Download practice test 4a

M 107 TEST 4 SOLUTIONS
1. (5p) Convert the following angles in radians and express your answer in multiple of π :
a) θ = 1 → θ ( Rad ) =
π
180
1
1 π
θ ( Rad )
=
⇒ θ ( Rad ) =
180
180
π
25
π
b) θ = 250 → θ ( Rad ) =
18
250 θ ( Rad )
250π 25π
=
⇒ θ ( Rad ) =
=
180
180
18
π
2. (5p) Convert the following angles in degree and write your answer in decimal form and then in DMS
form:
a) θ ( Rad ) =
7π
'
''
→ θ = .................... = ........................
3
θ = 420 b) θ ( Rad ) = 1
θ = (
180
π
) = 57.29 = 57 17' 44' '
3. (10p) New York and Los Angeles are 2450 miles apart. Find the angle that the arc between these two
cities subtends at the center of the earth. Write your answer to the nearest thousandth of a radian and
Transform in Degree Minutes Seconds.
Assume that the radius is 3960 miles.
θ (rad ) =
D 2450
180
≈ 34.4088
=
= 0.618( rad ) ⇒ θ = 0.618
R 3960
π
4. (10p) Without using a calculator, you must clearly show how you arrive at your answer, find the exact
value of:
π
3
7π 
 = − cos = −
 6 
6
2
a) cos 
b) cot 
4π 

 3 
= cot
π
3
=
1
3
 7π 
2  7π 
 − sec  −
 =
4


 3 
c) csc  −
2 −4
7π
1
1
1

)=
=−
=−
= + 2
7π
7π
1
4
sin(− )
sin( )

4
4

2
 7π 
2  7π 
 ⇒ csc  − 4  − sec  − 3  = 2 − 4




7π
1
1
1
sec(− ) =
=
=
=2 

7π
7π
π
3
cos(− ) cos( ) cos( )

3
3
3
csc(−
Pg 1
5. (5p) Find the length of the arc subtended by a central angle of 12 0 23’ 44” if the radius of the circle is
3 meters. Write your answer to the nearest thousandth.
θ = 12 0 23’ 44”
θ = 12.3955555
12.3955555
s
12.3955555
=
⇒
s
=
6
π
≈ 0.6490
360
2π ⋅ 3
360
6.(10p) Use the fundamental trigonometric identities and/or the complementary angle theorem
(cofunctions) to find the exact value of each of the following:
sin 500 cot 40 sin 40 − cos 40 cos 40 − cos 40
=
=
=0
sin 400
sin 40
sin 40
b) 1 + tan 2 50 − csc 2 850 = 0
because 1 + tan 2 5 = sec 2 5 = csc 2 5
a) cot 400 −
7. (10p) Without using a calculator, you must clearly show how you arrive at your answer, find the exact
value of the following:
π 
cos
π
2 = 0 =0
a) cot   =
1
 2  sin π
2
1
 9π 
b) csc 
= 2
=
 4  sin 9π
4
1
1
c) sec 4200 =
=
=2
cos 420 cos 60
5
8. (10p) If csc θ = − ,
2
and
−
π
2
< θ < 0 , find the exact value of the remaining trigonometric
values of angle θ .
csc θ = −
5
2
⇒ sin θ = −
2
5
x>0
x 2 + 2 2 = 5 2 ⇒ x 2 = 21 ⇒ x = ± 21 ⇒ x = 21
cos θ =
tan θ = −
21
5
π
θ in (− ,0) ⇒ sec θ =
5
2
21
2
21
⇒ cot θ = −
2
21
Pg 2
9. (10p) Find the amplitude, period, phase shift and then graph for 2 periods the following
trigonometric function:


π
− 3.
4 
π
a) f ( x ) = 2 cos 3 x +
A=2 T=
2π
3
Φ=


12
b) f ( x ) = −3sin  2 x −
A=3
T= π
π
3 
Φ = - π /6
Pg 3
10.(10p)

 3 
 4 
a) Find the exact value of sin cos −1  −   . Draw a sketch to support your working.


cos −1  −

3
 =θ
4
π
θ in [ , π ]
2
4
7
θ
−3
sin θ =
7
7
 3
⇒ sin[cos−1  − ] =
4
4
 4

 x 
tan  cos −1    , 0 ≤ x ≤ 3 as an algebraic expression
 3 

b) Write the trigonometric expression
in “ x ”.
0≤ x ≤3⇒0≤
x
 x π
≤ 1 ⇒ 0 ≤ θ = cos −1   ≤
3
3 2
3
9 − x
θ
2
x
tan θ = +
9 − x2
x
0<x≤3
11. (15p) Establish the following identities:
a)
csc x − 1 1 − sin x
=
csc x + 1 1 + sin x
1
1 − sin x
−1
1 − sin x sin x
1 − sin x
LHS = sin x
= sin x =
=
= RHS
1
1 + sin x
sin
x
1
+
sin
x
1
+
sin
x
+1
sin x
sin x
Pg 4
b)
1 − cos θ
= (csc θ − cot θ ) 2
1 + cos θ
2
2
cos θ   1 − cos θ 
(1 − cos θ ) 2
 1
−
RHS = 
 =
 =
sin θ 2
 sin θ sin θ   sin θ 
1 − cos θ 1 − cos θ 1 − cos θ (1 − cos θ ) 2 (1 − cos θ ) 2
=
=
LHS =
=
= RHS
1 + cos θ 1 + cos θ 1 − cos θ
1 − cos 2 θ
sin 2 θ
c)
cos θ + 1 1 + sec θ
=
cos θ − 1 1 − sec θ
1
cos θ + 1
cos θ = cos θ = cos θ + 1 cos θ = cos θ − 1 = LHS
RHS =
1
cos θ − 1
cos θ cos θ − 1 cos θ − 1
1−
cos θ
cos θ
1+
Pg 5