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## Explanation & Answer

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Solutions

1. Consider an angle 5/4π rad

a) Sketch the angle in standard position

Solution

π

5 π/4

6 π/4

b) Determine 2 conterminal angles

Solution

I.

II.

5 π/4 +2 π =13 π/4

5 π/4 - 2 π=-3 π/4

c) Convert the angle to degree measures

Solution

5 π/4 * 180/ π =225°

2. A track moving at a rate of 90km/h and the diameter is 1.25m. Find the angular velocity of

the wheel in radians

Solution

The linear speed of wheel = 90km/h,

90km/h= (90*1000)/(3600) v=25m/s

radius of wheel path r = (1.25/2)=0.625

Angular Speed ω =v*r

=25*0.625

= 15.625 rad/sec

3. Find the exact values of the six trigonometric functions of the angle shown in the figure

Solution

When point (-1,4) i.e. (x=-1,y=4) is on the terminal side of an angle θ in standard position the

trigonometric functions are defined using radius(difference between the terminals) and the

origin:

radius r =√ (1)2+(4)2

=√ 17

1. Sin θ=y/r in this case sin θ= 4/√ 17

2. Cos θ=x/r in this case cos θ= -1/√ 17

3. Tan θ= y/x in this case tan θ= 4/-1 =-4

4. Cosec θ= r/y = √ 17/4

5. Sec θ=r/x= √ 17/-1

=-√ 17

6. Cot θ=x/y= -1/4

=-1/4

4. Given Tan θ= 7/2 and θ is an acute angle , find the other trigonometric

functions of θ

Solution

Tan θ=opposite/adjacent

Opp=7

Adj=2

I.

Sin θ=opp/hypotenuse

Hypotenuse==√ (7)2+(2)2

=√ 53

=7/√ 53

II.

Cos θ=adj/hyp

=2/ √ 53

5. Determine the reference angle θ’ of the angle θ=225 and sketch θ’ and θ in standard

position

Solution

Reference angle of 225°

255 lies in the third quadrant hence the formula is

=360°−ac

Ac=225°

=360-225

θ ‘=135°

to pie radian =225*π/180

=1.25 π rad

=5/4 π rad

θ ‘=135* π/180

=0.75 π rad

=3/4 π rad

Sketch

3π/4 (θ’)

π

5π/4(θ)

6 π/4

6. Determine the quadrant in which θ lies if Sec θ0

Solution

Sec θ=1/cos θ

And what is required 1/cos θ 0 and this will occur in quadrant 1&3

Hence they overlap in quadrant 3

7. Find two exact values of θ in degrees if cos θ=-√2/2

Solution

From special triangle

Cos θ= 1/√2

=45°

Hence cos245=1/2=2/4=√2/2

Therefore the angles lies between 2nd and 3rd quadrant since cos is negative

180-45=135°

180+45=225°

8. Use calculator to approximate 2 values of θ radians if θ =csc 1.030

Solution

csc θ = 1.03

csc θ=1/sin θ

sin θ = 1/1.03 = 0.971

θ = 1.33

also you can compute

π-1.33 = 1.8127

θ = 76.14°

or

θ= 103.86°

9. Find the remaining trigonometric functions of θ given that cos θ= =3/5 and sin θ >0

Solution

Cos θ=-3/5

Then θ is in quadrant, if x and y are positive numbers we will have

(-x,y) is a point in quadrant 2.

hence (r cos(θ), r sin(θ) )= (-x,y)

sin(θ) = y/r

cos(θ) = -x/r = -3/5

x = 3, r = 5

r2= x2+ y2

25 = 9 + y2

=4

sin(θ) = y/r = 4/5

cos(θ) = -x/r = -3/5

tan(θ) = sin(θ)/cos(θ) = -4/3

cot(θ) = 1/tan(θ) = -3/4

sec(θ) = 1/cos(θ) = -5/3

csc(θ) = 1/sin(θ) ...