t_2=4/9+2k/3 The girls want to find when the seat will be 34 meters above the ground for the first time. Therefore, the least value of t should be found. Note that when k=0, t_1 has the smallest value out of all. It can be concluded that they will be 34 meters above the ground after 29 of a minute. Use a conversion factor of 60sec1min to rewrite 29 minutes in seconds.

2/9 min* 60sec/1min

▼

Simplify

MoveRightFacToNum

a/c* b = a* b/c

2min/9 * 60sec/1min

MultFrac

Multiply fractions

2min* 60sec/9* 1min

a/b=.a /3min./.b /3min.

2* 20sec/3* 1

Multiply

40sec/3

CalcQuot

Calculate quotient

13.333333... sec

RoundDec

Round to 1 decimal place(s)

≈ 13.3 sec

Therefore, Magdalena and Paulina will be 34 meters high after only 13.3 seconds of riding on the Ferris wheel.

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