2.1 Trigonometric
Definition: For any acute angle f), (00
Hyp
<
f)
<
Functions of Acute Angles
900), in standard position
Opp
f)
Adj
1. sinB =Opp
2.
Hyp
cscB= Hyp
Adj
4.
Ad'
cosB=-lJHyp
3. tanB= Opp
Adj
5.secB= Hyp
Adj
6.
cotB
Ad"
= _lJ_
Opp
Examples.
be a right triangle with b
1. Let~ABC
trigonometric
= 7 and c = 12, labeled
in the usual manner.
Find the six
function values of angle A.
B
c=12
AF--
-----'C
b=7
Solution:
a 2 + b2 = c'
a2
+72
=122
a2 +49 = 144
c
a2 =144-49
= 12
a2 =95
a=J95
b=7
. A .Opp
J95
=--=-Hyp
12
SIn
cscA= Hyp =~=
Opp J95
Adj
cosA=--=Hyp
12J95
95
7
12
secA = Hyp = 12
Adj
7
_ Opp _ J95
tan A -----Adj
7
cot A = Adj
Opp
=_7_= 7J95
J95 95
2. Let6ABC
be a right triangle with
trigonometric
a
=
5 and b
= 9, labeled
in the usual manner.
Find the six
function values of angle B.
B
c
a=S
A
C
b=9
2
2
Solution: a + b = c'
52 + 92 = c2
25+81 = c2
c =~106
sinA= Opp =_9_=
Hyp ~106
9M6
106
cosA = Adj = _5_ = 5M6
Hyp ~106
106
cscA = Hyp = ~106 = 12../95
Opp
9
95
tan A
see A = Hyp = ~106
Adj
5
0
Derivation of the Trigonometric Function Values of 30 , 450, and 600
Suppose
h2 + 12
tJ. ABC is an equilateral
triangle with sides of length 2.
= 22
h2 + 1 = 4
h2 =3
h=J3
sin 60° =
°
1
cos600 =-,
2
J3
2 '
2
csc60 =-=--
J3
. 300 =1
Sill
2'
2J3
3'
tan600 =
°
sec60 =2
J3
cos300 =-,
2
csc300 = 2, see 30° = ~
J3
=
2J3
3'
J3
cot 600 = _1_ =
,
J3
J3
3
= Opp = 2.
Adj
5
Adj
cotA=--=Opp
5
9
SupposeD
ABC is an isosceles right triangle with legs of length
B
oM.=1
A~C
b=l
a2 +b2
e+
12
=c
= c2
2
c2 =2
c=J2
sin 450 = _1_
csc45°
J2
= J2,
= J2
cos-l S" = _1_ =
J2
see 45° = J2,
2'
J2
tan450 = 1
2'
cot 45° = 1
3. Find the lengths of the unknown sides of the given figure .
. 450 =a
Solution:
Sill
q
J2 a
---
2
7 7-13
-13 3
qJ2 =2a
m=-=-sin600
2
=2
q=-a
J2
a
-13
7
2
a
-=-
2J2
q=--a
2
-I3a = 14
14-13
a=--
q
q=J2a
q = J214-13 = 14.J6
3
3
14-13
n=a=--
3
3
Cofunction Identities: For any acute angle A
B
c
sin A
=
a and cos(900 - A)
c
=!!.c
so sin A = cos (90° - A)
= cos (90° - A)
cos A
= sin (90° - A)
tan A
= cot (90° - A)
cscA = sec(900 - A)
see A
= csc(900
cot A
= tan (90° - A)
sin A
sin 27° in terms of its cofunction.
4. Write
sin 27° = cos (90° - 27° )
Solution:
= cos63°
5. Solve
sec( B + 10°) = csc( 2B + 20°) for B
B + 10° + 2B + 20° = 90°
3B+300 =90°
3B
= 60°
B=20o
6. Solve
sin (B - 20° ) = cos ( 2B + 5° )
B-200 +2B+5° =90°
3B-15° = 90°
3B = 105°
B=35°
- A)