Download Trig Diagnostic - Horace Mann Webmail

A Review of the
Trigonometry of Algebra II
7-1 Measurement of Angles
1 A) Convert –80˚ to radians in terms of π
5π
B) Convert 12 radians to degrees.
C) Convert –18˚ to radians
2 Find two angles, one positive and one negative, which are co-terminal with the given angle.
A) –100˚
5π
B) 12 rds
3 pp.261, 262 1, 5, 9, 13, 17, 19, 25
7-2 Sectors of Circles
4 A sector of a circle is the region bounded by a central angle and the intercepted arc.
For example, sector ABD, as pictured below, is the region bounded by segments AB and
AD and arc BD.
B
s

A
D
r
A) A sector of a circle has a central angle of 2 radians and a radius of 5 inches. Find the
length of the sector's arc.
B) The angle of a sector is 100˚ and the sector's arc is 5 inches. Find the radius of the sector
to the nearest thousandth of an inch.
1
6 A sector of a circle has a central angle of π/2 radians and a radius of 2 inches. Find the length
of the intercepted arc to the nearest thousandth.
7-3 The Sine and Cosine Functions
9 Name the quadrant described: cosW > 0 and tanW < 0
10 Evaluate with no calculator:
5π
A) cos(0)
B) sin( 2 )
π
7π
C) tan (– 2 ) D) sin(– 2 )
π
π
11 Find x, in terms of π, if one knows that x lies in the interval (2 , π): cosx = –cos 15
12 pp. 272-274: 1-17 (odd) 21, 27, 33-41 (odd)
7-4 Evaluating and Graphing Sine and Cosine
13
14
A) Graph y = cosx for one period if the left end-point of the domain is π/2.
1
B) How many solutions does the equation x = cosx have for the domain of part A?
2
π
A) Graph y = sinx on the domain [– 2 , 2π]
B) i) T/F: sin1.1 > sin1.3
ii) T/F: sin1.6 > sin1.7
iii) T/F: sin3.5 > sin4
15 We say that both the sine and cosine are periodic (or cyclic) functions because their graphs
keep repeating. What is the length of one period (or cycle) of y = sinx?
16 What is the amplitude of y = sinx?
17 pp. 279-281: 1, 3, 11-17(odd)
7-5 The Other Trigonometric Functions
2 5
18 If W is an acute angle and tanW = 3 then find the exact answer of secW.
2
19 If cscW = 3 then express tanW in radical form knowing that W is in quadrant II
7π
B) sec( 2 )
20 Evaluate without the calculator: A) sec(π)
11π
21 Express in terms of a reference angle: A) cot 14
4π
B) csc 5
22 Find the exact values without the calculator:
π
A) tan 3
π
B) sec3
9π
C) cot 4
23 pp. 285-286: 1-7 (odd), 13-17 (odd), 23-27 (odd)
8-1 Simple Trigonometric Equations
24 Find the values of x, in terms of π, in the interval [0,2π] for which
6sinx – 3 3 = 0
25 Solve for x on the interval [0,2π] in terms of π. No calculator.
A) 2cosx – 3 = 0
B) 5cscx – 3 = 7
C) tanx = – 1
D) cscx = –1
2 3
E) cscx = – 3
26 pg. 299: 7-21 (odd)
8-2 Sine and Cosine Curves
27 Graph each of the following for one period beginning at x = 0.
A) y = cos2x
B) y = cos 2πx
3
5π
C) tan 6
28 Graph the following for one period beginning at x = –π/2
1
y = sin2 x
29 pp. 305-306: 1-19 (odd)
8-4 Relationships Among the Functions
30 Each of the following can be simplified to one of the six trig. functions. Simplify completely,
A)
2 – 2cos2x
2sinx cosx
B)
cosx (secx + 1) (secx – 1)
sinx
31 Simplify completely:
A) cotx + tanx
B)
(sinx + cosx)2 – 2sinxcosx
cosx
sinx
1+cosx
C) 1+cosx + sinx
32 Simplify completely:
secy – tany
A) 1–siny
tanx
1+secx
B) 1+secx + tanx
9-1 Solving Right Triangles
33 pp. 334-337: 1-9 (odd), 15, 16, 23, 33, 35
4
9-2 The Area of a Triangle
34 Find the area, to the nearest thousandth, of ∆WRT if w= 4, t = 8 and R = 98˚
35 pp. 342-344: 1-15 (odd), 19
Challenges. Entire solutions are in the back, not just the answers.
1
p265
9
You should wind up with an equation that you can solve on the calculator.
2
Find the quadrant that z is in if
cscz < 0 and cos(–z) < 0
3
Evaluate without the calculator:
sec(
– 35π
6 )
4
p286
5
Solve for x in terms of π on the interval [0,2π]:
3secx
+5=3
3
6
p305
7
8
18 but just find cot
16
cscx
cscx
Simplify: cscx –1 + cscx + 1
(Don't forget that there are three Pythagorean Identities)
p353
13
5
Answers
1A) -4π/9
1B) 75˚
1C) -0.314 rds.
4A) 10 inches
4B) 2.865 inches
2A) 260˚,–460
2B)
29π/12,
19π/12
–
3) back of text
6) 3.142 in.
9) IV
10A) 1
10B) 1
14π
11) 15
12) text
13A)
13B) zero
10C) undef.
10D) 1
3
2
1
.5š
1š
1 .5š
2š
2 .5š
-1
-2
-3
14A)
14B) i. F
ii. T
iii. T
iv. F
3
2
1
-.5 š
.5š
1š
1 .5š
-1
-2
-3
15) 2π
29
18)
3
21B) csc


16) 1
19) –

5
17) text
8
8
21C) tan

6
20A) –1
20B) undef.
22A)
22B) 2
3



6
3π
21A) – cot14
22C) 1
23) text
π 11π
24) π/3, 2π/3 25A) 6 , 6
26) text
25B)
 5
,
6 6
3π 7π
3π
25C) 4 , 4 25D) 2
27A)

2
1
.1š
.2š
.3š
.4š
.5š
.6š
.7š
.8š
.9š
.2
.3
.4
.5
.6
.7
.8
.9
-1
-2
27B)
2
1
.1
-1
-2
28) P=4 π; Amp=1
29) text
30A) tanx
31A)
30B) tanx
1
sin x cos x
32A) secy
 33) text
31B) secx
31C) 2cscx
32B) 2cscx
34) 15.844 sq. units
35) text
7
4π 5π
25E) 3 , 3
Solutions to Challenge Problems
1)
1.5cm or 2 cm
2)
III
Step #1 cscz < 0 means that z is in III
or IV
y
x
x
Step #2
Perimeter is 2x+y Thus, 2x+y =7
Circumference of entire circle is 2πx
y
Sector is
of the circle's area πx2
2πx
y
Thus, we have 2πx . πx2 = 3
y
Thus, 2 x = 3
but y = 7–2x
(7–2x)
Thus, 2
x = 3 on calculator gives
1.5 or 2
z
–z
Angles
z and –z will have points on the circle
which have the same x-coordinates.
Therefore, cos(–z) = cosz
Step #3 cos(–z) < 0 becomes
cosz <0
thus z is in II or III
Step #4 Putting Steps 1 and 3 together,
z must be in quad III
12π
π
3) 2/ 3 ; Keep adding 2π until you get rid of negative. I keep adding 6 till I get to 6 and
π
2
sec6 =
3
4) 2 6
π
3π
2 < x < 2 means that the choices for x are II and III quadrants
but csc is negative
Thus, x must be in III; cot must be positive
Use a right triangle whose hypotenuse is 5 and where a leg is 1.
5π 7π
5) 6 , 6
2 3
3
Multiply both sides by 3 and solve for sec to get secx = – 3 Thus, cosx = –
2 3
3
Rationalize denominator to get: cosx = – 2
8
6) amp is 5.
Period is 2
y = –5cos(πx)
Period is 2 can be seen by noting that one cycle goes from one max to the next max and that the
length of this cycle is 2.
I know answer uses cos because y-intercept is not 0
y = A cos(kx)
Amp is 5. Thus A is 5 or –5. I know A is negative because y-intercept is negative. Thus A = –5
2π
k =2
Solve to get k =π
7) 2sec2x
Combine:
cscx(cscx+1)+cscx(cscx–1)
=
(csx–1)(cscx+1)
csc2x + cscx + csc2x – cscx
=
csc2x–1
2csc2x
cot2x
2
2
= sin 2 x
cos x
sin2 x
=
2
= 2sec2x
cos2x
8) 9.74 cm and 13.31 cm
We are given the following parallelogram:
B
C
6
A
D
70 °
10
To find diagonal BD work in ∆ABD, law of cosines deals with SAS. Let a be the measure of BD
2
a 2  36  100 – 2(6)(10)cos70 o
Thus a  94.9575828 Thus, a  9.74
Now work in ∆ACD, let d = measure of AC
DC= 6 D= 180–70 =110
2
2
o
d  36  100 – 2(6)(10)cos110 Thus, d  177.0424172 Thus, d  13.31



9





Trig Review: Chapter 12
1) Draw an angle of 550˚ in standard position 11) Solve for x on the interval 0˚ ≤ x ≤ 360˚
2 3
A) tanx = – 3
B) cscx = – 3
2) Give the measure of an angle in standard
position that has the following two
C) tanx = 1
D) sinx = 1
properties
i) it is co-terminal to the 550˚ angle
12) Solve for x, to the nearest degree, on the
ii) the measure is positive but smaller than
interval 0˚ ≤ x ≤ 360˚
550˚
A) secx = – 2.5
B) cotx = 4
3) Give the measure of an angle in standard
position that has the following two
properties
i) it is co-terminal to the 550˚ angle
ii) the measure is negative
13) Find the measure of angle E to the nearest
degree where F is a right angle.
G
5.3
4) Draw the angle of #3.
5) Assume all numbers are in degrees.
We have discussed the co-function:
cosx = sin(90–x)
cscx = sec(90–x)
cotx = tan(90–x)
E
14) Find AD to the nearest unit if we know
that C is a right angle and A, D and C are
collinear.
B
A) Write the remaining three co-function
formulas.
A
B) Solve for x where x is the measure of
an acute angle: cos48 = sinx
C) If tanx = K then express the cot(90–x)
in terms of K
6) Sample: Write sin179˚ as a function of an
acute angle:
The answer is sin1˚
Write each of the following as a function
of an acute angle.
A) sec(127˚)
B) cot(190˚)
C) sin(302˚)
F
4
23o
38o
D
6
C
15) In ∆ABD, a=5, b = 7 and d = 10.5. find the
measure of angle D to the nearest degree
16)In ∆ABD, a = 5, b = 7 while D = 26˚.
Find d to the nearest tenth.
17) In∆ABD, A = 34˚, D = 50˚ while b =
7. Find d to the nearest tenth.
Answers
2)190˚ 3)–170˚ 5B)42 C) K
6A) –sec53˚ B) cot10˚ C) –sin58˚
2
7)A)=sec150˚ = –
B)= tan300˚ = – 3
7) Evaluate. Your answer must be exact.
3
A) sec(870˚)
B) tan(–420˚)
5
22
10) A) –
B)
3 11A) 120˚,300˚
22
8) Explain why csc(450˚) =1
11B) 240˚,300˚ 11C) 45˚,225˚
11D)
90˚
9) Explain why cot(–180˚) is undefined
12A) 114˚, 246˚ B) 14˚, 194˚
5
10) If 90˚ < x < 270˚ and if cscx = –
then 13 41˚ 14) 5 15) 121˚ 16) 3.3 17) 5.4
3
find A) secx
B) cotx
10
Review of Chapter 13
Please do in your blue book or follow your teacher’s instructions. Show all work
and give good explanations. Remember that you will find this useful when it is
returned next year.
18)
Change 100˚ to radians, in terms of π
19)
Change 5 radians to degrees.
20)
Evaluate. Your answer should be left in radical form.
A) sec
5π
6
B) csc
3π
2
C) tan(–
π
)
4
21)
Solve for x on [0,2π). Your answer should be in radians, in terms of π.
3
2
A) tanx = – 3
B) sinx = – 2
C) cosx = –1
22)
Graph for one period and state the amplitude:
1
A) y = 3sin(2x)
B) y = –2cos(2 x)
23)
Simplify completely:
A) sinx + cosxcotx
Answers
5π
18) 9
900
19)  π  ˚
 
2 3
20) A) – 3 B) –1 C) –1
5π 11π
5π 7π
21) A) 6 , 6
B) 4 , 4
B)
secx– cosx
tan2
C) π
22) check on the calculator
23) A) cscx
B) cosx
11