Download ****Exam 4 Review Questions sec 5.3 to 6.3 rev0612 6th ed Math

****Exam 4 Review Questions
sec 5.3 to 6.3 rev0612
incomplete-check even-odd functions- parametric equations
Math 170
6th ed
Objectives: Evaluate trigonometric functions for double angles and half angles,
sum and difference functions.
Express product form of sum and difference formulas and vice versa.
Solve trigonometric equations for multiple and half angles. Find particular
solutions and all solutions for one cycle
Prove trigonometric identities involving multiple angle and half angle formulas.
Selected Practice Test Questions
a) Given the value of trigonometric functions, evaluate for double angles
3
in QIII, Find sin 2A, cos 2A, tan 2A, cot 2A
5
24
7
24
ans. sin 2A=
cos 2A=
tan 2A=
25
25
7
1
1. sin A = -
2. cos A =
10
ans. sin 2A=-
cot 2A=
7
24
in QIV, Find sin 2A, cos 2A, tan 2A, cot 2A
3
4
cos 2A= 5
5
tan 2A=
3
4
cot 2A=
4
3
b) Find exact values ( without using a calculator)
3
5
1. sin( sin-1 ( ) – tan-1 (3))
3
5
2. cos( arctan(3)- arcsin( ))
ans. ans
9
5 10
13
5 10
=-
=
9 10
50
13 10
50
c) Write an expression that involves x only (assume x is positive) if
1. x=5 tan θ in the expression
θ
2
-
sin 2θ
4
2. x=3 sin θ in the expression 2 θ + tan 2 θ
ans.
1
x
5x
(tan-1 - 2
)
2
5 x + 25
-1
ans. 2(sin
x
x 9 − x2
+
)
3
9 − 2x 2
1
d) Prove the identity:
1. 2 cot (2x) = cot(x) – tan(x)
2. cos2 y=
1 + cos(2 y )
2
3. 2 sin4 x + 2 sin2 xcos2 x= 1- cos 2x
4. tan 2t =
2 cot(t )
csc 2 (t ) − 2
5. cos 3 θ = 4 cos3 θ - 3cos θ
e) Given the value of the trigonometric function of an angle, evaluate for the
half angle.
3
in QIII, Find sin
5
A
A
3
ans. sin =
cos = 2
2
10
1. sin A = -
A
A
A
A
, cos , tan , cot
2
2
2
2
A
1
tan = - 3
2
10
cot
A
1
=2
3
4
3
in QII and sin B = in Q l ,
5
5
A
A
Find sin , cos , csc 2A, cos (A+B), sin (A+ B),sin (A-B), cos (A-B)
2
2
A 2
A
1
25
24
cos =
csc 2A= , cos (A+B)= ,
ans. sin =
2
2
24
25
5
5
2. sin A = =
sin(A+ B)=
7
25
sin (A-B)=1
cos (A-B)= 0
f) Use half-angle formulas to evaluate
1. cos 105 o
3. tan
π
12
ans: -
1− 3
2
2. sin 75 o
ans:
2+ 3
2
ans: 2 - 3
2
g) Prove the identity: Use half angle formulas
1. sin2
x csc( x) − cot( x)
=
2
2 csc( x)
2. cos2
3. tan
t
tan(t ) + sin(t )
=
2
2 tan(t )
x
x
- cot = -2cot x
2
2
h) Rewrite each expression from product from to sum and differences or from
sum and differences to product
1. 10 sin 5x sin 3x
ans. 5 ( cos 2x – cos 8x)
2. cos 3 π sin π
ans . 0
3. cos 5x – cos 3x
ans. – 2 sin 4x sin x
4. sin 75 o – sin 15 o
ans.
2
2
i) Prove the Identities Use sum to product, product to sum formulas
sin 3 x + sin x
cos 3 x − cos x
sin 5 x + sin 3 x
2. tan 4x =
cos 3 x + cos 5 x
1. – cot x =
j) Solve the trigonometric equations: for 0 o ≤ θ < 360 o or radians
1. 2 cos θ + 3 = 0
ans. θ = 150 o , 210 o
5π 7π
,
6
6
2. 5 cos θ + 12 = cos θ
ans. θ =
3. ( cos x – 1) (2 cos x – 1) = 0
ans. x = 0,
4 . ( 2sin θ – 3 ) (2 sin θ – 1) = 0
5. 2 sin2 θ - 7 sin θ
=-3
3
,
5π
3
2π π 5π
, ,
3
3 6
6
π 5π
ans. θ = ,
6
6
ans. θ =
π
π
,
3
k) Solve more complicated trigonometric equations: Use double or half angle
formulas for 0 o ≤ θ < 360 o or radians
1. 2 sin θ + sin2 θ = 0
2. cos 2 θ - cos θ - 2 = 0
3. sin x – cos x = 2
4 . cos
5. sin
θ
2
θ
2
- cos θ = 0
+ cos θ = 0
6. 13 cot θ + 11 csc θ = 6 sin θ
ans. θ = 0 o , 180 o
ans. θ = π
ans. x = 135 o
ans. θ = 0 o , 240 o
ans. θ = π
ans. θ =120 o , 240 o
l) Solve trigonometric equations: Use multiple angles
for 0 o ≤ θ < 360 o or radians
1. sin3 θ = -1
ans. θ =
π
2
,
7π 11π
,
6
6
2. 2cos 2 θ - cos 2 θ - 1 = 0
ans. θ = 0 o , 60 o , 120 o and
All possible solutions θ = 0 o+ 180 o k , θ = 60 o+ 180 o k, θ = 120 o+ 180 o k
3. sin 2x cos 3x + cos 2x sin 3x= - 1
ans. All possible solutions
x=
3π
2π
+
k
10
5
4 . tan2 (3 θ ) = 3
ans. All degree solutions 0 o ≤ θ < 360 o
(Add 60 o to the previous angle until 360 o exceeded)
o
o
o
o
θ =20 + 60 k , θ = 40 + 60 k
k=0
20 o
40 o
k=1
80 o
100 o
k=2
140 o
160 o
k=3
200 o
220 o
k=4
260 o
280 o
k=5
320 o
340 o
4
5. tan2 (3 θ ) = 3
(Change
π
3
k to
6. sin2 (4 θ ) = 1
(Change
π
2
k to
ans. All radian solutions 0 ≤ θ < 2 π
3π
3π
k and add
to the previous angle until 2 π exceeded)
9
9
π
π
2π
π
θ = + k, θ =
+ k
9
3
9
3
π
2π
k=0
9
9
4π
5π
k=1
9
9
7π
8π
k=2
9
9
10π
11π
k=3
9
9
13π
14π
k=4
9
9
16π
17π
k=5
9
9
ans. All radian solutions 0 ≤ θ < 2 π
4π
4π
k and add
to the previous angle until 2 π exceeded)
8
8
π
π
3π
π
θ = + k, θ =
+ k
8
2
8
2
π
3π
k=0
8
8
5π
7π
k=1
8
8
9π
11π
k=2
8
8
13π
15π
k=3
8
8
5