1 · Web viewMore Trigonometric Equations (Trigonometry, Day 8) In this handout, we will practice a...

of 13/13
Algebra 2/Pre-Calculus Name__________________ More Trigonometric Equations (Trigonometry, Day 8) In this handout, we will practice a variety of equation solving techniques. All problems should be done without the use of a calculator. 1. Consider each of the following equations. Find all solutions. a. Solve . b. You should have found that or . Now solve the equation .
  • date post

    24-Sep-2020
  • Category

    Documents

  • view

    2
  • download

    0

Embed Size (px)

Transcript of 1 · Web viewMore Trigonometric Equations (Trigonometry, Day 8) In this handout, we will practice a...

1

Algebra 2/Pre-Calculus

Name__________________

More Trigonometric Equations (Trigonometry, Day 8)

In this handout, we will practice a variety of equation solving techniques. All problems should be done without the use of a calculator.

1.Consider each of the following equations. Find all solutions.

a.Solve

sin(θ ) = 12

sin(q)=

1

2

.

b.You should have found that

θ = 30! +360!N

or

θ =150! +360!N

.

Now solve the equation

sin(2θ ) = 12

sin(2q)=

1

2

.

c.Since

2θ = 30! +360!N

or

2θ =150! +360!N

, we can conclude that

θ =15! +180!N

or

θ = 75! +180!N

. (Notice that the dividing by 2 turns

360!N

into

180!N

.)

Use a similar approach to solve the equation

sin(5θ −10) = 32

sin(5q-10)=

3

2

.

d.Here’s the solution to the last problem:

5θ −10 = 60! +360!N

or

5θ −10 =120! +360!N

5θ = 70! +360!N

or

5θ =130! +360!N

θ =14! + 72!N

or

θ = 26! + 72!N

Now solve the equation

cos(2θ +8) = − 12

cos(2q+8)=-

1

2

.

e.You should have found that

θ = 56! +180!N

or

θ =116! +180!N

Now solve the equation

cos(θ + 20) = 22

cos(q+20)=

2

2

. Note: Answers are provided at the end of this problem.

f.Solve the equation

tan(3θ +12) = 3

tan(3q+12)=3

.

g.Solve the equation

sin(2θ +14) = 0

sin(2q+14)=0

.

h.Solve the equation

tan(5θ − 40) = −1

tan(5q-40)=-1

.

i.Solve the equation

sin(2θ ) =1

sin(2q)=1

.

j.Solve the equation

tan(θ − 4) = 0

tan(q-4)=0

.

Some answers e.

θ = 25! +360!N

or

θ = 295! +360!N

f.

θ =16! +120!N

or

θ = 76! +120!N

g.

θ = −7! +180!N

or

θ = 83! +180!N

h.

θ = 35! + 72!N

or

θ = 71! + 72!N

i.

θ = 45! +180!N

j.

θ = 4! +360!N

or

θ =184! +360!N

2.In this problem, we will explore different ways of writing the answer for a trigonometric equation.

a.Jermaine and Connor were both trying to solve the equation

sin(θ + 200) = 12

sin(q+200)=

1

2

. Jermaine found that

θ = −170! +360!N

or

θ = −50! +360!N

whereas Connor found that

θ =190! +360!N

or

θ = 310! +360!N

. Who was right?

b.You should have found that Jermaine and Connor both had correct ways of writing the solution. What value of N would Jermaine need in order to get a solution of

190!

? What value of N would Connor need in order to get a solution of

190!

?

c.Explain why Connor's solutions and Jermaine's solutions actually give us the same values.

Answers a. They are both right. b. Jermaine:

N =1

N=1

. Connor:

N = 0

N=0

. c. When the N-values in Jermaine's solutions are 1 larger than the N-values in Connor's solutions, they both get the same numbers.

3.In this problem, we will learn how to find solutions for a trigonometric equation on a given interval.

a.Solve

sin(θ ) = 12

sin(q)=

1

2

. Find all solutions for

0! ≤θ < 360!

.

b.You should have found that

θ = 30!

or

θ =150!

. Now solve the equation

sin(2θ ) = 12

sin(2q)=

1

2

. Again, find all solutions for

0! ≤θ < 360!

.

c.Ted said that the equation in part b had two solutions:

15!

or

75!

, but Kaitlin said that it actually had four solutions:

15!

,

75!

,

195!

, or

255!

. Who is right? Explain. Note: Remember, you can check each solution by plugging in back into the equation.

d.You should have found that all four of Kaitlin's solutions worked. But how did she find the solutions that Ted missed?

To answer this question, solve the equation

sin(2θ ) = 12

sin(2q)=

1

2

again. This time, find all solutions (not just the ones on

0! ≤θ < 360!

).

e.You should have found that

θ =15! +180!N

or

θ = 75! +180!N

. What values do you get when

N = 0

N=0

? What values do you get when

N =1

N=1

? When

N = 2

N=2

?

f.Here's what you should have found in the last problem:

N = 0

N=0

:

θ =15!

or

θ = 75!

N =1

N=1

:

θ =195!

or

θ = 225!

N = 2

N=2

:

θ = 375!

or

θ = 435!

Now let's return to the original problem: Explain how you can find all solutions for the equation

sin(2θ ) = 12

sin(2q)=

1

2

on

0! ≤θ < 360!

.

4.The approach we used to solve the last problem is challenging, so we will practice it. Solve

sin(3θ −18) = − 32

sin(3q-18)=-

3

2

. Find all solutions for

0! ≤θ < 360!

. Note: The solution is provided below, so check your work after you complete the problem.

Solution:

sin(3θ −18) = − 32

sin(3q-18)=-

3

2

3θ −18 = 240! +360!N

or

3θ −18 = 300! +360!N

3θ = 258! +360!N

or

3θ = 318! +360!N

θ = 86! +120!N

or

θ =106! +120!N

N = 0

N=0

:

θ = 86!

or

θ =106!

N =1

N=1

:

θ = 206!

or

θ = 226!

N = 2

N=2

:

θ = 326!

or

θ = 346!

Thus, the equation has six solutions:

86!

,

106!

,

206!

,

226!

,

326!

,

346!

.

5.Solve each of the following equations. Find all solutions for

0! ≤θ < 360!

. Remember: No calculators!

a.

cos(2θ −10) = 12

cos(2q-10)=

1

2

b.

sin(3θ ) = 22

sin(3q)=

2

2

c.

tan(2θ − 40) = 3

tan(2q-40)=3

d.

cos(2θ ) = −1

cos(2q)=-1

Answers a.

35!

,

155!

,

215!

,

335!

b.

15!

,

45!

,

135!

,

165!

,

255!

,

285!

c.

50!

,

140!

,

230!

,

320!

d.

90!

,

270!

_1392739075.unknown
_1392742429.unknown
_1392742978.unknown
_1392789167.unknown
_1392789410.unknown
_1392789497.unknown
_1392807127.unknown
_1392807423.unknown
_1392807422.unknown
_1392789513.unknown
_1392789427.unknown
_1392789289.unknown
_1392789304.unknown
_1392789185.unknown
_1392743472.unknown
_1392786274.unknown
_1392786508.unknown
_1392743622.unknown
_1392786086.unknown
_1392743551.unknown
_1392743157.unknown
_1392743186.unknown
_1392743423.unknown
_1392743443.unknown
_1392743451.unknown
_1392743433.unknown
_1392743382.unknown
_1392743172.unknown
_1392743137.unknown
_1392743146.unknown
_1392743130.unknown
_1392742546.unknown
_1392742771.unknown
_1392742923.unknown
_1392742952.unknown
_1392742961.unknown
_1392742941.unknown
_1392742779.unknown
_1392742559.unknown
_1392742572.unknown
_1392742585.unknown
_1392742745.unknown
_1392742579.unknown
_1392742565.unknown
_1392742553.unknown
_1392742522.unknown
_1392742535.unknown
_1392742540.unknown
_1392742528.unknown
_1392742509.unknown
_1392742515.unknown
_1392742502.unknown
_1392741161.unknown
_1392741230.unknown
_1392742388.unknown
_1392742415.unknown
_1392742382.unknown
_1392742041.unknown
_1392741180.unknown
_1392741199.unknown
_1392741175.unknown
_1392739481.unknown
_1392740929.unknown
_1392740985.unknown
_1392741091.unknown
_1392741144.unknown
_1392741031.unknown
_1392741054.unknown
_1392740998.unknown
_1392740959.unknown
_1392739513.unknown
_1392740917.unknown
_1392739496.unknown
_1392739475.unknown
_1392739134.unknown
_1392739158.unknown
_1392737454.unknown
_1392737699.unknown
_1392738070.unknown
_1392738978.unknown
_1392739004.unknown
_1392738250.unknown
_1392737810.unknown
_1392737988.unknown
_1392737777.unknown
_1392737543.unknown
_1392737584.unknown
_1392737614.unknown
_1392737566.unknown
_1392737485.unknown
_1392737530.unknown
_1392737468.unknown
_1392737288.unknown
_1392737373.unknown
_1392737408.unknown
_1392737423.unknown
_1392737438.unknown
_1392737392.unknown
_1392737325.unknown
_1392737359.unknown
_1392737310.unknown
_1360990184.unknown
_1360990543.unknown
_1392737248.unknown
_1392737264.unknown
_1360990597.unknown
_1360990629.unknown
_1360990462.unknown
_1360990503.unknown
_1360990355.unknown
_1360989458.unknown
_1360989660.unknown
_1360989375.unknown