Calculus 2nd Edition Rogawski

SECTION 1.4 Trigonometric Functions 29
The first equation gives 6x = 2πk or x = (π/3)k and the second equation gives 2x =
π + 2πk or x = π/2 + πk. We obtain eight solutions in [0, 2π) (Figure 14):
x = 0,
π
3 , 2π
3 , π, 4π
3 , 5π
3
and x = π 2 , 3π
2
sin θ 1
θ 1
y
y = sin 4x + sin 2x
θ 1
θ 2 = −θ 1
θ 2 = θ 1 + π
1
0
−1
π
3
π
2
2π
3
π
4π
3
3π
2
5π
3
2π
x
−sin θ 1
FIGURE 13 sin θ 2 = − sin θ 1 when
θ 2 = −θ 1 or θ 2 = θ 1 + π.
FIGURE 14 Solutions of sin 4x + sin 2x = 0.
EXAMPLE 5 Sketch the graph of f (x) = 3 cos ( 2 ( x + π 2
))
over [0, 2π].
CAUTION To shift the graph of y = cos 2x Solution The graph is obtained by scaling and shifting the graph of y = cos x in three
to the left π/2 units, we must replace x by
x + π 2 to obtain cos ( 2 ( x + π )) steps (Figure 15):
2 . It is
incorrect to take cos ( 2x + π ) • Compress horizontally by a factor of 2: y = cos 2x
2 . ( (
• Shift to the left π/2 units: y = cos 2 x + π ))
2
( (
• Expand vertically by a factor of 3: y = 3 cos 2 x + π ))
2
3
y
Compress
horizontally by
a factor of 2
3
y
Shift left
π/2 units
3
y
Expand
vertically by
a factor of 3
3
y
1
−1
π
2
π
2π
x
1
−1
π
2
π
2π
x
1
−1
π
2
π
2π
x
1
−1
π
2π
x
−3
−3
−3
−3
FIGURE 15
(A) y = cos x
(B) y = cos 2x
(periodic with period π)
π
(C) y = cos 2(x +
2
)
π
(D) y = 3 cos 2(x +
2 )
The expression (sin x) k is usually denoted
sin k x. For example, sin 2 x is the square of
sin x. We use similar notation for the other
trigonometric functions.
Trigonometric Identities
A key feature of trigonometric functions is that they satisfy a large number of identities.
First and foremost, sine and cosine satisfy a fundamental identity, which is equivalent to
the Pythagorean Theorem:
sin 2 x + cos 2 x = 1 1
Equivalent versions are obtained by dividing Eq. (1) by cos 2 x or sin 2 x:
tan 2 x + 1 = sec 2 x, 1 + cot 2 x = csc 2 x 2