Chapter 7

Trigonometric Functions

This chapter starts with graphs of sine functions. Then graphs of cosine, tangent and other trigonometric functions are shown. After that, the inverse of those functions and their graphs come. The essential part of this chapter is the differentiation of trigonometric functions.

7.1 Graphs of Sine Functions

Graph of the Sine Function y = sin x

fig 7.1-1

Domain: The set ℝ of all real numbers.       Range: {y| -1 < y < 1}
$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\pi ,0) (2\pi , 0)\\ \text{maximum point:} (\dfrac{\pi}{2},1) \quad \text{minimum point:} (\dfrac{3\pi}{2},-1) \end{cases} } $$ Periodic function: If f(x) = f(x + p) where p is a positive real number, then the function F is called a periodic function. If p is the smallest such number, then p is the period of function f. The sine function y = sin x is a periodic function with period 2π, since 2π is the smallest positive real number such that sin x = sin (x = 2x). The amplitude of a periodic function is half the difference between the maximum and the minimum values. So the amplitude of the sine function y = sin x is 1 - (-1)⁄2 = 1.

Graph of the Sine Function y = a sin bx, a > 0, b > 0

In the following example, we will see how to obtain the graph of y = a sin bx, where a > 0 and b > 0, from the graph of y = sin x.

Example 1.
From the graph of y = sin x, draw step-by-step transformation graph to get the graph of y = 2 sin π⁄2 x.
Solution
Method 1
$$y = \text{sin}\, x \xrightarrow[\text{scale factor 2}] {\text{vertical scaling}} \, y = \text{2 sin}\, x $$
fig 7.1-2

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor 2}]{\text{vertical scaling}} \, y = \text{2\, sin} \, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{2 sin} \dfrac{\pi}{2} x $$
fig 7.1-3

Method 2
$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{sin} \dfrac{\pi}{2} x $$
fig 7.1-4

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{sin}\, \dfrac{\pi}{2} \, x = \xrightarrow[\text{scale factor 2}]{\text{vertical scaling}} \, \text{2 sin} \dfrac{\pi}{2} x $$
fig 7.1-5


From the graph of y = sin x, the graph of y = a sin bx, a > 0, b > 0, can be obtained as
$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor }a]{\text{vertical scaling}} \, y = a\, \text{sin} \, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{b}} ] {\text{horizontal scaling}} y = a\, \text{sin} \, bx $$ or
$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{b}} ] {\text{horizontal scaling}} \, y = \text{sin} \, bx \, \xrightarrow[\text{scale factor }a]{\text{vertical scaling}} y = a\, \text{sin} \, bx $$
y = sin x → y = a sin bx
(x,y) → (x⁄b, ay)
fig 7.1-6

$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } $$

Graph of the Sine Function y = -a sin bx, a > 0, b > 0

$$y = a\, \text{sin}\, bx\, \xrightarrow{\text{Reflection on the}\, x\text{-axis}}\, y = -a\, \text{sin}\, bx $$
fig 7.1-7

Example 2
Draw the graph of y = 2 sin π⁄2 x and y = -2 sin π⁄2 x.
Solution
fig 7.1-8

Note that y = a sin (-b)x = -a sin bx, so no need to consider b < 0.


Graph of the Sine Function y = a sin b(x - h) + k

$$\small{y = a \, \text{sin}\, bx\, \xrightarrow[\text{vertical translation}\, k\, \text{units}] {\text{horizontal translation}\, h\, \text{units}} \, y}$$ $$\small{= a \, \text{sin} \, b(x - h) + k }$$

y = sin x → y = a sin bx → y = a sin b(x - h) + k
(x, y) → (x⁄b, ay) → ( x⁄b + h, ay + k)

Key points on the midline y = k

Maximum and minimum points
fig 7.1-9


Example 3.
Draw the graph of y = 3 sin 2 (x - 1) + 4.
Solution
Midline: y = 4,
Amplitude: 3,
Period: π
Five key points: (1, 4), (π⁄4+1, 7), (π⁄2+1, 4), (3π⁄4+1, 1), (π+1, 4)
fig 7.1-10


Example 4.
Draw the graph of y = -3 sin &pi(x + 1) - 2.
Solution
Midline: y = -2,
Amplitude: 3,
Period: 2
Five key points: (-1, -2), (- 1⁄2, -5), (0, -2), (1⁄2, 1), (1, -2)
fig 7.1-11


Exercise 7.1
1. From the graph of y = sin x, draw step-by-step transformation graphs to get the graph of y = -3 sin π⁄3 x.
Solution
$$y = \text{sin}\, x \xrightarrow[\text{scale factor 3}] {\text{vertical scaling}} \, y $$ $$= 3 \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{3}{\pi} }]{\text{horizontal scaling}} y $$ $$= 3 \text{sin}\, \dfrac{\pi}{3}\, x \, \xrightarrow[\text{the}\, x{-axis}]{\text{Reflection on}} y $$ = -3 sin π⁄3 x
(x, y) ⟶ (x, 3y) ⟶ (3x⁄π, 3y) ⟶ (3x⁄π, -3y)
(0, 0) ⟶ (0, 0) ⟶ (0, 0) ⟶ (0,0)
(π⁄2, 1) ⟶ (π⁄2, 3) ⟶ (3⁄2, 3) ⟶ (3⁄2, -3)
(π, 0) ⟶ (π, 0) ⟶ (3, 0) ⟶ 3, 0)
(3π⁄2, -1) ⟶ (3π⁄2, -3) ⟶ (4.5, -3) ⟶ (4.5, 3)
(2π, 0) ⟶ (2π, 0) ⟶ (6. 0) ⟶ (6, 0)
fig 7.1-b1

fig 7.1-b2

fig 7.1-b3

2. Draw the graph of (a) y = 1⁄2 sin x
(b) y = sin 4x.
Solution
(a) y = 1⁄2 sin x
Comparing with y = a sin bx,
a = 1⁄2,   b = 1
$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{1} ,0), (\dfrac{2\pi}{1} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2(1)},\dfrac{1}{2}) \quad \text{minimum point:} (\dfrac{3\pi}{2(1)},- \dfrac{1}{2}) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\pi ,0), (2\pi , 0)\\ \text{maximum point:} (\dfrac{\pi}{2}, \dfrac{1}{2}) \quad \text{minimum point:} (\dfrac{3\pi}{2},- \frac{1}{2}) \end{cases} } $$
Domain: R
Range: {y | - 1⁄2 ≤ y ≤ 1⁄2}
Period: 2π
Amplitude: 1⁄2
x-intercept: (0,0), (π,0), (2π, 0)
Maximum point: (π⁄2, 1⁄2)
Minimum point: (3π⁄2, - 1⁄2)
fig 7.1-b4

(b) y = sin 4x
Comparing with y = a sin bx,
a = 1,   b = 4
$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{4} ,0), (\dfrac{2\pi}{4} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2(4)},1) \quad \text{minimum point:} (\dfrac{3\pi}{2(4)},-1) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{4} ,0), (\dfrac{\pi}{2} , 0)\\ \text{maximum point:} (\dfrac{\pi}{8}, 1) \quad \text{minimum point:} (\dfrac{3\pi}{8},-1) \end{cases} } $$
fig 7.1-b5


3. Draw the graph of (a) 2 sin (a) y = π⁄4 x
(b) y = -sin πx.
Solution
(a) y = π⁄4 x
Comparing with y = a sin bx,
a = 2,   b = π⁄4
$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{\dfrac{\pi}{4}} ,0), (\dfrac{2\pi}{\dfrac{\pi}{4}} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2(\dfrac{\pi}{4})},2) \quad \text{minimum point:} (\dfrac{3\pi}{2(\dfrac{\pi}{4})},-2) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (4 ,0), (8 , 0)\\ \text{maximum point:} (2, 2) \quad \text{minimum point:} (6,-2) \end{cases} } $$
fig 7.1-b6

(b) y = -sin πx
Comparing with y = a sin bx,
a = -1,   b = π
$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{\pi} ,0), (\dfrac{2\pi}{\pi} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2(\pi)},-1) \quad \text{minimum point:} (\dfrac{3\pi}{2(\pi)},-(-1)) \end{cases} } $$ $$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (1 ,0), (2 , 0)\\ \text{maximum point:} (\dfrac{1}{2}, -1) \quad \text{minimum point:} (\dfrac{3}{2}, 1) \end{cases} } $$
fig 7.1-b7

4. Draw the graph of (a) y = 2 sin π⁄3(x-2) + 1
(b) y = -2 sin 1⁄2(x+1) + 2.
Solution
(a) y = 2 sin π⁄3(x-2) + 1
Comparing with y = a sin b(x-h) + k,
a = 2,   b = π⁄3,   h = 2,   k = 1
Key points on the midline y = k
(h, k), (π⁄b+h, k), (2π⁄b+h, k)
Maximum and minimum points
(π⁄2b+h, a+k), (3π⁄2b+h, -a+k)

Key points on the midline y = k
(2, 1), (π⁄ π⁄3 +2, 1), (2π⁄ π&frasl3 +2, 1)
Maximum and minimum points
(π⁄2 (π&frasl3) +2, 2+1), (3π⁄2 (π&frasl3) +2, -2+1)

 Key points on the midline y = k
(2, 1), (5, 1), 8, 1)
Maximum and minimum points
(7⁄2, 3), (13⁄2, -1)

Five key points: (2, 1), (5, 1), (8, 1), (7⁄2, 3), (13⁄2, -1)
fig 7.1-b8


(b) y = -2 sin 1⁄2(x+1) + 2.
Comparing with y = a sin b(x-h) + k,
a = -2,   b = 1⁄2,   h = -1,   k = 2
Key points on the midline y = k
(h, k), (π⁄b+h, k), (2π⁄b+h, k)
Maximum and minimum points
(π⁄2b+h, a+k), (3π⁄2b+h, -a+k)

Key points on the midline y = k
(-1, 2), (π⁄ 1⁄2 -1, 2), (2π⁄ 1&frasl2 -1, 2)
Maximum and minimum points
(π⁄2 (1&frasl2) -1, -2+2), (3π⁄2 (1&frasl2) -1, -(-2)+2)

 Key points on the midline y = k
(-1, 2), (2π-1, 2), 4π-1, 2)
Maximum and minimum points
(π-1, 0), (3π-1, 4)

Five key points: (-1, 2), (2π-1, 2), (4π-1, 2), (π-1, 0), (3π-1, 4)
fig 7.1-b9


5. Show that y = sin bx is an odd function.
Solution
Let f(x) = y = a sin bx
f(-x) = a sin b(-x)
= a sin (-bx)
= - a sin bx

-f(-x) = -(- a sin bx) = a sin bx
f(x) = -f(-x)
∴ y = a sin bx is an odd function.