Ek
Primary Trigonometric Ratios and the CAST Rule
1. An exact value f r a trig’ )Il( ulletric ram ) is given
for each angle. l)etermiue the exact values ol
the other two priiiiary mrigonoiiletric ratios.
a) sinE)
13’
=
ranx
d) sinx
a) cscx
4,o°oo°
b) cosO
c)
5. An exact value for a reciprocal trigonometric
ra Ui) 5 give II k r each angle. I )e te rii inc the
exact values
the other two recipn)cal
rrit4uuitilcmric ratios.
=
_,
=
-,
900
x
150°
cnex
=
ci) csc6
r
ci
90°
270°
x
2. Use the CAST rule to determine the sign of
each value. Then, use a ca Icu Ian w to eva lu ate
each trigonometric ratio, ii uncled tc four
dcci ma! places.
6. Use a calculator to evaluate each trig’ nlonlemric
ratio, rounded to Ii ur decuiial places.
a) csc35°
b) cos .56°
c) cot 25°
ci)
ci) csc 121°
e) cos
f) tan
e) set l1_30
h) cos31S°
f) cotll.i°
1570
-
g) csc3ll°
CONNECTIONS
hi sec3SS°
Recall that the niemory device
CAST shows which trigonornetnc
ratios are positive in each quadrant.
A
ill
S
SUit’
(I
C
7. i; a calculator n determine angle x, rounded
to the nearest degree.
a) cscx
=
b) sec x
=
tIIircTlt
3. Use a calculator n determine angle x. rot! ndetl
to the nearest degree.
b) cusx
=
S..346
d)
ccisx
-
t).4.5
Reciprocal Trigonometric Ratios
Note: If von have not worked with reciprocal
rigr ml) metric rat i is he k ire, refer to tile
Preretjtiisite Skills Appendix oi page 454.
4. I )eterniine the reciprocal ol each luiliher.
a)
2
MHR
.3
.
c)4
Advanced R.,rc:cns
.
c)
1.2.5
12
cotx3.14l6
d) secx
=
—1.32
Exact Trigonometric Ratios of Special Angles
=
c) tatix
90° S x S I 50°
5fl° S 05270°
c) raii75°
Ohs
91)0
b) st..I6 0
g) sin 302°
200
——,
a) sin 5°
a) sinx
00 x
ii) secO
O 3600
2O°
,
CJ’apt&’
1
8. Use appropriate special right triatlgles to
tleterluille exact values [or tine priniary
trig( Jul )nletric ratii 5.
1)
a)
30’
b)
45’
c)
60
sinO
cosO
tanO
9. a) Use a unit circle, similar to the une shown, to
represent an angle of 45° in standard position
with
point
I’(x, v) as the point
of intersection
Distance Between Two Points
11. Use the dista ice formula
xj2 ÷ (3,
d =
—
of the terminal arm and the unit circle.
—
distance between the points
to
iii
determine
the
each pair.
a) A(I0, 8) and 13(6,5)
b) C(—5, —3) and DE, 2)
c) E(—X. 4) and F(7, 12)
d) G(—3, —6) and 11(3,2)
Product of Two Binomials
12. Expand and simplify each product.
Create a right triangle that will allow you to
determine the primary trigonometric ratios
of 45°. Label the cuordmates of point P
450)
using the exact values for (cos 45°, sin
b) Determine exact expressions for the lengths
of the sides of the triangle.
c)
Use the side lengths to determine exact values
450
for the reciprocal trigonometric ratios of
10. Extend your diagram from question 9 to help
you determine exact primary and reciprocal
trigonometric ratios for 135°, 225°, and 315°
a) (a
b) (c
b)(a
d)(c
+
c) (2x
+
dl (sinx
+
—
y)(3x
--
b)
d)
—
2y)
cosy)(sinx
+ cosy)
Trigonometric identities
13. Use a calculator to verify that the Pythagorean
identity sinx + cos2x = I is true for x = 20°.
x = 130°. x = 200°, and x = 305°.
14. Use a calculator to verily that the quotient
identity tan x
.
-
x
=
110°, x
=
sinx. true for x
is
cos x
232°, and x =
=
=
40°,
3550
SPROBLEM
<
\Vhar do a civil engineer designing highway on-ramps, a pilot turning an aircraft
onto the final approach for landing, and a software designer developing the code for
engineering design software have in common? All of them use angles and trigonometry
to help solve the problems they encounter. The engineer applies trigonometry to
determine the length and the angle of inclination ol the on-ramp. The pilot selects
the angle of hank for the aircraft from tahles based on the tangent ratio to ensure a
proper alignment with the runway at the end
of the turn. The software designer employs
trigonometry to realistically render a
three—dimensional world onto a two—dimensional
computer screen. As you work through this
chapter, you will apply trigonometry to solve
problems relating to the transportation industry.
Pterecuisite S<iIIs
.
MHR
201
7. a) ma>anitim apirnsiinarcIv ( 2.25, 964) minim.i
appriixiriitc!
3.14. l4.I6. 26.6l. 70.80)
b) lieruce,:
I
CHAPTER 4
—
1)4, -—14.16) and 12.25, 9.64) apprilNimarely
2.61; between (12.25, 9.64), and (26.61, 70.80),
x:matelv 5.60 c) x
32
8. Answers may vary. Sample answers:
y=2x(x+7)(x- .1)2;y= tx(x+7)(x 3(2
—
Prerequiske Skills, pages 200—201
1,a)cosU= t,tanO= 2 b)sinO
-
c)sinx
=
a ppm
9. v = k Cv — 2 )1)x
5). Aiisvers ‘‘a vary. Sn (lip) e 1 iiswers:
2Cv — 2)2x r 5). y-iiirercept ID; y
y
-3x 21(.v + 5::.
y-intercepr —6))
1O.a)4x_5x1_6x+200(2xfl)(2x -4x±Jl)
it.
=
5x2 —2$
—
11. a) 3$ I’)
=
(x
—
2B3x’ + 6x2 + 7x + 14),x #2
9
12. a) No, h) Yes.
13.k =
14.aHx_3Hx+3x+9:b)Cv 2);2xrSx +3)
15.a)x= 4,rx
I t,rx
S
-4
4 + v’3I
vu
b)x = 3 nrx = ——— —
r,rx =
—%
S
20.(xeF(,x
,
v-intercept
-
ii) [6siz
l},{veE,yI)
(ye ft y 3);x-inrercepr
x 0 2},
: a svm pLc ites x
x < 2.x > 2;
decreasing
‘fT
c)cscx
=
a
= --.cotx
‘5
—,seex
=
b(
Sill
()
-.
tan))
=
= —i—:. tan I)
=
=
(1
_._!_._,
x < 2;
sin U =
y
1;
-
T:ca
increasing a-
—i——.
ens U
.
=
,
tan U
‘4+-
U se slope
> 2
i:”o\
-
ft
i 0
I .x
0);
3.
nil
=
_j
b)x
.
35° =
C
‘fT
iv + 6
x
I
—
Inrx>2
H1
c)c45°
.
=
=,nanI35=
‘fT.
sec
35°
v’i,
sinUS’
_Jici)s3I5°
sec22S°
=
=
SPLv)<5}
Advanced Func nuns
Answers
I,
=
—,
-
,
cot225°
—,tan3I5°
=
=
I,
—I,
cscilS°
vi.see3IS
vi. L(it315 = —I
11. a) Sb)) 3 ci 17 d) ID
2ah1:2b)e—dc)6x2—xy—2y2
dl siix+2s,nx enss + ct,yy
3
ci The prolix is a (ways less that: %50h))).
26. Since I represents tinle. I
0: 1 * II) becanse rite
tIeil(iiiiiii.lttir e.inniit be zero.
=
tan 225°
=
—,
135°
7i,cnt
4.1 Radian Measure, pages 202—210
b){xe(5,x 0}{1’)x)cft
2
6
3
b) A c) A d) A
12
1$
24
36
J. a)
b) Mi
t d) :!i
2
4
4. a)
b) A) Ad) A
S
12
2(1
60
a)
A
f
*‘‘°=
ese22S°;
=
\/i,e:ir45°
vi.sec4s°
ens22S°
=
MHR
v
sin 225°
—
546
=
V2
10. sin 35°
ese
-
23. a) x = —2.2 b) x
2. 5
24.a)x< 2.7Snrx> 2b)—Ix
25. a) is::coom: — -i
-
/
x-intercepr,
=
I
=
I -,
1.2690
=
-
0
0: p sin ye
slopex<—3.—3<x<{);negarivesliipeo<x<3.x>i;
decreasingo<x<3,— l<x<0;incre,isingx< 3,x >3
ii)
21/(x)
a
9. a) r-—3r
(I
= 2,
LELJ
b) I)
ft x * 3, x .1), (ye
v-i itercept — : asvnpri ires x =
U
12
——.corU
—jd)secO’ ———,enr
vi
ens))
a
$
—
=
0.9205
6.a) 1.7434 b)
1.2361 c) 2.1445 d) l.1792e)
I) 1.0724 q)
I. c890 h) 1.003 $
7. a) 53° b) 54° c) lS°d) 139°
8. a) sin I)
_.g
I_i
-,
—j b)cscO
‘5
=
‘s-2
(xc
4,nanU
Ii
15
——,tanx
a
16.a)xs)orx6b)x< 3or—2<x<2
17. ironi 0mm to 10 pin
18. A
19. a) 1(x) - 0 b) 1(x) —* 0 c) [)x) —
d) /(x) —* —x
21. a)) xE
=-
4
‘4
—. d)ci:sx
2. a) 0.25$$ b) 1)5592 c) 3.7121 d) 0984$ e)
I) 22475 g) 0.$48() h) 0.978 I
3. a) 11°b) 65° C) 83° d) I
4. a) - b) 3 c)
d)
cx
b) 3x1
5
7
3—-,cosx