180 Chapter 7. Trigonometric Functions
Terminology A circle with radius 1 is called aunit circle. The circle with radius 1 and center at the origin is calledthe unit circle.
Definition The angle determined by an arc of length 1 along the circum- ference of a unit circle is said to be of measureone radian.
1 1
1 rad.
length=1
Figure 7.2
Since the circumference of a unit circle has length 2π, there are 2πradians in one revolution. Therefore, we have 360◦ =2πradians. The conversion between degrees and radians is given by
d◦=d× π
180 radians Thus, we have 90◦ = π
2 (radian) and 60◦= π
3 (radian) for example.
RemarkIn calculus, it is more convenient to consider angles in radians and the unitradianis usually omitted.
Exercise 7.1
1. Convert the following degree measures to radians:
(a) 270◦ (b) 210◦
(c) 315◦ (d) 750◦
2. Convert the following radian measures to degrees:
(a) π
6 (b) 3π
4
(c) 5π
2 (d) 7π
7.2. Trigonometric Functions 181
follows:
tanx = sinx
cosx provided that cosx,0, cotx = cosx
sinx provided that sinx,0,
secx = 1
cosx provided that cosx,0,
cscx = 1
sinx provided that sinx,0.
Note Since sinx= 0 if and only ifx=kπfor some integerkand cosx=0 if and only ifx= kπ
2 for some odd integerk, it follows that
dom (tan) = dom (sec) = R\n
± π
2,±3π
2, . . .o , dom (cot) = dom (csc) = R\n
±π,±2π, . . .o .
RemarkBelow we will discuss some results for the sine, cosine and tangent functions. The secant function will only be used in an identity and a formula for differentiating the tangent function. The cotangent and cosecant functions will not be used in this course.
Properties
(1) The sine and cosine functions are periodic with period 2π, that is,
sin(x+2π) = sinx for allx∈R, cos(x+2π) = cosx for allx∈R.
(2) The tangent function is periodic with periodπ, that is,
tan(x+π) = tanx for allx∈dom (tan).
(3) The sine function and the tangent function areodd functionsand the cosine function is aneven function, that is,
sin(−x) = −sinx for allx∈R, cos(−x) = cosx for allx∈R,
tan(−x) = −tanx for allx∈dom (tan).
(4) From (3), we see that the graphs of the sine function and tangent function are symmetric about the origin and the graph of the cosine function is symmetric about they-axis. See Figures 7.4(a), (b) and (c).
2p 4p
-2p -4p
-1
1 y=sinx
Figure 7.4(a)
182 Chapter 7. Trigonometric Functions
2p 4p
-2p -4p
-1
1 y=cosx
Figure 7.4(b)
- p 2
p 2
y=tanx
Figure 7.4(c)
Remark The graph of the cosine function can be obtain from that of the sine function by moving it π
2 units to the left. This is because
cosx=sin x+ π
2
for allx∈R.
CAST RuleThe signs of sine, cosine and tangent in each quadrant can be memorized using the following rule:
S A T C
whereCstands forcosine,Aforall,SforsineandT fortangent—for example,Cin the 4th quadrant means that if x is an angle in the fourth quadrant, then cosxis positive and the other two values sinx and tanxare negative.
Sine, Cosine and Tangent of some Special Angles sin 0 = 0
sinπ
6 = 1
2 sinπ
4 = 1
√2 sinπ
3 =
√3 2 sinπ
2 = 1
cos 0 = 1 cosπ
6 =
√3 2 cosπ
4 = 1
√2 cosπ
3 = 1
2 cosπ
2 = 0
tan 0 = 0 tanπ
6 = 1
√3 tanπ
4 = 1
tanπ
3 = √
3 tanπ
2 undefined
The values of the sine, cosine and tangent functions at the special angles can be obtained by drawing appropriate figures or triangles. For example, we can use Figures 7.5(a), (b) and (c) to find the values of the trigonometric functions for angles with size π
2, π
4 and π
6respectively.
1 1
π 2
P(0,1)
Figure 7.5(a)
1 1
π 4
P √1
2,√1
2
Figure 7.5(b)
1 1
π 6
P √12, √12
Figure 7.5(c)
7.2. Trigonometric Functions 183
An Important Identity (Py) sin2x+cos2x=1
ExplanationThe result is called an identity because it is true for all x∈R.
Note that sin2x=(sinx)2etc.
Proof LetP(a,b) be the point on the unit circle corresponding to angle x.
By definition, we have
sinx=b and cosx=a.
The required result then follows sincea2+b2=1.
1 1
cosx
sinx P
x
Figure 7.6
Another Identity
(Py1) 1+tan2x=sec2x, x∈R\
± π
2,±3π
2, . . . Proof 1+tan2x = 1+ sin2x
cos2x Definition of tan
= cos2x+sin2x cos2x
= 1
cos2x Identity (Py)
= 1 cosx
2
= sec2x Definition of sec
More Identities
sin π
2−x
= cosx sin π
2+x
= cosx sin(π−x) = sinx sin(π+x) = −sinx sin 3π
2 −x
= −cosx sin 3π
2 +x
= −cosx sin(2π−x) = −sinx
cos π
2−x
= sinx cos π
2+x
= −sinx cos(π−x) = −cosx cos(π+x) = −cosx cos 3π
2 −x
= −sinx cos 3π
2 +x
= sinx cos(2π−x) = cosx The above identities can be derived using appropriate figures. For example,
• from Figure 7.7(a), we get sin π
2 −x
=a=cosx and cos π
2 −x
=b=sinx.
184 Chapter 7. Trigonometric Functions
• from Figure 7.7(b), we get
sin(π−x)=b=sinx and cos(π−x)=−a=−cosx.
1 1
π 2−x
x P2(b,a)
P1(a,b)
Figure 7.7(a)
1 1
π−x x
P1(a,b) P2(−a,b)
Figure 7.7(b)
RemarkThe above identities can be memorized in the following way:
f
odd multiple ofπ
2
±x
is ±g(x) f
even multiple ofπ
2
±x
is ±f(x)
where (f,g)=(sin,cos) or (cos,sin) and the sign can be obtained by the CAST rule.
As illustrations, we describe how to obtaining the identities for sin(π−x), cos(π−x), sin 3π
2 +x
and cos 3π
2 +x .
• Note thatπ= 2·π
2 is an even multiple of π
2. According to the second form (the trigonometric functions are unchanged),
(1) sin(π−x) is either sinx or −sinx (2) cos(π−x) is either cosx or −cosx
To determine the correct sign, we assume thatxbelongs to the 1st quadrant and so (π−x) belongs to the 2nd quadrant. According to the CAST Rule, sin(π−x) is positive and cos(π−x) is negative (and sinx and cosxare positive). Thus we have
(1c) sin(π−x)=sinx (2c) cos(π−x)=−cosx
• Note that 3π
2 is an odd multiple of π
2. According to the first form (the two trigonometric functions are switched),
(3) sin 3π
2 +x
is either cosx or −cosx (4) cos 3π
2 +x
is either sinx or −sinx
To determine the correct sign, we assume that xbelongs to the 1st quadrant and so 3π
2 +x
belongs to the 4th quadrant. According to the CAST Rule, sin 3π
2 +x
is negative and cos 3π
2 +x
is positive (and sinxand cosxare positive). Thus we have
(3c) sin 3π
2 +x
=−cosx
7.2. Trigonometric Functions 185
(4c) cos 3π
2 +x
=sinx
RemarkSince the values of the sine and cosine functions atπ
2orπor3π
2 can be found easily, the above identities can also be derived using the following results, calledcompound angle formulas.
Compound Angle FormulasLetAandBbe real numbers. Then we have
sin(A+B) = sinAcosB+cosAsinB cos(A+B) = cosAcosB−sinAsinB sin(A−B) = sinAcosB−cosAsinB cos(A−B) = cosAcosB+sinAsinB
RemarkThe formulas for sin(A−B) and cos(A−B) can be deduced from that for sin(A+B) and cos(A+ B) respectively. This is because sin(−x)=−sinxand cos(−x)=cosxfor all x∈R. Moreover, since sin π
2−x
= cosxand cos π
2−x
=sinx, the formula for sin(A+B) can be deduced from that for cos(A+B) and vice versa.
However, the proof for either formula is very tedious and thus is omitted.
Continuity of sin and cos The sine and cosine functions are continuous onR, that is, for everya∈R, we have
limx→asinx = sina limx→acosx = cosa
Reason Ifxis close toa, then the pointQlying on the unit circle that corresponds toxis close to the pointPthat corresponds toa.
1
1 P Q
x a
Figure 7.8
An Important Limit (sin) lim
x→0
sinx x =1
Proof First we consider right-side limit. Letx be small positive (0 < x < π
2).
Consider the triangles4OABand4OACand the sectorOABshown in Figure 7.9.
Note that
area of4OAB = 1
2·1·1·sinx = sinx 2 area of sectorOAB = 1
2·12·x = x 2 area of4OAC = 1
2·1·AC = tanx 2
O A B
x
C
1
Figure 7.9
Since4OAB⊆sectorOAB⊆ 4OAC, it follows that sinx
2 < x
2 < tanx 2 . Dividing each term by sinx
2 (which is positive), we get 1< x
sinx < 1 cosx,
186 Chapter 7. Trigonometric Functions
which, by taking reciprocal, yields
1> sinx
x >cosx. (7.2.3)
By the continuity of the cosine function (at 0), we have
x→0limcosx=cos 0=1. (7.2.4)
Letting x→0+, by (7.2.3) and (7.2.4) together with the Sandwich Theorem (which is also valid for limits at a point and one-sided limits), we get
x→0+lim sinx
x =1.
Since sinx
x is an even function, that is,sin(−x)
−x = sinx
x for allx,0, it follows that lim
x→0−
sinx
x =1. Therefore, we have lim
x→0
sinx
x =1.
Remark
• The result means that ifxis small, then sinxis approximately equal tox.
• Ifxis in degrees, the result is different: lim
x→0
sinx◦
x = π
180.
-10 -5 5 10
-0.2 0.2 0.4 0.6 0.8 1
y= sinx x
Figure 7.10
The following result will be used in deriving the formula for d dxsinx.
A Limit Result (cos−1) lim
h→0
cosh−1
h =0
ExplanationTo get the limit, we try to “cancel” the factorhin the denominator. However, the numerator is not a polynomial. Instead of making a factorhin the numerator, we try to make a factor sinhand apply (sin).
Proof lim
h→0
cosh−1
h = lim
h→0
(cosh−1)(cosh+1)
h(cosh+1) Note: cosh+1,0 ifh≈0
= lim
h→0
cos2h−1 h(cosh+1)
= lim
h→0
−sin2h
h(cosh+1) Identity (Py)
= lim
h→0
"
(−sinh)· sinh
h · 1
cosh+1
#
= lim
h→0(−sinh)×lim
h→0
sinh h ×lim
h→0
1
cosh+1 Limit Rule (La5)
= (−sin 0)·1· 1 cos 0+1
Continuity of sin & cos and Limit Result (sin)
= 0.