Unit 1 Maths Methods - School For Excellence School For Excellence 2017 Succeeding in the VCE – Unit 1 Mathematical Methods Page 5 The coordinate axes divide the unit circle into

Unit 1 Maths Methods

succeeding in the vce, 2017

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The School For Excellence 2017 Succeeding in the VCE – Unit 1 Mathematical Methods Page 1

CIRCULAR FUNCTIONS

(TRIGONOMETRY)

MEASURING ANGLES In previous years, we used to measure angles in degrees. Another unit of measure for

angles is called the radian ( )c . The radian is defined as the angle that results when the length of the arc of a circle is equal to the radius of that circle.

As the circumference of a circle is r2 , there are 2 radians in a full circle.

Therefore, 1 radian o

c

180)1( and 1 degree (1o )

180

radians.

Units: Radians: c or nothing. For example: 66

c

.

Degrees: o For example: o120


ANGLE CONVERSIONS

Degrees Radians

To convert degrees to radians, multiply by o

c

180

.

To convert radians to degrees, multiply by c

o

180

.

QUESTION 1 – EXAM 1

(a) Convert 30 to radians.

30180

co

6

c

(b) Convert 4

to degrees.

180

4 45

180

180


QUESTION 2 – EXAM 1 Express the following angles in radian measure.

(a) 60

(b) 240

(c) 420

(d) 390 QUESTION 3 – EXAM 1 Write the following angles in degree measure.

(a) 3

2

(b) 6

7

(c) 5

9

(d) 12

7


THE UNIT CIRCLE The principles and definitions in trigonometry are based on the UNIT circle which is a circle

with centre )0,0( and a radius of 1 unit. The equation of this circle is 122 yx . Domain: 1 1x Range: 1 1y

If is an angle measured anti clockwise from the positive direction of the X axis:

sin represents the y coordinate of a point on the circle.

cos represents the x coordinate of a point on the circle.

tan represents the gradient of the radius line that passes through a point that lies on

the circle sin

tancos

.


The coordinate axes divide the unit circle into four quadrants. These quadrants are numbered in an anti-clockwise fashion from the positive direction of the X-axis.

NOTE: Positive angles – Move in an anticlockwise direction from the positive X axis. Negative angles – Move in a clockwise direction from the positive X axis.

As the radius of the unit circle is one, the maximum and minimum values of sin and cos are 1 i.e. All values of sin and cos must lie between -1 and 1

inclusive.

1st Quadrant 2nd Quadrant

3rd Quadrant 4th Quadrant


EXACT VALUES IN THE FIRST QUADRANT

Exact values for 3

0, , , 22 2

are determined by reading values off the horizontal and

vertical axes.

QUESTION 4 – EXAM 1 Find exact values for each of the following expressions:

(a) 0902

cos 0

atcoordinatex

(b) 12702

3sin 0

atcoordinatey

(c) undefinedatcoordinatex

atcoordinatey

0

1

90

90

2tan

0

0

(d) 02702

cos 0

atcoordinatex


EXACT VALUES BASED ON THE AXES

Using the Unit Circle, the following exact values may be obtained:

sin 0 0 cos 0 1 tan 0 0

sin 12

cos 02

tan2

undefined

sin 0 cos 1 tan 0

3sin 1

2

02

3cos

3

tan2

undefined

sin 2 0 cos 2 1 tan 2 0


EXACT VALUES BASED ON TRIGONOMETRIC RATIOS

Exact values for 3

,4

,6

are determined by using trigonometric ratios i.e. SOHCAHTOA.

ehypothenus

oppositesin

ehypothenus

adjacentcos

adjacent

oppositetan

Using trigonometric ratios (SOHCAHTOA), the following exact values are obtained:

Angle ( ) 0

6

4

3

2

sin

0

2

1

2

2

2

3

2

4

2

cos

4

2

3

2

2

2

1

2

0

2

tan 0

1

3 1

3

1

Undefined

Note: In the first examination paper for Mathematical Methods, the Examiners will assume that the exact values above are known. Learn these values off by heart.

1st Quadrant 2nd Quadrant

3rd Quadrant 4th Quadrant


QUESTION 5 – EXAM 1 Complete the unit circle below.


SYMMETRY PROPERTIES OF THE UNIT CIRCLE

The symmetry properties of the Unit Circle are used to find expressions for trigonometric functions in terms of first quadrant angles. Once an expression is written in terms of a first quadrant angle, we can solve equations using exact values.

SUPPLEMENTARY ANGLES These rules allow you to write trigonometric expressions in terms of:

acute exact value 2 acute exact value

As the unit circle is symmetrical about the X and Y axes, sin, cos and tan have the same numeric values in each quadrant. The only difference in the solutions lies with the sign in each quadrant i.e. whether the solution is positive or negative. For example: In quadrant 1, sin is positive and cos is positive (as the x and y coordinates are

positive). Tan will also be positive since T has a positive y value. In quadrant 2, the point Q has a negative x value and a positive y value. Therefore,

sin is positive and cos is negative. The line QS can be extended until it meets the tangent at V. The y value of this point

is negative and therefore, tan is negative in the forth quadrant.

Quadrant 2 Quadrant 1

Quadrant 3 Quadrant 4

Q

R S

V

T

P


Once the first quadrant values of a trigonometric function are known, we can use the symmetry properties of the unit circle to find the corresponding values in the other quadrants.

SUMMARY OF SIGNS (CAST)

1st quadrant: All trigonometric values are positive. 2nd quadrant: Only Sin is positive. 3rd quadrant: Only Tan is positive. 4th quadrant: Only Cos is positive.

THE SINE FUNCTION As lengths above the X axis are signed positive, sin is positive in the 1st and 2nd quadrants. As lengths below the X axis are signed negative, sin is negative in the 3rd and 4th quadrants. By symmetry:

sin(180 ) sino

sin)sin(

sin(180 ) sino sin)sin(

sin(360 ) sino

sin)2sin(


THE COSINE FUNCTION As lengths to the right of the Y axis are signed positive, cos is positive in the 1st and 4th quadrants. As lengths to the left of the Y axis are signed negative, cos is negative in the 2nd and 3rd quadrants. By symmetry:

cos(180 ) coso

cos)cos(

cos(180 ) coso

cos)cos(

cos(360 ) coso

cos)2cos(

THE TANGENT FUNCTION As the gradient of the radius line is positive in the 1st and 3rd quadrants, tan is positive in these quadrants. As the gradient of the radius line is negative in the 2nd and 4th quadrants, tan is negative in these quadrants. By symmetry:

tan(180 ) tano tan)tan(

tan(180 ) tano

tan)tan( tan(360 ) tano

tan)2tan(


NEGATIVE ANGLES

cos(2 ) cos)cos(

sin(2 ) sin)sin(

tan(2 ) tan)tan(

INVERSE OPERATIONS

1sin undoes sin i.e. 1sin (sin )x x

1cos undoes cos i.e. 1cos (cos )x x

1tan undoes tan i.e. 1tan (tan )x x


QUESTION 6 – EXAM 1 If tx )sin( write the following expressions in terms of t .

(a) sin(180 )o x

(b) sin(360 )o x (c) )sin( x


QUESTION 7 – EXAM 1 If 5.0sin and cos 0.9 , find the exact value of: (a) )sin( )sin( sin 5.0 (b) )sin( )sin( sin 5.0 (c) )2cos( (d) )tan(



If 5

12tan state the exact value for the following expressions:

(a) )2cos( (b) )tan( (c) )sin(


(d) )sin( (e) 6tan (f) )sin()cos(



(a) If sin sin6

x

and

2x

, find the value of x .

(b) If sin sin6

x

and

3

2x



(c) If sin sin6

x

and

32

2x



COMPLEMENTARY ANGLES

To simplify ratios of trigonometric expressions with the angles

2 and

2

3

(where represents an acute angle), we apply complementary rules.

Consider the points ( )P and 2

P

.

By symmetry, the x and y coordinates of ( )P are equal to the y and x coordinates of

2P

. Therefore:

cos

2sin

sin

2cos

Similarly:

cos

2sin

sin

2cos

Other Complementary Rules:

cos

2

3sin

sin2

3cos

cos

2

3sin

sin2

3cos


QUESTION 10 – EXAM 1 & 2 If tx )sin( and ux )cos( write the following expressions in terms of t and u . (a) )sin( x txx sinsin (b) )2sin( x txx sin2sin (c) )cos( x

(d)

x

2sin

(e)

x

2

3sin

QUESTION 11 – EXAM 1 & 2

Given that 3

cos 0.587810

, find sin5

using complementary rules.

Solution


FINDING EXACT VALUES OF TRIGONOMETRIC EXPRESSIONS To find exact values of trigonometric expressions, write expressions in terms of an acute

angle such as ,2 3

,

4

,

6

, and apply the rules derived from the symmetry properties of

the unit circle. METHOD: Step 1: Identify the quadrant in which the angle lies.

Step 2: Write the given expression in terms of a 1st quadrant angle.

Step 3: Write the appropriate quadrant rule and solve.

To write an expression whose angle is based on

2 in terms of a first

quadrant angle, apply the following rule:

tan

cos

sin

2

tan

cos

sin

To write an expression whose angle is based on

2

32

in terms of a first

quadrant angle, apply the following rule:

cos

sin

2

32

sin

cos

Use CAST to determine the sign of the answer.

Use CAST to determine the sign of the answer.


QUESTION 12 – EXAM 1 & 2

Show that the exact value of 5 2

sin4 2

.

Solution Identify the quadrant in which the angle lies:

o225180

4

5

As 225 lies between 180 and 270, the angle lies in the third quadrant.

Write the given expression in terms of a 1st quadrant angle:

4sin

4

5sin

Write the appropriate quadrant rule and solve:

tan

cos

sin

2

tan

cos

sin

sinsin

4sin

4sin

1 2

22

Use CAST to determine the sign of the answer


QUESTION 13 – EXAM 1 Find exact values of the following expressions:

(a)

3

2tan

3tan

3tan

3

(b) 225sin )45180sin( oo

2

2

2

1

2

2

(c)

3

5cos

32cos

3cos

2

1

(d) 315cos


(e)

3

2cos

(f)

2

5sin

(g)

4sin

(h)

4

3tan

(i)

6

11cos


(j)

3

5sin

(k)

4

13tan

(l)

4

15cos

(m)

6

7sin



If 4

3cos and

1sin

5 , find the exact value of the following expressions:

(a) )cos(

(b) sin( )

(c) cos2

Solution


QUESTION 15 – EXAM 1 Calculate the exact values of the following expressions without using a calculator:

(a)

6tan

4cos

3sin

(b) 150cos150sin2300sin QUESTION 16 – EXAM 2

Show that

4

7cos

6

7sin

4sin

3

2cos

0 .

Solution


THE FUNDAMENTAL TRIGONOMETRIC IDENTITY By applying the Pythagorean Identity on the Unit Circle, the following relationship is obtained:

122 yx Substituting ysin and xcos , we obtain the following rule:

1cossin 22 This statement is true for all values of , and is known as an identity. This identity may be used to find the value of one trigonometric expression (such as cos) given the value of a different trigonometric expression (such as sin).



If 17

15cos and

2

find sin .

Solution Step 1: Use the appropriate identity to find a solution for the unknown trigonometric expression. As 1cossin 22

117

15sin

22

1289

225sin 2

2sin289

64

sin17

8

Step 2: Determine the correct sign by observing the quadrant in which the solution is to lie. Since lies in the second quadrant, sin is positive.

17

8sin

We can also use triangles to help us solve these types of problems.

A knowledge of the following triads will assist in the construction of triangles:

3, 4, 5 5, 12, 13 8, 15, 17 7, 24, 25

Whichever technique is used, careful consideration must be given to the quadrant in which the solution lies. Make sure that you assign the correct sign (positive or negative) by considering CAST.



If 9

5cos and

2

find tan .

Solution Step 1: Draw a triangle in the relevant quadrant and label according to the given ratio. Step 2: Find the missing side using Pythagoras’s Theorem.

5659 22 x

Step 3: Find the required ratio taking the signs of the quadrant into consideration.

9

5cos

5

142

5

56tan

ADJ

OPP

9

-5



(a) If 7

4sin x find xcos .

(b) If 3

2cos

x find xsin where

x2

.


(c) If 12

5cos x , find xsin given that

22

3 x .


QUESTION 20 – EXAM 1 Find:

(a) cos if 5

3sin and

2

.

(b) sin if 13

3cos and

22

3 .


(c) tan if 13

5sin

when

2

3 .


GRAPHS OF TRIGONOMETRIC FUNCTIONS

THE SINE FUNCTION


THE COSINE FUNCTION

THE TANGENT FUNCTION


SOLVING TRIGONOMETRIC EQUATIONS

If we were asked to find x such that 2

1sin x , then from your table of values (or memory)

you would say that 306

orx

.

We could also use the inverse sine function on our calculator to obtain the same solutions:

1(0.5)6

Sin

In both cases, only one of the possible solutions will have been identified. For most trigonometric expressions, there will always be two solutions for each cycle of that trigonometric function.

For example: Given 2

1sin x ,

6

x or

6

5.

The total number of solutions for a trigonometric expression depends upon the given period and domain. To find the solutions for a given application: Step 1: Use exact values or the calculator to find the first quadrant angle. Step 2: Apply the relevant quadrant rules to determine the solutions across one period. Step 3: Add and subtract the period to each of the solutions, until all solutions across the specified domain are obtained.

x

y

-1

0

1


METHOD:

Step 1: Write all expressions in terms of one trigonometric function. Step 2: Transpose the given equation so that the trigonometric expression (and the angle) is on one side of the equation, and the constants are located on the other side of the equation. Step 3: Use the sign in front of the constant on the right hand side to determine the quadrants in which the solutions are to lie (use CAST). Step 4: Calculate the first quadrant solution. If the exact value cannot be determined:

Press Inverse Sin, Cos or Tan of the number on the right hand side of the equation (but ignore the sign). (Ensure that the calculator is in Radian Mode).

For example: )(1 signtheignorebutequationofRHSonnumberSin Step 5: Solve for the variable (usually x or ). Let the angle equal the rule describing angles in the quadrants in which the solutions are to lie. Note: First Quadrant Angle = FQA Let angle = FQA if solution lies in 1st Quadrant.

Let angle = FQA if solution lies in 2nd Quadrant.

Let angle = FQA if solution lies in 3rd Quadrant.

Let angle = FQA2 if solution lies in 4th Quadrant.

1st Quadrant

Rule: FQA 2nd Quadrant

Rule: - FQA

3rd Quadrant

Rule: + FQA 4th Quadrant

Rule: 2 - FQA


Step 6: Evaluate all possible solutions by observing the given domain. This is accomplished by adding or subtracting the PERIOD (P) or (T) to each of the solutions, until the angles fall outside the given domain.

For sine and cosine functions: variable the of front in number The

Period2

For tangent functions: variable the of front in number The

Period

Always look closely at the brackets in the given domain and consider whether the

upper and lower limits can be included in your solutions.

DO NOT discard any solution until the final step.

Step 7: Eliminate solutions that do not lie within the given domain. Note: Given an inequation – solve the equation without the inequality and then reason from the graph. Students may also solve trigonometric equations by rearranging the domain.



Solve the following equation for x : 3sin2 x , ]4,0[ x . Solution Transpose the given equation so that the trigonometric expression (and the angle) is on one side of the equation, and the constants are located on the other side of the equation:

2

3sin x

Calculate the first quadrant solution: 1st Quadrant Angle: 32

31

Sinx

Use the sign in front of the constant on the right hand side to determine the quadrants in which the solutions are to lie: Solutions are to lie in the quadrants where sine is positive i.e. the 1st and 2nd quadrants: Solve for the variable (usually x ). Let the angle equal the rule describing angles in the quadrants in which the solutions are to lie:

3

x and AngleQuadrantstx 1

3

3

2

3

2,

3:

xx

Evaluate all possible solutions by adding and/or subtracting the PERIOD (P) or (T) to each of the calculated answers observing the given domain:

3

62

1

2 T

3

8,

3

7,

3

2,

3:

xx

CAS Application:

Using a CAS calculator, sketch 2sin 3y x and find the X intercepts. The answers should be the same as the solutions above.

S A T C



Solve 2 sin 1x , ]2,2[ .

Solution



For the equation 2sin 2 3x , the sum of the solutions in the interval [0, 2 ] is equal to:

A 2

3

B 4

3

C 11

6

D 2

E 5



Solve 2cos 3x , ]2,0[ x . Solution


QUESTION 25 – EXAM 2 The number of solutions of the equation cos(3 ) 0.5x between 0x and 2x is equal to: A 0 B 1 C 2 D 3 E 4



Solve the equation 3 tan 2 1 0x over the domain [ , ] . Solution

© The School For Excellence 2017 Succeeding in the VCE – Unit 1 Mathematical Methods Page 1

SUCCEEDING IN THE VCE 2017

UNIT 1 MATHEMATICAL METHODS

STUDENT SOLUTIONS

FOR ERRORS AND UPDATES, PLEASE VISIT

WWW.TSFX.COM.AU/VCE-UPDATES QUESTION 2

(a) 3

(b) 3

4

(c) 3

7

(d) 6

13

QUESTION 3 (a) o120

(b) o210

(c) o324

(d) o105 QUESTION 4 (i) (b) '9166 o

(c) o18.229 (ii) (b) c87.1

(c) c39.2 QUESTION 6 (a) t (b) t (c) t QUESTION 7 (c) 0.9

(d) 5

9


QUESTION 8

(a) 13

5

(b) 5

12

(c) 13

12

(d) 13

12

(e) 5

12

(f) 169

60

QUESTION 9 (a) (b) (c) QUESTION 10 (c) uxx coscos

(d) uxx

cos

2sin

(e) uxx

cos

2

3sin


QUESTION 11 5878.0 QUESTION 13

(d) 2

2

(e) 2

1

(f) 1

(g) 2

2

(h) 1

(i) 2

3

(j) 2

3

(k) 1

(l) 2

1

(m) 2

1

QUESTION 14

(a) 4

3coscos

(b) 1sin sin

5

(c) 1

cos sin2 5

QUESTION 15

(a) 4

2

(b) 0


QUESTION 16

QUESTION 19

(a) 7

33cos x

(b) 3

5sin x

(c) 12

119

QUESTION 20

(a) 5

4

(b) 13

2

(c) 5

12

QUESTION 22 4

7,

4

5,

4,

4

3 x

QUESTION 23 Answer is E

QUESTION 24 6

7,

6

5 x


QUESTION 26

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Unit 1 Maths Methods - School For Excellence School For Excellence 2017 Succeeding in the VCE – Unit 1 Mathematical Methods Page 5 The coordinate axes divide the unit circle into