Math 12 - 4-Trigonometry

Math 12 - 4.1-Angle Measurement in Radians, Degrees and Turns

1 full rotation = 360° = 2π Radians 12

rotation = 180° = π Radians

14

rotation = 90° = π2

Radians 16

rotation = 60° = π3

Radians

18

rotation = 45° = π4

Radians 112

rotation = 30° = π6

Radians

r

r

AB

a

+θ-θ

Arc length-a travels along the circumference of a circle of radius -r for acentral angle θ in radian a=θr

A co-terminal angle in degree --> θ±n360 ° , n∈N

A co-terminal angle in radians --> θ±n2 π , n∈N

Exp. 4.1.1 Convert each angle to a number of rotations and then draw it in standard position of a circle.

(a) 2 π3

(b) 160° (c) 15turns (d) – 320° (e) −7 π

8 (f) −3

5turns

Solution:

(a) 2 π3=180 °

π x 2 π3=120 °=1 turn

2πx2 π

3=+1

3turn→

(b) 160 °= π180 °

x 160 °=8 π9=1 turn

360 °x160 °=+4

9turn→

(c) 15turn= 360 °

1 turnx 1

5turn=72°=( 2π

1 turn) x 1

5turn= 2π

5

(d) −320 °= π180°

x(−320 °)=(−16π)

9=1 turn

360 °x(−320 °)=−8

9turn→

r

A

2 π3

r

A

-320°

r

A

72°

r

A

160°

(e) (−7π)

8=180 °

π x(−7π)

8=−157.5°=1 turn

2πx(−7π)

8=−7

16turn→

(f) −35

turn= 360 °1 turn

x(−35

turn)=−216 °=( 2π1 turn

) x (−35

turn)=−6π5

r

A

−7 π8

r

A

−35

turns

Exp. 4.1.2 Given thedomainof −1080°≤θ≤1080° , find the co-terminal angles of 160° and express

them in general form ?

Solution: Given thedomainof −1080°≤θ≤1080° , the co-terminal angle of θ is θ±n360 °

θ ( ° ) n θ±n360 ° Is the angle in the domain

−1080°≤θ≤1080° ?

160 -4 -1280 No

160 -3 -920 Yes

160 -2 -560 Yes

160 -1 -200 Yes

160 0 160 Yes

160 1 520 Yes

160 2 880 Yes

160 3 1240 No

So the co-terminal angle of 160° is 160 °±n360° where n∈N ,−2≤n≤2

Exp. 4.1.3 Given the domainof −6π≤θ≤6π , find the co-terminal angles of 25π and express

them in general form ?

Solution: Given thedomainof −6π≤θ≤6π , the co-terminal angle of θ is θ±n2π

θ( Radian )

n θ±n2π Is the angle in the domain

−6π≤θ≤6π ?

25π

-4 −385

πNo

25π

-3 −285

πYes

25π

-2 −185

πYes

25π

-1 −85

πYes

25π

0 25π

Yes

25π

1 125

πYes

25π

2 225

πYes

25π

3 325

πNo

So the co-terminal angle of 25π is 2

5π±n2πwhere n∈N ,−4≤n≤3

Exp. 4.1.4 A big circle round about with a radius of 15 m., Jennifer walks along the edge from point A to B as shown in the figure. What is the distance AB ?

Solution: Arc length-a travels along the circumference of a circle

of radius -r for a central angle θ in radian a=θ r

θ in radian is (20 °+40 °) x π180 °

=π3

So

The distance AB=a=π3x15m=5πm=5x3.14168=15.7m

Exp. 4.1.5 John wants to find out the distance PQR along the circumference of a big circle with a radius of 30 m. He draws the diagram as shown in the figure. Can you help John

calculate the distance PQR?

Solution: Arc length-a travels along the circumference of a circle

of radius -r for a central angle θ in radian a=θ r

θ in radian is (360°−120 °) x π180 °

=4π3

So

The distance PQR=a=4π3

x 30m=40x3.14168=125.66 m

r = 15 m.

A

r = 15 m.

B

20°

40°

r = 30 m.

Pr = 30 m.

R

Q

120°

Math 12 - 4.2-The Unit Circle – Trigonometric Ratios

A unit circle is a circle that has the radius of 1, from the general circle equation x2+ y2=r2 , replace a

radius with 1, so x2+ y2=1

Trigonometric Ratios Definition Reciprocal

sine sin θ= yr

1cscθ

cosine cosθ= xr

1secθ

tangent tanθ= yx

1cotθ

cosecant cscθ= ry

1sin θ

secant sec θ= rx

1cosθ

cotangent cotθ= xy

1tanθ

R=1 unitr

S(x, y)

θ

sin, csc --> +

cos, sec --> +tan, cot --> +

ALL --> +

R=1 unit

θ= π6

S( , )12

√32

R=1 uni

t

S( , )1√2

1√2

θ= π4

R=1

uni

t

θ= π3

S( , )12

√32

Exp. 4.2.1 On a unit circle as shown in the figure, point S(0.91, 0.42) is on the circumference. Determine

sin θ ,cosθ , tanθ , cscθ , secθ ,cot θ ?

Solution:

Find the Trigonometric Ratios

sin θ= yr=0.42

1=0.42

cosθ= xr=0.91

1=0.91

tanθ= yx=0.42

0.91=0.4615

cscθ= ry= 1

0.42=2.3809

sec θ= rx= 1

0.91=1.0989

cotθ= xy=0.91

0.42=2.1666

R=1 unit

S(0.91, 0.42)

θ

Exp. 4.2.2 Find the exact value from the unit circle of

(a) csc ( 43

π)

(b) sec(−32

π)

(c) cot(−0.8turn)

(d) sin (109

π)

(e) cos(−23

π)

(f) tan (−2.25 turns )

Solution:

(a) csc( 43π) , 4

3π → 180°

π x 43

π=240 °

From the Unit Circle --> cscθ= ry= 1

sin θ --> csc( 4

3π)= 1

sin( 43π)

= 1

( yr)

Draw the angle in the Unit Circle and then find y.

csc( 43π)= 1

sin( 43π)

= 1

(−0.8661

)=1.1547

r=1

43π →240 °

(b) sec (−32

π) , −32

π→ 180 °π x(−3

2π)=−270 °

From the Unit Circle --> sec θ= rx= 1

cosθ --> sec (−3

2π)= 1

cos(−32

π)= 1

( xr)

Draw the angle in the Unit Circle and then find y.

sec (−32

π)= 1

( xr)= 1

(01)→ It can ' t be defined.

(c) cot(−0.8turns ) , (−0.8 turn)→ 360 °1 turn

x (−0.8 turn)=−288 °

From the Unit Circle --> cot θ= xy= 1

tanθ --> cot(−0.8turns )= 1

tan(−0.8turns )= 1

( yx)

Draw the angle in the Unit Circle and then find x, y.

cot(−0.8turns )= 1

(0.950.3

)=0.3154

r=1

−32

π→−270°

r=1

−0.8 turns→−288°

(d) sin (109

π) , 109

π=π+19

π→180°+(180 °π x

19

π)=200 °

From the Unit Circle --> sin θ= yr

Draw the angle in the Unit Circle and then find y.

sin (109

π)= yr=−0.3

1=−0.3

(e) cos(−23

π) , −23

π→ 180°π x (−2

3π)=−120 °

From the Unit Circle --> cosθ= xr

Draw the angle in the Unit Circle and then find y.

cos(−23

π)= xr=−0.5

1=−0.5

r=1

109

π→ 200 °

r=1

−23

π→−120°

(f) tan (−2.25 turns ) , (−2.25 turns )→ 360°1 turn

x (−2.25 turns)=−810°=−((2X360 ° )+90 °)

From the Unit Circle --> tan θ= yx

Draw the angle in the Unit Circle and then find x, y.

tan (−2.25 turns )= yx= 1

0→ It can' t be defined.

r=1

Exp. 4.2.3 Given trigonometric ratios, determine the angles (rounding to 2 decimal places) ?

(a) sin θ=0.6428

(b) cosθ=0.7071

(c) tanθ=−4.7046

Solution:

(a) sin θ=0.6428 , sin θ= yr

--> draw the angle in the Unit Circle, y=0.6428, r=1

Then measure the angle θ -->

θ≈40.76 °

(b) cosθ=0.7071 , sin θ= xr

-->

draw the angle in the Unit Circle, x=0.71,r=1

Then measure the angle θ -->

θ≈45°

Y=

0.65

r=1

θ

r=1

θX=0.71

(c) tanθ=−4.7046 , tanθ= yx

-->

Find the possible proportion between y and x that satisfies

the equation in the Unit Circle,

the possible y is 0.94 and x is 0.2

draw the angle in the Unit Circle, y=−0.94, x=0.20

Then measure the angle θ -->

θ≈78°

r=1

θ

Math 12 - 4.3-Trigonometric Equations – How to solve it?

- Find solution for trigonometric ratio.- Determine all solutions in a given domain, such as 0°≤θ≤360 °

- Write all possible solutions in general form of

A co-terminal angle in degree --> θ±n360 ° , n∈N

A co-terminal angle in radians --> θ±n2 π , n∈N

It's good to remember that :

• sin θ , cscθ are positive−Q1 ,Q2

• cosθ , secθare positive−Q1 ,Q4

• tan θ ,cotθ are positive−Q1 ,Q3

sinθ , cscθ are positive.

cosθ , secθare positive.

tanθ ,cot θ are positive.

ALL are positive.

Exp. 4.3.1 Given first-degree trigonometric equations, find the angles (rounding angles to the nearest tenth of a degree ) ?

(a) sin θ−1=2−3sinθ ,0 °≤θ≤360 °

(b) 7cosθ+5=2+3cosθ ,0 °≤θ≤360 °

(c) 5tanθ−7=−8+3tanθ ,−π≤θ≤0

Solution:

(a) sin θ−1=2−3sinθ ,0 °≤θ≤360 ° --> sin θ= 34=sine48.59 ° →θ=48.6°

(b) 7cosθ+5=2+3cosθ ,0 °≤θ≤360 ° --> cosθ=−34

=cos138.59° →θ=138.6 °

(c) 5tanθ−7=−8+3tanθ ,−180 °≤θ≤0 ° --> tan θ=−12

=tan(−26.565°)→θ=−26.6°

Exp. 4.3.2 Given second-degree trigonometric equations, find the angles (rounding angles to the nearest tenth of a degree ) ?

(a) 4(cosθ)2=1,0 °≤θ≤360 °

(b) 2(sin θ)2=1,0 °≤θ≤360 °

(c) ( tan θ)2−tanθ=6,−π≤θ≤0

Solution:

(a) 4(cosθ)2=1,0 °≤θ≤360 ° --> cosθ=±12

If cosθ= 12

→cosθ=cos60 °=cos300 °→θ=60° ,300 °

So θ=60 ° ,300 ° , where0≤θ≤360°

(b) 2(sin θ)2=1,0 °≤θ≤720 ° --> sin θ=± 1√2

If sinθ= 1√2

→ sin θ=sin45°=sin135 °→θ=45° ,135°

So θ=45°+n360 ° , where 0≤θ≤720° ,n∈N and

θ=135 °+n360° ,where 0≤θ≤720 ° ,n∈N

(c) ( tan θ)2−tanθ=6,−π≤θ≤0 --> ( tan θ)2−tanθ=6→( tanθ−3)( tanθ+2)=0

If tanθ=3→ tan θ=tan ( 25

π)→θ= 25

π→θ∉the domain of −π≤θ≤0

If tanθ=−2 → tan θ=tan (−67360

π)=tan (−293360

π)→θ=−67360

π ,− 293360

π→θ∈the domainof −π≤θ≤0

So θ=−67360

π ,− 293360

πwhere−π≤θ≤0

Math 12 - 4.4-Trigonometric Functions – How to Graph it

In the form of y=asin bθ , y=acosbθ , where

a Vertical stretch by the factor of |a|

bHorizontal stretch by the factor of

1∣b∣

Amplitude max. value−min. value2

Period 2π∣b∣

y=sin θ , y=cosθ

y=sin θ y=cosθ

Domain {θ : [θ∈R } {θ : [θ∈R }

Range {y : [−1≤ y≤1, y∈R } {y : [−1≤ y≤1, y∈R }

The maximum value 1 1

The minimum value -1 -1

The amplitude 1 1

The period 2π 2π

The y-intercept 0 1

The θ - intercepts −2π ,−π ,0,π ,2π , −5π2

,−3π

2,−π

2, π

2,3π2

,5π2

y=cos θ

y=sin θ

Am

plit

ude

period

Exp. 4.4.1 Sketch the graphs of the functions ?

(a) y=cosθ ,0≤θ≤4π (b) y=sin θ ,0≤θ≤4π

Solution: (a) y=cosθ ,0≤θ≤4π , find θ , y=cosθ then sketch the graph as shown.

θ Radians y=cosθ

0 0 1π4 0.79 0.7

π2 1.57 0

3π4

2.36 -0.71

π 3.14 -1

5π4

3.93 -0.71

6π4

4.71 0

7π4

5.5 0.71

2π 6.28 1

9π4

7.07 0.71

10 π4

7.85 0

11 π4

8.64 -0.71

12 π4

9.42 -1

13 π4

10.21 -0.71

14 π4

11 0

15 π4

11.78 0.71

4π 12.57 1

(b) y=sin θ ,0≤θ≤4π , find θ , y=sin θ then sketch the graph as shown.

θ Radians y=sin θ

0 0 0π4 0.79 0.71

π2 1.57 1

3π4

2.36 0.71

π 3.14 0

5π4

3.93 -0.71

6π4

4.71 -1

7π4

5.5 -0.71

2π 6.28 0

9π4

7.07 0.71

10 π4

7.85 1

11 π4

8.64 0.71

12 π4

9.42 0

13 π4

10.21 -0.71

14 π4

11 -1

15 π4

11.78 -0.71

4π 12.57 0

Exp. 4.4.2 Given the function y=−2sin2 θ ,0≤θ≤4π

(a) What is the period ? (b) What is the amplitude ?

Solution: Sketch the graph of y=−2sin2θ ,0≤θ≤4π as shown in the figure.

(a) The period is ¶(b) The amplitude is 2 units

Period=¶

Am

plit

ude

=2

Exp. 4.4.3 Given the function y=−3cos4θ ,0≤θ≤2π

(a) What is the period ? (b) What is the amplitude ?

Solution: Sketch the graph of y=−3cos4θ ,0≤θ≤2π as shown in the figure.

(a) The period is π2

(b) The amplitude is 3 units

Period=¶/2 Am

plit

ude=

3 un

its

Math 12 - 4.5-Transformations of Trigonometric Functions.

In the form of y=asin b(θ−c)+d , y=a cosb(θ−c)+d , where

a Vertical stretch by the factor of |a| .The amplitude is changed to |a|.

a < 0 --> Horizontally reflected in the x-axis.

bHorizontal stretch by the factor of

1∣b∣

The period is changed to 360°∣b∣

or 2π∣b∣

radians

b < 0 --> vertically reflected in the y-axis.

c Horizontal translation – phase shiftc > 0, the graph shifts to the right c units.c < 0, the graph shifts to the left c units.

d Vertical translationd > 0, the graph shifts up d units.

d < 0, the graph shifts down d units.

d =max. value+min. value

2

The sinusoidal axis It is defined by the line y = d

max. value−min. value2

Domain {θ : [θ∈R }

Range {y : [min. value≤ y≤max.value , y∈R }

Order oftransformations 1. Horizontal stretches and reflections, by the factor of

1∣b∣

2. Vertical stretches and reflections, by the factor of ∣a∣

3. c – horizontal translation and d – vertical translation.

Exp. 4.5.1 Given the function y=sin (θ−π2)−2

(a) Sketch the graph of the function ? (b) What are the Domain and Range ?

Solution: (a) Sketch the graph of y=sin (θ−π2)−2

(b) The Domain is {θ : [θ∈R }

The Range is {y : [−3≤ y≤−1 , y∈R }

y=sin(θ−π2)−2

y=sin θ

Exp. 4.5.2 Given a graph as shown in the figure, determine the equation of the graph?

Solution: It looks like a sine function equation, assume y=asin b(θ−c)+d

aVertical stretch ∣a∣= 1

4, horizontally reflecting in the x-axis --> a=−1

4

bHorizontal stretch by 1

∣b∣= 1

3, The period is changed to 2π

∣b∣= 2π

3

c No horizontal translation

d Vertical translation = 0.5, the graph shifts down 0.5 units --> d = - 0.5

The sinusoidal axis The line y = d max. value+min. value

2=

−0.25+(−0.75)2

=−0.5

Domain {θ : [θ∈R }

Range {y : [−0.75≤ y≤−0.25 , y∈R }

So y=asin b(θ−c)+d → y=(−14

)sin 3(θ)− 12, where {θ : [θ∈R } ,{y : [−0.75≤ y≤−0.25 , y∈R }

Period

Am

plit

ude

Math 12 - 4.6-Trigonometric Functions – Tangent Function

y=tan θ

The period 180 ° ,π

The vertical asymptotesThe discontinuity θ=π

2,32

π ,52

π , .... ,(2n−1)

2πwhere n∈Z

Domain {θ : [θ≠nπ+π2,θ∈R ,n∈Z }

Range {y : [ y∈R}

X – intercepts θ=π ,2π ,3π , .... , nπwhere n∈Z

At the Unit Circle

tan θ= yx=The slope= sin θ

cosθ

Exp. 4.6.1 Sketch the graph of y=tan θ ,−π≤θ≤π

Solution: y=tan θ ,−π≤θ≤π , find y=tan θ then sketch the graph as shown.

θ Radians y=tan θ θ Radians y=tan θ θ Radians y=tan θ

−π -3.14 0 −5π12

-1.31 -3.73 π2

1.57 328149

−5π6

-2.62 0.58 −π3

-1.05 -1.74 π2

1.57 -328149

−2π3

-2.09 1.73 −π6

-0.52 -0.58 19 π36

1.66 -11.43

−98 π180

-1.71 7.11 0 0 0 5π9

1.75 -5.67

−19 π36

-1.66 11.43 π6

0.52 0.58 2π3

2.09 -1.73

−π2

-1.57 328149 π3

1.05 1.74 5π6

2.62 -0.58

−π2

-1.57 -328149 5π12

1.31 3.73 π 3.14 0

−17 π36

-1.48 -11.43 17 π36

1.48 11.43

Exp. 4.6.2 An AIR Defence station aims the STA missile to attack the intruder DRONE as shown in the figure. The intruder DRONE flies at the constant attitude at 2,000 m. High. Describe the distance d in term of ɵ ?

Solution: As the DRONE approaches the station horrizontally, the relation between d and ɵ is described

tan θ= d2000

then sketch the graph as shown.

From the graph, the bigger the angle ɵ, the shorter the distance dWhen the angle ɵ approaches to 90°, the distance changes rapidly.

d

2000

0 m

ɵ

π2

−π2

π4−π

4

Math 12 - 4.7-Trigonometric Functions – Equations and Graphs

To find the the graph that fit characteristics of the specific phenomenon,the following sinusoidal equations are very useful.

y=a cosb(θ−c)+d

ory=asin b(θ−c)+d

The parameters a, b, c, d can be adjusted to fit the equations for each phenomenon such as- Wave shapes of ocean waves, sound waves, heart beats

-Circular motion of ferris wheel-Swing of pendulum-Pistons of machine

Exp. 4.7.1 Solve these equations over the intervals by graphing ?

(a) sin θ=√32

,−3π≤θ≤3π

(b) cos3θ= 12,−360 °≤θ≤360°

(c) 1=5sin [3(θ−1)]+3,−2π≤θ≤2π

Solution:

(a) sin θ=√32

,−3π≤θ≤3π --> sketch the graphs of y=sin θ and y=√32

From the graphs --> θ=−5π3

,− 4π3

, π3,2π3

,5π3

,8π3

,

y=sin θ

y=√32

(b) cos3θ= 12,−360 °≤θ≤360° --> sketch the graphs of y=cosβ ,β=3θ and y=1

2

From the graphs --> β=3θ=−300 °→θ=−100 °

β=3θ=−60° →θ=−20 °

β=3θ=60° →θ=20°

β=3θ=300 °→θ=100°

y=cosβ ,β=3θ

y=12

(c) 1=5sin [3(θ−1)]+3,−2π≤θ≤2π --> −25

=sin [3(θ−1)] ,−2π≤θ≤2π -->

sin β=−25

,β=3(θ−1) ,−2π≤θ≤2π

--> sketch the graphs of y=sinβ ,β=3(θ−1) and y=−25

From the graphs --> β=3(θ−1)=−78

π →θ= 1724

π

β=3(θ−1)=−18

π →θ= 2324

π

β=3(θ−1)=98

π→θ=3324

π

β=3(θ−1)= 158

π→θ= 3924

π

y=sin β ,β=3(θ−1)

y=−25

Exp. 4.7.2 Given the average temperatures of 12 months in Chamonix, France. Find the sinusoidal function equation of these temperatures ?

Month (°C) Month (°C) Month (°C)

January -20.5 May 5 September 6.6

February -15 June 9.7 October 1.9

March -9.5 July 12.4 November -11.5

April -2 August 10.6 December -14.9

Solution: Sketch the graph of Month(x axis) – Temperature(°C – y axis)

The graph looks like a cos function, assume y=a cosb( x−c)+d ,a<0

∣a∣=max y value−min y value2

=12.4 °C−(−20.5° C )

2=16.45° C≈ 33

2

The estimated Period: 12

3.1416π≈3.8π =

2π∣b∣

--> ∣b∣= 2π3.8π

≈ 12

The phase shift c , notice at the minimum Jan- -20.5C – it shifts to the right almost 13π≈1unit

c = 1, d=max yvalue+min y value2

=12.4° C+(−20.5° C )

2=−4° C

S0 y=−332

cos(12( x−1))−4 --> sketch this grpah and compare with the first graph which is from the

values from the table as hsown in the next figure.

y=−332

cos(12(x−1))−4

Math 12 - 4.8-Trigonometric Functions

Equations from Pythagorean Theorem

Reciprocal and Quotient Equations

Equations from Pythagorean Theorem

x2+ y2=r2 where x=r cosθ , y=r sinθ

Replace a radius with 1 in the unit circle. --> x2+ y2=1, x=cosθ , y=sinθ

(sin θ)2+(cosθ)2=1

(sinθ)2

(cosθ)2 +(cosθ)2

(cosθ)2 =1

(cosθ)2 →( tanθ)2+1=(secθ)2

(sinθ)2

(sinθ)2 +(cosθ)2

(sinθ)2 =1

(sin θ)2 →(cotθ)2+1=(csc θ)2

Reciprocal Equations

sin θ= 1cscθ

, cosθ= 1sec θ

, tan θ= 1cotθ

R=1 unit

S(x, y)

R

θ

Quotient Equations

tan θ= sin θcosθ

, cotθ= cosθsin θ

Sum Equationssin (θ+β)=sin θ cosβ+cosθsinβ , cos(θ+β)=cosθcosβ−sinθ sinβ ,

tan (θ+β)=tan θ+tanβ

1−tanθ tanβ

Difference Equationssin (θ−β)=sin θ cosβ−cosθsinβ , cos(θ−β)=cosθcosβ+sin θsinβ ,

tan (θ−β)=tan θ−tanβ

1+tanθ tanβ

Equal-2 Angles Equations

sin (2θ)=2sin θcosθ , cos(2θ)=(cosθ)2−(sin θ)2=1−2(sin θ)2=2(cosθ)2−1 ,

tan (2θ)= 2tanθ1−(tanθ)2

Useful Exact Trigonometric Valuesθ sinθ cosθ tanθ

0 0 1 0

30° , π6

12

√32

1√3

45° , π4

1√2

√2 1

60° , π3

√32

12

√3

90° , π2

1 0 Un-defined

180 ° ,π 0 -1 0

270 ° ,32

π-1 0 Un-defined

360° ,2π 0 1 0

Exp. 4.8.1 Verify that (cotθ)2−(cscθ)2=−1 can be derived from (sinθ)2+(cosθ)2=1 ?

Solution: (sinθ)2+(cosθ)2=1→(sinθ)2

(sinθ)2+(cosθ)2

(sin θ)2 =1

(sin θ)2 →1+ 1

(tanθ)2=1

(sin θ)2 -->

1+ 1(tanθ)2=

1(sin θ)2 →(cotθ)2−(cscθ)2=−1

Exp. 4.8.2 Given ( secθ)2−(cscθ)2=3 ,θ=π3

, Is this equation true ?

Solution: ( sec θ)2−(cscθ)2=3 ,θ=π3

→ 1(cosθ)2−

1(sin θ)2=3 ,θ=π

3

θ=π3,cos π

3= 1

2,sin π

3=√3

2 -->

1

(cos π3)

2−1

(sin π3)

2=3 → 1

( 12)

2−1

( √32

)2=

83≠3

( sec θ)2−(cscθ)2=3 ,θ=π3

→This is equationis not TRUE.

Exp. 4.8.3 Given (secθ)(cotθ)(cscθ)( tanθ)

, find the non- permissible θ ?

Solution: Simplify (sec θ)(cotθ)(cscθ)( tanθ)

→(sin θ)

(cosθ)( tanθ)2 → 1( tanθ)

, tanθ must not be 0.

So θ≠nπ ,where n∈ I

Exp. 4.8.4 Simplify (a)(cot π

3)(csc π

3)

(csc π3)

2

−(3sec π3)

(b) (cos π6)(cos π

5)+(sin π

6)(sin π

5)

(c) csc 2θ

sec θ−cscθ (d)

tan 150 °−sec 120 °csc150 °+cot120°

Solution:

(a) (cot π

3)(csc π

3)

(csc π3)

2

−(3sec π3)→

1( tan π

3)

1(sin π

3)

( 1sin π

3

)2

−(31

cos π3

) , tan π

3=√3,sin π

3=√3

2,cos π

3=1

2 -->

(cot π3)(csc π

3)

(csc π3)

2

−(3sec π3)→

1( tan π

3)

1(sin π

3)

( 1sin π

3

)2

−(31

cos π3

)→

1(√3)

1

(√32

)

( 1√32

)2

−(3 112

)=−1

7

(b) (cos π6)(cos π

5)+(sin π

6)(sin π

5)=cos( π

6−π

5)=cos(− π

30)

(c) csc2θ

sec θ−cscθ→

1sin 2θ

1cosθ

− 1sin θ

→

12sinθ cosθsin θ−cosθsin θ cosθ

→ 12(sin θ−cosθ)

(d) tan 150 °−sec 120 °csc150 °+cot120°

→tan (180 °−30° )− 1

cos(90°+30°)1

sin (180 °−30° )+ 1

tan(180°−60° )

→

tan180 °−tan30 °1+tan180 ° tan30 °

− 1cos 90° cos30°−sin 90° sin 30°

1sin 180° cos 30°−cos180 ° sin 30°

+ 1tan 180 °−tan 60 °

1+tan 180 ° tan 60°

→

(0)−( 1√3

)

1+(0)( 1

√3)− 1

(0)(√32

)−(1)( 12)

1

(0)( √32

)−(−1)(12)+ 1

(0)−√3

1+(0)√3

→→−( 1

√3)+2

2− 1√3

→ →→→1

Exp. 4.8.5 Prove that the given equations are true for all permissible θ

(a) ( secθ)2−1=(tanθ )2 (b) sec2θ+1= tan 2θ

tanθ

(c)cos2θ( secθ)2

2(( cscθ√2

)2

−1)=(tanθ)2

(d) cosθ−sin θ=(1−2(sin θ)2)cosθ

1−(sin θ)2+sin θcosθ

Solution:

(a) ( secθ)2−1=(tanθ )2

( secθ)2−1= 1

(cosθ )2−1=1−(cosθ)2

(cosθ )2 =(sin θ)2

(cosθ )2=(tanθ)2 ,where θ∈R

(b) sec2θ+1= tan 2θtanθ

sec2θ+1= 1cos 2θ

+1=1+cos 2θcos 2θ

=1+2(cosθ)2−1

(cosθ)2−(sin θ)2=2

(cosθ)2

(cosθ)2−(sin θ)2

(cosθ)2

= 2

1−( tanθ)2 →

2 tanθ(1−(tan θ)2) tanθ

= tan2θtanθ

whereθ≠n+π2, n∈R

(c) cos2θ( secθ)2

2(( cscθ√2

)2

−1)=( tan θ)2

cos2θ( secθ)2

2(( cscθ√2

)2

−1)=

cos2θ(sec θ)2

2( 1

(√2 sin θ)2−1)=

cos2θ(sec θ)2(√2 sin θ)2

2(1−(√2sin θ)2)

=2cos2 θ

(sin θ)2

(cosθ)2

2(cos2θ)=( tanθ)2 ,θ∈R

(d) cosθ−sin θ=(1−2(sin θ)2)cosθ

1−(sinθ)2+sin θcosθ

(1−2(sin θ)2)cosθ1−(sin θ)2+sin θcosθ

= cos2θcosθ(cosθ)2+sin θcosθ

= cos2θcosθcosθ(cosθ+sin θ)

=(cosθ)2−(sin θ)2

(cosθ+sin θ)

(cosθ)2−(sin θ)2

(cosθ+sin θ)=

(cosθ+sin θ)(cosθ−sin θ)(cosθ+sinθ)

=(cosθ−sin θ) , whereθ∈R

Exp. 4.8.6 Given the domain of 0≤θ≤2π , solve each equation for the θ ?

(a) 6(cosθ)2−5cosθ+1=0 (b) cosθ=2(sin θ)2−1 (c) 3(sin θ)2+4sinθ=−1

(d) (cosθ)2+2sinθ=2 (e) ( tan θ)2−2 tanθ=−1 (f) sin 2θ=2cosθcos 2θ

Solution:

(a) 6(cosθ)2−5 cosθ+1=0=(6(cosθ)+1)(cosθ−1)→ cosθ=−16

,1 -->

cosθ=1=cos(2n π)→θ=2n π ,where n=0,±1

cosθ=−16

=cos(2n π± 59

π)→θ=2n π±59

π ,where n=0,±1

(b) cosθ=2(sinθ)2−1=−cos2θ=−(2(cosθ)2−1) -->

2(cosθ)2+cosθ−1=0=(2cosθ−1)(cosθ+1)→ cosθ=12,−1

cosθ=12=cos(nπ±π

3)→θ=(nπ±π

3) ,where n=0,2

cosθ=−1=cos(π)→θ=π

(c) 3(sin θ)2+4sinθ=−1→3(sin θ)2+4sinθ+1=0=(3sin θ+1)(sin θ+1)

3sinθ+1=0→ sin θ=−13

≈sin (π+π9)=sin (2π−π

9)→θ≈10

9π ,

179

π

sin θ+1=0→ sin θ=−1=sin ( 54

π)→θ=54

π

(d) (cosθ)2+2sinθ=2→1−(sinθ)2+2 sinθ−2=0 →(sin θ)2−2sinθ+1=0=(sinθ−1)(sin θ−1)

(sin θ−1)=0→ sin θ=1=sin ( π4)→θ=π

4

(e) ( tan θ)2−2 tan θ=−1→(tanθ)2−2 tanθ+1=0=( tanθ−1)( tanθ−1)

tan θ−1=0→ tanθ=1=tan (π+π4)=tan ( π

4)→θ= 5

4π , π

4

(f) sin 2θ=2cosθcos 2θ=2sin θcosθ=2cosθ(1−2(sin θ)2)

sinθ=(1−2(sin θ)2)→2(sinθ)2+sinθ−1=0=(2sinθ−1)(sinθ+1)

2sinθ−1=0→ sin θ= 12=sin π

6=sin (π−π

6)→θ=π

6,56

π

sin θ+1=0→ sin θ=−1=sin ( 32

π)→θ= 32π