Suggested+Solutions+VG MVG+Test+V2+Ch5+MaDNVC06+Trigonometry

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Suggested solutions VG/MVG Level Test V2 Ch5 MaDNVC06 trigonometry Sjödalsgymnasiet

Chapter Test: Ch5 trigonometry; MATHEMATICS COURSE D;

Spring 2008: MaDNVC06

Suggested solutions VG-MVG Level test.

Please note that you have to try to solve the problems by yourself before

checking your solutions against mine. My solutions are just suggestedones. Usually there are more than one methods of solving a givenproblem.

Warning: Just reading the solutions can never replace your own strugglein solving a given problem. By just reading the solutions you may not beable to understand the mathematics of the problem deep enough andtherefore, just going through my solutions may not help you to solve asimilar problem by yourself. Have Fun!

Behzad

InstructionsTest period 8:15-10:45

Tools Formula sheet, ruler and graphic calculator.

The test For most items a single answer is not enough. It is also expected

• that you write down what you do

• that you explain/motivate your reasoning

• that you draw any necessary illustrations.

After every item is given the maximum mark your solution can receive. [2/3] means that the item can

give 2 g-points (Pass level) and 3 vg-points (Pass with distinction level).

Items marked with give you a possibility to show MVG-quality (Pass with special distinction

quality). This means that you use generalised methods, models and reasoning, that you analyse your

results and account for a clear line of thought in a correct mathematical language.

Problem 15 is a larger problem which may take up to 45 minutes to solve completely. It is important

that you try to solve this problem. A description of what your teacher will consider when evaluating

your work is attached to the problem. This problem is heavily graded. Even in G-level there are a lot

of moments of the problem that you can solve.

Try all of the problems. It can be relatively easy, even towards the end of the test, to receive some

points for partial solutions. A positive evaluation can be given even for unfinished solutions.

Mark limits The test gives totally at the most 50 points, out of which 30 vg-points and 7¤ MVG-pints. To pass thetest you must have at least 15 points and to get the test character Pass with distinction (VG) you must

have at least 30 points out of which at least 10 points on Pass with distinction level. Excellence(MVG) requires 35 points out of which at least 20 VG points and excellent quality presentation of the

solutions ¤¤¤¤. Have fun!

Note: Only those problems checked here will be graded!

Problem 1 2 3 4 5 6 7 8 9 10 11 12

Grade

©[email protected] Not for sale. Free to use for educational purposes.☺ 1/11





1. For what values of x the function1cos

32

−

+=

x

x y is not defined? [0/1]

Suggested solutions: Answer: ...2,,1,02 =∀≠ nn x π

1cos

32

−

+

= x

x y is not defined if the denominator of the function is zero, i.e. if 01cos =− x .

Therefore:

01cos ≠− x

1cos ≠ x

( )1cos1−≠ x

Answer: ...2,,1,02 =∀≠ nn x π

2. If the point )2,5− A is on the terminal side of the angle α find the exact values of:

a. α sin [0/1]

b. α cos [0/1]c. α tan [1/0]

Suggested solutions:

The radius of the “unit circle” may be calculated using the information about the coordinates

of the point )2,5− A

( ) 394525 22

2 =⇔=+=+= r r Using the properties of the unit circle that sinus is the y-

component and cosine is the x-coordinates of the point on the unit circle, we may conclude

that:

a. Answer:3

2sin =α [0/1]

b. Answer:3

5−=α [0/1]cos

c. Answer:5

2−=

cos

sintan =

α

α α

[1/0]

Of course, the value of the tangent could directly be calculated from the information regarding

the coordinates of the point on the unit circle.: tangent is simply the ratio of the y-component

of the point and that of its x-component:





3. In the triangle ABC illustrated blow angle C B ∠=∠ 2 . If the relationship between the

length of two corresponding sides are 7:5 . The figure is not properly scaled

a. Find the angles of the triangle. [1/2]

b. If mCB 0.12= find lengths of the sides of

the triangle. [1/1]

c. Find the area of the triangle. [1/1]Suggested solutions:

c

C

b

B sinsin=

x x /=

/ 5

sin

7

2sin α α

x x /=

/ 7

cossin2

5

sin α α α

7

cos2

5

1 α =

°≈⎟ ⎠

⎞⎜⎝

⎛ =⇔= − 57.45

10

7cos

10

7cos 1

α α

°=≡∠ 57.45α B , °=°×==∠=∠ 15.9157.45222 α BC ,

°=°×−°=−−°=∠ 28.4357.4531802180 α α A

Use again the sine law to calculate the length of each side of the triangle:

c

C

b

B

a

A sinsinsin==

m x x

5.238.43sin5

57.45sin12

5

57.45sin

12

38.43sin=

°⋅

°⋅=⇔

°=

°

mm x AB 5.125.255 =×== m x AC 5.177 ==

The area of the triangle may be calculate using the area formula:2

sin Abc Area =

( ) 2222 98.745.29976.119976.112

28.43sin75

2

sinmm x

x x Abc Area ===

°⋅⋅==

Answer: °=∠ 3.43 A , °=≡∠ 6.45α C , °=∠ 2.91 B , m AB 5.12= , m AC 5.17= ,2

75 m Area ≈

4.

Find the exact value of °

75cos and°

75sin by using the fact that it may be expressed as:( )°+° 45 .=° 30cos75cos

Note: You are not allowed to use a calculator to find the exact value of °45cos . You may,

however, use the trigonometric table in the yellow pages. [1/3]


( ) ( ) ( ) ( ) ( )°°−°°=°+°=° 45sin30sin45cos30cos4530cos75cos

4

26

2

2

2

1

2

2

2

375cos

−=⋅−⋅=°

( ) ( ) ( ) ( ) ( )°°+°°=°+°=° 45sin30cos45cos30sin4530sin75sin

4

26

2

2

2

3

2

2

2

1

75sin

+

=⋅+⋅=° Answer: 4

26

75cos

−

=° , 4

26

75sin

+

=°

A

C

x7

α 2

x5

α

B

mCB 0.12=





5. Find all solutions of the equation

33

25cos2 =⎟

⎠

⎞⎜⎝

⎛ −⋅

π x . [1/2]

Suggested solutions: Answer: n x ⋅+−=

5

2

15

π π and n x ⋅=

5

2π ,...3,12,1,0=∀ n

33

25cos2 =⎟

⎠

⎞⎜⎝

⎛ −⋅

π x

2

3

3

25cos =⎟

⎠

⎞⎜⎝

⎛ −

π x

n x ⋅+=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =− −

π π π

262

3cos

3

25 1

⎪⎪⎩

⎪⎪⎨

⎧

⋅+=⋅++=⇔⋅+=⋅+−=−

⋅+=⋅++=⋅++=⇔⋅+=−

nn xnn x

nnn xn x

π π

π π π

π π

π π

π π

π π

π π π

π π π

π π π

26

152

6

11

3

252

6

112

62

3

25

26

52

66

42

63

252

63

25

Answer:

⎪⎪⎩

⎪⎪⎨

⎧

⋅+=

⋅+=

5

2

2

5

2

6

n x

n x

π π

π π

6. Find all values of a for which the equation

5

32sin

a x

+= has a solution. [2/2]

Suggested solutions: Answer: 13

7≤≤− a

Using the fact that 1sin1 ≤≤− x

15

321 ≤

+≤−

a

Multiply all sides by 5:

5325 ≤+≤− a

Subtract 2 to all sides:

2532225 −≤+−≤−− a

337≤≤−

a Divide all sides by 3:

Answer: 13

7≤≤− a





7. Show thatα

α α

2cos1

2sintan

+= by the use of

a. the properties of the unit circle. [0/2]

b. other trigonometric relationships. [0/1]


a. Using the properties of the unit circle, we maycomplete the figure by making a normal from the

point B to C . The angle BOC is α 2 , and using

the fact that in the unit circle illustrated below:

α 2sin= BC and α 2cos=OC we may write:

α

α α

2cos1

2sin

1tan

+=

+=

OC

BC

b. Using the following trigonometric relationships:

α α α cossin22sin =

1cos22cos 2 −= α α

andα

α α cossintan = we may start from the RHS of the

equation above and derive the formula above:

α α

α

α

α α

α

α α

α

α tan

cos

sin

cos2

cossin2

1cos21

cossin2

2cos1

2sin22

==/

/=

/−+/=

+

8. Without the aids of a calculator find the exact value of the expression below:

°+°+°++°+°+°= 90sin85sin80sin...15sin10sin5sin222222

x [1/3/¤]

Suggested solutions: Answer: 5.990sin85sin...10sin5sin 2222 =°+°++°+°= x

Using the trigonometric relationships: ( ) ( )α α cos90sin =− , 1cossin 22 =+ α α , 190sin2 = ,

5.02

1

4

2

2

245sin

2

2 ===⎟⎟ ⎠

⎞⎜⎜⎝

⎛ = and coupling the symmetrical terms from both ends, i.e.:

( ) ( ) ( ) (5cos85sin5cos590sin =⇔=− ), and 15cos5sin85sin5sin 2222 =+=+

( ) ( ) ( ) ( )10cos80sin10cos1090sin =⇔=− , and 110cos10sin80sin10sin 2222 =+=+







5.02

1

4

2

2

245sin

2

2 ===⎟⎟ ⎠

⎞⎜⎜⎝

⎛ = and 190sin 2 =

Answer: 5.990sin85sin80sin...15sin10sin5sin

222222

=°+°+°++°+°+°= x

α A

B

O1 α cos

α sin

α A

B

O1 α cos

α sin

α 2

α cos

α

1

C





9. The quadrangle ABCD is circumscribed by a circle as illustrated below. The angle

°= 90 ADC , the lengths of the sides a AB 4= and

a BC 3= . Find the area of the quadrangle ABCD in

terms of a . [2/4/¤]

Suggested solutions: Answer: 274.14 a Area ABCD ⋅=

The angle °=°=∠ 1352

270 ABC 1]

due to the fact that the angle

[0/

This is ABC ∠ stands on

the major arc AC and therefore, the angle ABC ∠ is

subtended by the major arc AC . In the sam the

major arc

e way

AC subtends the angle °=∠ 270 ADC at the

center. The angle subtended by an er of a

circle is twice the angle subtended at the circumference

by the same angle.

The side

arc at the cent

AC in the triangle ABC may be calculated using the cosine law:

( ) ( ) ( )( °−+= 135s43243 22 aa AC )co2 aa [0/1]

a AC a ⋅≈⇔ 48.67 2 aaa AC ≈°−+= 9.41135cos24169 2222

On the other hand in the right-triangle ADC the side AC in terms of the radius of the circle

r may be calculated using the Pythago eory:ras th

2222 ⋅=⇔+= r AC r r C A [1/0]

Therefore ar ar a 58.448.6

≈⇔⋅≈⇔⋅ [0/1] r AC 2

48.62 ≈⋅=

On the other hand the area of the quadrangle ABCD may be expressed as the sum of the areas

ADC and ABC

222

74.1424.449.102

135sin43

2 aaaaar r

Area ABCD ⋅=⋅+⋅=⋅⋅

+=⋅

[0/1]

Answer: 274.14 a Area ABCD ⋅= [1]

0. Find all values of

1 α for which [2/2/¤]

a. the products of the solutions of the following equation is the largest:

( ) 0sin4cos21322 =−−− α α x x

the sum of the solutions of the fo b. llowing equation is the largest:

( ) 0sin4cos213 22 =−−− α α x x

Suggested solutions: Answer: n⋅°+°= 18090α ; ,...2,1,07236 =∀⋅°+°= nnα

t of the solutions of the equationThe largest value of the produc α 2sin4− is 4− . This is

due to the fact that α sin is a bounded function, i.e. 1sin1 ≤≤− α . Therefore, the product

of the roots of the equation is maximum if n⋅°+°=⇔= 1809011sin2α α .±=⇔ sinα

he largest value of the sum of the solutions of the equationT ( )α 5cos213 − is: 9+ . This

corresponds to the minimum value of α cos which is 15cos −=α :

( ) ( )( ) ( ) 93312135cos213 ==−−=− α

,...2,1,0360180515cos =∀⋅°+=⇔−= nnα α

,...2,1,05

360

5

180=∀⋅

°+= nnα Answer: ,...2,1,07236 =∀⋅°+°= nnα

A

B

D

a3

a4

C





11. Show that if °<<° 18090 x :

9cos

12

sin

12 >⎟

⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ +

x x. [2/2/¤]

uggested solutions:

at

S

Due to the fact th °<< 180 x°90 , i.e. it is in the second quadrant, 1sin0 << x and

1cos0 −>> x . Therefore 1> andsin

1

x1

cos

1>−

x. Therefore 3

sin

12 >⎟

⎠

⎞⎜⎝

⎛ +

xand

3cos

2 >− x

. There1

fore: 33cos

2 ×>⎟ ⎠

⎜⎝

−⎟ ⎠ x x

or 1

sin

12

⎞⎛ ⎞⎜⎝

⎛ + 9

sin2 >⎟

⎠⎜⎝

+ x

ethod:

cos

12

1 ⎞⎜⎝

⎛ −⎟

⎠

⎞⎛

xQED.

Alternative m

Expand the LHS of the equation 9cos

12

sin

12 >⎟

⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ +

x x

Due to the fact that °<<°90 180 x , i.e. it is in the second quadrant, 1sin0 << x and

1cos0 −>> x . Therefore 1sin x

1 > and 1cos

1 >− x

.

122cossin

224 ++−+−

x x 4

11

sincos+>

x x

9cos

11

sinsin

2

cos

2>−+−

x x x x 4

9cos

12

sin

12 >⎟

⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ +

x xQED.




12. THE ROAD SIGN [Open Solutions Part NPMaDHT97 2/ 4/ ¤¤¤¤¤]

The lower part of a 2.0 m high road sign is situated 4.0 m above the eye-level of the motorist

in the figure above. The sign is hard to read from a long distance, as it is from too short a

distance.

Examine how the sight angle θ varies when the car is approaching the sign on a straightmotorway.

At what distance is the sight angle as large as possible?

The sign is readable when the sight angle is greater than 1°. Find out for how long the sign is

readable.

When assessing your work your teacher will consider:

• if the method you have used is reasonable

• if your calculations are correct

• what conclusions you have made from your investigation

• how plain and pronounced your presentation is, and

• what kind of mathematical knowledge you have shown to possess.


α

θ

m4

m2

m x A B

C

D

We may label the vertices of the triangles as

below.

In the triangle ABC :

x4tan =α ⇔ ⎟

⎠ ⎞⎜

⎝ ⎛ = −

x4tan 1α

In the triangle ABD :

( ) x

6tan =+θ α ⇔ ⎟

⎠

⎞⎜⎝

⎛ =+ −

x

6tan 1

θ α ⇔

α θ −⎟ ⎠

⎞⎜⎝

⎛ = −

x

6tan 1

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ = −−

x x

4tan

6tan 11

θ

We may plot this function:






0

2

4

6

8

10

12

0 25 50 75 100 125 150

x m

a n g l e d

e g r e e

°θ

0,21 1,00

4,90 11,54

114,36 1,00

m x °θ

Using a graphic calculator we may plot the function and find its maximum point graphically.

Answer: The maximum value of the angle θ is °≈°= 5.1154.11θ and it occurs at

mm x 59.4 ≈= . The sign is not readable at distances larger than 114 meter from the sign.

At this distance the angle drops below its lower limit and sign therefore can not be seen

clearly.

Second Method: “Linda’s method”In the right triangle : ABC Δ 222 4 x AC +=Similarly the right triangle ABDΔ :

222 6 x AD +=Using the cosine law:

Abccba cos2222 −+=

in the triangle ABC Δ

θ cos22 222 ⋅⋅⋅−+= AD AC AD AC

( ) ( ) ( )( ) θ cos462644 22222222 ⋅++−+++= x x x x

( )( ) 482436162cos462222222

+=−++=⋅++ x x x x θ

( )( )2222

2

46

24cos

x x

x

++

+=θ

( )( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

+= −

2222

21

46

24cos

x x

xθ

α

θ m4

m2

m x A B

C

D

0

2

4

6

8

10

12

0 25 50 75 100 125 150

°θ

0,210 1,001

4,900 11,537

114,360 1,000

m x °θ

( )( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

+= −

2222

21

46

24cos

x x

xθ

0,210 1,001

4,900 11,537

114,360 1,000

m x °θ





Third Method: Analytical method, using approximations to find the upper and lower limits.

Maximum value can not be calculated using this method:

When

rad rad rad 180180

tan180

1π π π

θ ≈⎟ ⎠

⎞⎜⎝

⎛ ⇔=°=

x

4tan =α and ( )

x

6tan =+θ α

( )1tantan1

1tantan1tan

⋅−

+=+

α

α α

⎟

⎠

⎞⎜

⎝

⎛ ⋅−

+=⎟

⎠

⎞⎜⎝

⎛ +

rad

rad

rad

180

tan1

180tan

180tan

π α

π α

π α ⇔

⎟

⎠

⎞⎜

⎝

⎛ ⋅−

+=

rad

x

rad x

x

180

41

180

4

6

π

π

⇔

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ ⋅− rad

xrad

x x 180

4

180

41

6 π π ⇔ 2

180

64

45

66 x x x

π π +=− ⇔ 0

45

62

180

6 2 =+−π π

x x ⇔

0243602 =+− x xπ

⇔ 24180180

2

−⎟ ⎠

⎞⎜⎝

⎛ ±=

π π x ⇔ m x 21.01 = and m x 1142 =

Answer: The road sign will not be readable when the driver is further than m x 1142 = or

closer than m x 21.01 = to it.

Fourth method: Analytical method to find the maximum value of the angle.

Later on, in the next chapter, we will find out that the differential of the function

( ) x A x f 1

tan−⋅= with respect to x is ( )2

1 x

A x f

+=′

( ) ( )ax A x f 1

tan−⋅= ⇔ ( ) ( ) y A x f 1

tan−⋅≡ ⇔ ( )( )22

11 ax

Aa

y

Aa

dx

dy

dy

df

dx

df x f

+=

+=⋅==′

( ) ( )ax A x f 1

tan−⋅= ⇔ ( )( )2

1 ax

Aa x f

+=′

Earlier, in the first method, we found out that ⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ = −−

x x

4tan

6tan 11

θ

Using the differentiation law above, we may therefore write:

Lets make a change of variable and define a new variable such that 11 −=≡ x x

y . If we

differentiate it with respect to x , it will result in: 2−−= xdx

dy.

( ) ( y y 4tan6tan11 −− −=θ )

( ) ( )2

2

2

222 161

4

361

6

161

4

361

6 −− −⋅+

−−⋅+

=⋅+

−⋅+

=⋅= x y

x ydx

dy

ydx

dy

ydx

dy

dy

d

dx

d θ θ

36

6

16

41

16

41

36

61

161

41

361

62222

2

22

2

2

2

2

2 +

−

+

=

/

⋅

+

/⋅+

/

⋅

+

/⋅−=⋅

+

+⋅

+

−=

x x x x

x

x x

x

x x

x x

dx

d θ

α

θ m4

m2

m x A B

C

D




36

6

16

422 +

−+

= x xdx

d θ

The maximum value of the angle is associated with those x values such that 0=dx

d θ .

Therefore, we may equate the equation above to zero, solve the equation for x , put the value

in the equation and verify that the point is actually the maximum and not the minimum:

036

6

16

422

=+

−+ x x

36

3

16

222 +

=+ x x

( ) ( )16336222 +=+ x x

483722 22 +=+ x x

2448722 =−= x

m x 24±= ⇔ mm x 9.424 ≈= .

x m x 1= mm x 9.424 ≈= m x 10=

36

6

16

422 +

−+

= x xdx

d θ

Positive: +0=

dx

d θ

Negative: -

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ = −−

x x

4tan

6tan 11

θ °≈°= 5.1154.11θ

Maximum

0

2

4

6

8

10

12

0 25 50 75 100 125 150

x m

a n g l e

d e g r e e

°θ

0,21 1,00

4,90 11,54

114,36 1,00

m x °θ


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