unit 5 test 2006 solns - mr santowski's math...

Precalculus: Unit 5 Test Name: SOLUTIONS

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PART A – Calculator Active Multiple Choice: Second Chance. Use answer sheet.

For questions #1-3, refer tothe diagram at the right:

1. The minimum number of formulas requiredto generate all the coterminal angles to

θ1, θ2

, θ3, θ4

and θ5 is

A) 1 B) 2 C) 3 D) 4 E) 5

2. The equation that yields θ1 = 0 , θ3 =

π2

and θ5 =

3π2

as solutions is

A) cosθ = 1 B) cosθ = 0 C) cos2θ = 1 D) cos2θ = cosθ E) cos2θ = − cosθ

3. The expression that represents ONLY coterminal angles for θ3 =

π2

is

A) θ3 =π2

B) θ3 = 4n + 1( ) π2 ;n ∈I C) θ3 =nπ2

;n ∈I

D) θ3 = 2n + 1( ) π2 ;n ∈I E) θ3 = 4n + 3( ) π2 ;n ∈I

4. The Law of Sines is

A)

Asina

=B

sinb=

Csinc

C) c2 = a2 + b2 − 2ab ⋅ sinC

B)

asinA

=b

sinB=

csinC

D) sin2 A + sin2 B + sin2 C = a2 + b2 + c2

5. The Law of Cosines is

A)

Acosa

=B

cosb=

Ccosc

C) c2 = a2 + b2 − 2ab ⋅cosC

B)

acosA

=b

cosB=

ccosC

D) cos2 A + cos2 B + cos2 C = a2 + b2 + c2

θ5 =

3π2

θ2 =

π4

θ3 =

π2

θ4 =

3π4

θ1 = 0 16

16

103

94



6. The value of a in the triangle given below isA) 45.614B) 4.933C) 58.333D) 1.007

7. Which of the following is a true statement?A) The Cosine Law is a special case of the Sine Law.B) The Sine Law is a special case of Pythagorean Theorem (Law).C) The Pythagorean Theorem (Law) is a special case of the Cosine Law.D) None of the above statements is true.

8. A line has the equation y =

73

x , and makes an angle of θ with the x – axis. The

value of tanθ is

A) tan 7

3B)

73

C) 1.166 D) 66.801o

PART B – Calculator Active Free Response: Answer in the space provided.9. The function

f x( ) = 5sinx cos4 x − 10sin3 x cos2 x + sin5 x is given.

a) Sketch f x( ) in the above window with

x ∈ 0,2π⎡

⎣⎤⎦ ,

y ∈ −2,2⎡

⎣⎤⎦ and

x

scale=

π10

.

b) What is the period of the resulting sinusoid? The definition of PERIOD, T is the

number of radians to complete one cycle.

Since there are 5 cycles: 2π radians 1 cycle:

2π5

radians ∴T =

2π5

.

c) Write the equation of the sinusoid f x( ) in terms of sine or cosine:

y = sin 5x( )

d) State an identity, but do not prove, for 5sinx cos4 x − 10sin3 x cos2 x + sin5 x :

sin 5x( ) = 5sinx cos4 x − 10sin3 x cos2 x + sin5 x

70o 18o

a 15

4

4



10. Use the diagram to prove that tanθ =m1 −m2

1 +m1m2

, where m1 and m2 are the slopes of

the lines 1 and 2 as shown in the diagram below.

LS = tanθ

= tan α − β( )

=

tanα − tanβ1 + tanα tanβ

=

m1− m

2

1 + m1m

2

= RS QED

4

This proof can/should be donewithout a calculator. It is here,in the calculator section only fortime’s sake, so you can getstarted early, or in case youwant to use your calculator toexplore the question.

I will give you two hints, if youask, for a deduction of one pointeach. You may ask for one hintat a time.

Note that, for any angle measured instandard position (from the right side ofa horizontal line), that the tangent of theangle is the opposite over adjacent, whichis the rise over run, or slope of the line!

(see MC # 8)

Therefore:

tanα =

Δy1

Δx1

= m1 and

tanβ =

Δy2

Δx2

= m2

1 : y =m1x + b1

2 : y =m2x + b2

θ

x

y

βα



PART C – No Calculator Multiple Choice: Second Chance. Use answer sheet.

11. If the angle θ lies in quadrant II, then the angle 2θ lies in either quadrantA) I or II B) II or III C) III or IV D) IV or I

12. The solution to csc x = −1 is

A) x = 0 B) x =

π2

C) x = π D) x =

3π2

13. Which of the following is identically equal to sin π − x( )?

A) cos x B) sinx C) −cos x D) − sinx

14. Which of the following is NOT equivalent to cos 2θ( )?

A) cos2θ − sin2θ B) 2cosθ sinθ C) 2cos2θ − 1 D) 1 − 2sin2θ

15. The solution(s) to cos x = −

35

for x ∈ 0,2π⎡

⎣⎤⎦ are

A) x = cos−1 −

35

⎛

⎝⎜

⎞

⎠⎟ C)

x = π − cos−1 3

5

⎛

⎝⎜

⎞

⎠⎟ or

π + cos−1 3

5

⎛

⎝⎜

⎞

⎠⎟

B) x = cos−1 3

5

⎛

⎝⎜

⎞

⎠⎟ D)

x = π − cos−1 3

5

⎛

⎝⎜

⎞

⎠⎟ or

2π − cos−1 3

5

⎛

⎝⎜

⎞

⎠⎟

16. The domain of y = cos x is

A) x ∈ C) x ∈, 4n + 1( )π

2, 4n − 1( )π

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥,n ∈I

B) x ∈, x ≥ 0 D) x ∈, 4n − 1( )π

2, 4n + 1( )π

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥,n ∈I

17. The solutions to tan2 x − 1 = 0 , for x ∈ 0,2π⎡

⎣⎤⎦ are

A) x =

π4

, 5π4

B) x =

π2

C) x =

π4

, 3π4

, 5π4

, 7π4

D) x = 0,π,2π

40

40



18. The function with a period of

π3

is

A) y = cos 3x( ) B)

y = csc x

3

⎛

⎝⎜

⎞

⎠⎟ C)

y = tan 3x( ) D)

y = sin 3

2x

⎛

⎝⎜

⎞

⎠⎟

19. The equation that produces the solutions θ = 150o,330o is

A) sinθ =

12

B) secθ =

2 33

C) tanθ = −

33

D) cosθ = −

32

20. An identical way to express sec π

2− α

⎛

⎝⎜

⎞

⎠⎟ is

A) secα B) tanα C) cscα D) cosα

21. A simplification of tan π

2− θ

⎛

⎝⎜

⎞

⎠⎟ ⋅ tan −θ( ) is

A) 1 B) –1 C)

π2

D) π

22. The domain of validity of csc x =

1sinx

, for x ∈ 0,2π⎡

⎣⎤⎦ is

A) x ∈ 0,2π⎡

⎣⎤⎦, x ≠

π2

, 3π2

C) x ∈ 0,2π⎡

⎣⎤⎦, x ≠ 0, π

2,π, 3π

2,2π

B) x ∈ 0,2π⎡

⎣⎤⎦, x ≠ 0,π,2π D)

x ∈ 0,2π⎡

⎣⎤⎦

23. If f x( ) = g x( ) is an identity and

f x( )g x( ) = k , where k is a real constant, then which of

the following must be false?A)

g x( ) ≠ 0 B)

f x( ) = 0 C) k = 1 D)

f x( ) g x( ) > 0

24. The value of tan sin−1 7

25

⎛

⎝⎜

⎞

⎠⎟ is

A) undefined B)

247

C)

724

D)

2425



25. The correct half-angle identity for sinθ

2 is

A) ±

1 − cos2θ2

B) ±

1 − cosθ2

C) ±

1 + cos2θ2

D) ±

1 + cosθ2

26. The graph of y = sin2x for the graphing window of x ∈ 0,2π⎡

⎣⎤⎦ and

y ∈ −2,2⎡

⎣⎤⎦ is

A) B) C) D)

27. The solution(s) for sin2θ = 1 for θ ∈ 0,2π⎡

⎣⎤⎦ are

A) θ =

π2

B) θ =

π4

, 5π4

C) θ = 0,π,2π D) θ = 0, π

2,π, 3π

2,2π

28. For which values of x does sin2x = 2sinx ?

A) x ∈ B) x =

π2

, 3π2

C) x = 0,π,2π D) none

29. If

f 2x( ) = 2 ⋅f x( )1 − f x( )⎡

⎣⎢⎤⎦⎥

2, then

f x( ) must equal

A) sinx B) cos x C) tanx D) sin−1 x

30. The value of cos−1 cos −

π3

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥ is

A) −π3

B)

π3

C)

2π3

D)

4π3



PART D – No Calculator Free Response: Write answers in the space provided.

31. Evaluate cos165o exactly.

cos165o = cos 120o + 45o( )

= cos120o cos 45o − sin120o sin45o

= −12

⎛

⎝⎜

⎞

⎠⎟

22

⎛

⎝⎜⎜

⎞

⎠⎟⎟−

32

⎛

⎝⎜⎜

⎞

⎠⎟⎟

22

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=− 2 − 6

4

32. Evaluate the following exactly.

a) cot2 x − csc2 x2

, for any x ∈ domain of validity

=

cot2 x − cot2 x + 1( )2

= −12

b) sin π2− θ

⎛

⎝ ⎜ ⎞

⎠ ⎟ , given that secθ =

74

c) sin50o ⋅ sec 40o

= cosθ =

47

= sin50o ⋅csc50o = sin50o ⋅

1sin50o

= 1

33. Let θ = sec−1 5

4.

a) Rewrite the equation in trigonometric form.b) Label the triangle for θc) Determine cotθ :

d) Evaluate cot −sec−1 54

⎛

⎝ ⎜ ⎞

⎠ ⎟ exactly:

4

4

3

3

3

3

Triangle for #33(b):

a) secθ =

54

c)

43

d) −

43

θ

5

4

3

This means that as long as x doesnot make cotx or cscx undefined,that x can be any value.

This question was thrown in as a HINT throughout this test; every time you have an inversetrig function of a ratio, it may be (is!) helpful to rename it (inverse trig of ratio) as an angle

and draw a triangle! See #34a and #35.

What relationship existsbetween the angles of 40 degand 50 deg? What identitydoes this remind you of?

There are other choices for rewriting165 degrees. Two examples are:

(i) 210 deg – 45 deg(ii) 330/2 deg



34. Evaluate exactly. Answer ONLY ONE, either (a) or (b). You may answer the otherquestion for 2 marks bonus.

a) sin cos−1 3

5+ sin−1 3

5

⎛

⎝⎜

⎞

⎠⎟ b)

cos22.5o = cos 45o

2

⎛

⎝⎜

⎞

⎠⎟

Let α = cos−1 3

5 and

β = sin−1 3

5 = +

1 + cos 45o

2 Note: + ONLY

= +

22+ 2

2

2

Then sin cos−1 3

5+ sin−1 3

5

⎛

⎝⎜

⎞

⎠⎟

= +

2 + 22

⋅12=

2 + 22

= sin α + β( )

= sinα cosβ + cosα sinβ =

45

⎛

⎝⎜

⎞

⎠⎟

45

⎛

⎝⎜

⎞

⎠⎟ +

35

⎛

⎝⎜

⎞

⎠⎟

35

⎛

⎝⎜

⎞

⎠⎟

=

16 + 925

= 1

35. Show that sin 2sin−1 x( ) = 2x 1 − x2 .

Let θ = sin−1 x ⇒ sinθ =

x1

⇒

LS = sin 2sin−1 x( )

= sin 2θ( ) = 2sinθ cosθ

= 2 x

1

⎛

⎝⎜

⎞

⎠⎟

1 − x2

1

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 2x 1 − x2 = RS QED

36. Solve for x, x ∈ 0,2π⎡

⎣⎤⎦ : tanx = − 3

tanx = −

31

⇒ xref

=π3

and Quadrants: II and IV ∴x =

2π3

, 5π3

37. Solve for x, x ∈ : cos2x + cos x = 0

2cos2 x − 1( ) + cos x = 0

2a − 1( ) a + 1( ) = 0

Let a = cos x . ∴a = cos x =

12 ⇒ x

ref=π3

& Q: I, IV ⇒ x

0,2!⎡⎣

⎤⎦=π3

, 5π3

Then 2a2 + a − 1 = 0 a = cos x = −1 ⇒ x

0,2!⎡⎣

⎤⎦= π

3

3

3

3

3

3

5

5

α5

4

3β

5

4

3

θ

1

∴ 1 − x2

x Make sure to labelthe angle (that youintroduced) in the

triangle!

What do you notice aboutthe two angles, alpha andbeta? Is there another wayto determine that sine ofthe angle’s sum is 1? How?



38. Prove any TWO of the FOLLOWING THREE identities. You may do the third prooffor two marks bonus.

a)

sinθ1 + cosθ

=1 − cosθ

sinθ b) sinx tanx = secx − cosx c) csc4 x − cot4 x = 2csc2 x − 1

86

LS =

sinθ1 + cosθ

=

sinθ1 + cosθ

⋅1 − cosθ1 − cosθ

=

sinθ 1 − cosθ( )1 − cos2θ

=

sinθ 1 − cosθ( )sin2 θ

=

1 − cosθsinθ

= RS QED

LS = sinx tanx

= sinx sinx

cos x

=

sin2 xcos x

=

1 − cos2 xcos x

=

1cos x

−cos2 x

cos x = sec x − cos x = RS QED

Coterminal angles, for x ∈ :

x = 6n + 1( )π

3, 2n + 1( )π, 6n + 5( )π

3;n ∈I

BUT! The angles are evenly spaced, with

a spacing of 120o or

2π3

rad between

them. So, only one coterminal angleformula is required:

x =

π3+

2nπ3

= 2n + 1( )π3

,n ∈I . This

makes sense: these are the ODD

π3

’s!

A sketch of the angles instandard position is very helpful:

Solution:

LS = csc4 x − cot4 x

= csc2 x + cot2 x( ) csc2 x − cot2 x( )

= csc2 x + cot2 x( ) ⋅1

= csc2 x + csc2 x − 1( )

= 2csc2 x − 1 = RS QED



BONUS: Show work/answer on the answer sheet.

Solve 22 cos2x( )+1

− 5 ⋅2cos2x = −2 for x ∈ .

Let a = cos 2x( ) . (This is logical and useful because the compound variable expression

cos 2x( ) is repeated throughout the equation. Some people may start by letting α = 2x ,

which is perfectly acceptable, and perhaps a little more logical; see the note in the box.)

Then the equation becomes simpler: 22a +1 − 5 ⋅2a = −2

Now use exponent laws to rewrite: 22a ⋅2 − 5 ⋅2a = −2 ⇒ 2a( )2 ⋅2 − 5 ⋅2a = −2

So the equation looks like 2 2a( )2 − 5 2a( ) + 2 = 0 , and another substitution is in order:

Let u = 2a . Now the equation becomes: 2u2 − 5u + 2 = 0

2u − 1( ) u − 2( ) = 0

∴u =

12

or u = 2

Therefore 2a =

12⇒ a = −1 ⇒ cos 2x( ) = −1

2a = 2 ⇒ a = 1 ⇒ cos 2x( ) = 1 .

So we have that cos 2x( ) = ±1 . It may be helpful to let α = 2x .

Now cosα = ±1 ⇒α = 0,π,2π,3π,…⇒α = nπ,n ∈I

Therefore 2x = nπ ⇒ x =

nπ2

,n ∈I .

So the FINAL SOLUTION to 22 cos2x( )+1

− 5 ⋅2cos2x = −2for x ∈ is

x =

nπ2

,n ∈I

5

5

Do you know why/how?

This original functionon the LS is acomposition of severalother functions:

1) the outer function is a QUADRATIC2) next, exponential, base 23) then cosine function4) the inner-most is a linear function (multiplication by 2)

Each composition canbe easily dealt with byusing a substitution: aLET statement, andthen back-substitutingto solve for theoriginal variable.

unit 5 test 2006 solns - mr santowski's math...

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