CHAPTER 0 PRELIMINARIES
SECTION 0.7 Trigonometric Functions
Problem Set 0.7, Number 32 – 56.
- Find ientities analogous to the addition identities for each expression.
- Use the addition identity for the tangent to show that tan (t + π) = tan t for all t in the domain of tan t.
- Show that cos (x – π) = – cos x for all x.
- Suppose that a tire on a truck has an outer radius of 2.5 feet. How many revolutions per minute does the tire make when the truck is traveling 60 miles per hour?
- How far does a wheel of radius 2 feet roll along level ground in making 150 revolutions?
- A belt passes around two wheels, as shown in Figure 15. How many revolutions per second does the small wheel make when the large wheel makes 21 revolutions per second?
- The angle of inclination α of a line is the smallest positive angle from the positive x-axis to the line (α = 0 for a horizontal line). Show that the slope m of the line is equal to tan α.
- Find the angle of inclination of the following lines (see Problem 38).
- Let l1 and l2 be two nonvertical intersecting lines with slopes m1 and m2, respectively. If θ, the angle from l1 and l2, is not a right angle, then
- Find the angle (in radians) from the first line to the second (see Problem 40).
- Derive the formula A = ½ r2t for the area of a sector of a circle. Here r is the radius and t is the radian measure of the central angle (se Figure 17).
- Find the area of the sector of a circle of radius 5 centimeters and central angle 2 radians (see Problem 42).
- A regular polygon of n sides is inscribed in a circle of radius r. Find formulas for the perimeter, P, and area, A, of the polygon in terms of n and r.
- An isosceles triangle is topped by a semicircle, as shown in Figure 18. Find a formula for the area A of the whole figure in terms of the side length r and angle t (radians). (We say that A is a function of the two independent variables r and t.)
- From a product identity, we obtain
- The normal high temperature for Las Vegas, Nevada, is 55°F for January 15 and 105° for July 15. Assuming that these are the extreme high and low temperatures for the year, use this information to approximate the average high temperature for November 15.
- Tides are often measured by arbitrary height markings at some location. Suppose that a high tide occurs at noon when the water level is at 12 feet. Six hours later, a low tide with a water level of 5 feet occurs, and by midnight another high tide with a water level of 12 feet occurs. Assuming that the water level is periodic, use this information to find a formula that gives the water level as a function of time. Then use this function to ap- proximate the water level at 5:30 P.M.
- Circular motion can be modeled by using the parametric representations of the form x(t) = sin t and y(t) = cos t. (A parametric representation means that a variable, t in this case, determines both x(t) and y(t).) This will give the full circle for 0 ≤ t ≤ 2π. If we consider a 4-foot-diameter wheel making one complete rotation clockwise once every 10 seconds, show that the motion of a point on the rim of the wheel can be represented by x(t) = 2 sin(πt/5) and y(t) = 2cos(πt/5).
- The circular frequency v of oscillation of a point is given by v = (2π)/period. What happens when you add two motions that have the same frequency or period? To investigate, we can graph the functions y(t) = 2 sin(πt/5) and y(t)= sin(πt/5) + cos(πt/5) and look for similarities. Armed with this information, we can investigate by graphing the following functions over the interval [-5, 5]:
- We now explore the relationship between A sin(ωt) + B cos(ωt) and C sin(ωt + Φ).
- Graph the function f(x) = sin 50x using the window given by a y range of -1.5 ≤ y ≤ 1.5 and the x range given by
- Graph the function f(x) = cos x + (1/5) sin 50x using the windows given by the following ranges of x and y.
- Let f(x) = (3x + 2)/(x2 + 1) and g(x) = (1/100) cos (100x).
- Suppose that a continuous function is periodic with period 1 and is linear between 0 and 0.25 and linear between -0.75 and 0. In addition, it has the value 1 at 0 and 2 at 0.25. Sketch the function over the domain [-1, 1], and give a piecewise definition of the function.
- Suppose that a continuous function is periodic with period 2 and is quadratic between -0.25 and 0.25 and linear between -1.75 and -0.25. In addition, it has the value 0 at 0 and 0.0625 at ±0.25. Sketch the function over the domain [-2, 2], and give a piecewise definition of the function.
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(a)
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sin (x – y)
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(b)
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cos (x – y)
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(c)
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tan (x – y)
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💬 SOLUTION:
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(a)
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sin (x – y)
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= sin x cos (–y) + cos x
sin (–y)
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= sin x cos y – cos x sin
y
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(b)
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cos (x – y)
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= cos x cos (–y) – sin x
sin (–y)
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= cos x cos y + sin x sin
y
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(c)
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tan (x – y)
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= [tan x + tan (–y)]/[1 – tan
x tan (–y)]
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= [tan x – tan y]/[1 + tan x
tan y]
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💬 SOLUTION:
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tan (t + π)
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= [tan t + tan π]/[1 – tan t
tan π]
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= [tan t + 0]/[1 – (tan t) (0)]
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= tan t
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💬 SOLUTION:
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cos (x – π)
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= cos x cos (–π) – sin x
sin (–π)
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= – cos x – 0 · sin x
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= – cos x
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💬 SOLUTION:
s = rt = (2.5 ft)(2π rad) = 5π ft, so the tire goes 5π feet per revolution, or (1/5π) revolutions per foot.
[(1/5π)(rev/ft)][60(mi/hr)][(1/60)(hr/min)][5280(ft/mi)]
≈ 336 rev/min
💬 SOLUTION:
s = rt = (2 ft)(150 rev)(2π rad/rev) ≈ 1885 ft
💬 SOLUTION:
r1t1 = r2t2 ; 6(2π)t1 = 8(2π)(21)
t1 = 28 rev/sec
💬 SOLUTION:
Δy = sin α and Δx = cos α
m = (Δy)/(Δx) = (sin α)/(cos α) = tan α
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(a)
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y = √(3)
· x – 7
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(b)
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√(3) · x + 3y = 6
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💬 SOLUTION:
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(a)
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tan α = √3
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α =
π/3
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(b)
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(√3)x + 3y = 6
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3y = –(√3)x + 6
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y = –((√3)/3)x + 2; m = –(√3)/3
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tan α = –(√3)/3
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α = 5π/6
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BACA JUGA:
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θ = [m2 – m1]/[1 + m1m2]
Show this using the fact that θ = θ2 – θ1 in Figure 16.
💬 SOLUTION:
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m1 = tan θ1 and m2 = tan θ2
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||
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tan θ
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=
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tan (θ2 – θ1)
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=
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tan θ2 + tan (–θ1)
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1 – tan θ2 tan (–θ1)
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=
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tan θ2 – tan θ1
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1 + tan θ2 tan θ1
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=
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m2 – m1
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1 + m1m2
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(a)
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y = 2x,
y = 3x
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(b)
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y = (x/2),
y = –x
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(c)
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2x – 6y = 12, 2x + y
= 0
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💬 SOLUTION:
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(a)
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tan θ = [3 –
2]/[1 + 3(2)] = 1/7
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θ ≈ 0.1419
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(b)
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tan θ = [–1 –
(1/2)]/[1 + (1/2)(–1)] = –3
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θ ≈ 1.8925
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(c)
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2x
– 6y = 12
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2x
+ y = 0
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–6y
= –2x + 12y = –2x
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y =
(1/3)x – 2
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m1
= 1/3 ; m2 = –2
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tan θ = [–2 – (1/3)]/[1 + (1/3)(–2)] = –7
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θ ≈ 1.7127
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💬 SOLUTION:
Recall that the area of the circle is πr2. The measure of the vertex angle of the circle is 2π. Observe that the ratios of the vertex angles must equal the ratios of the areas. Thus,
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t
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=
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A
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, so
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2π
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πr2
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A
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=
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1
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r2t.
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2
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💬 SOLUTION:
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A
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=
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1
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(2)(5)2 = 25 cm2
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2
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💬 SOLUTION:
Divide the polygon into n isosceles triangles by drawing lines from the center of the circle to the corners of the polygon. If the base of each triangle is on the perimeter of the polygon, then the angle opposite each base has measure 2π/n. Bisect this angle to divide the triangle into two right triangles (See figure).
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sin (π/n) = (b/2r)
so b = 2r sin (π/n) and cos (π/n) =
(h/r) so h = r cos (π/n).
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P = nb
= 2rn sin (π/n)
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A = n
[(1/2) bh] = nr2 cos (π/n) sin (π/n)
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💬 SOLUTION:
The base of the triangle is the side opposite the angle t. Then the base has length 2r sin (t/2) (similar to Problem 44). The radius of the semicircle is r sin (t/2) and the height of the triangle is r cos (t/2).
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A
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=
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(1/2)[2r sin (t/2)][r cos (t/2)]
+ (π/2)[r sin (t/2)]2
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=
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[r2 sin (t/2) cos (t/2)]
+ [(πr2/2)
sin2 (t/2)]
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cos (x/2) cos (x/4) = ½ [cos ((3/4)x) + cos ((1/4)x)]
Find the corresponding sum of cosines for
cos (x/2) cos (x/4) cos (x/8) cos (x/16)
Do you see a generalization?
💬 SOLUTION:
cos (x/2) cos (x/4) cos (x/8) cos (x/16)
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=
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(1/2) [cos (3/4) x + cos (1/4) x] (1/2) [cos
(3/16) x + cos (1/16) x]
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=
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(1/4) [cos (3/4) x + cos (1/4) x] [cos
(3/16) x + cos (1/16) x]
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=
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(1/4) [cos (3/4) x cos (3/16) x + cos
(3/4) x cos (1/16) x + cos (1/4) x cos (3/16) x +
cos (1/4) x cos (1/16) x]
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=
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(1/4) [(1/2)(cos (15/16) x + cos (9/16) x)
+ (1/2)(cos (13/16) x + cos (11/16) x) + (1/2)(cos (7/16) x
+ cos (1/16) x) + (1/2)(cos (5/16) x + cos (3/16) x)]
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=
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(1/8) [cos (15/16) x + cos (13/16) x + cos
(11/16) x + cos (9/16) x + cos (7/16) x + cos (5/16) x
+ cos (3/16) x + cos (1/16) x]
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💬 SOLUTION:
The temperature function is T(t) = 80 + 25 sin [(2π/12)(t – (7/2))].
The normal high temperature for November 15th is then T(10.5) = 67.5 °F.
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BACA JUGA:
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💬 SOLUTION:
The water level function is F(t) = 8.5 + 3.5 sin [(2π/12)(t – 9)].
The water level at 5:30 P.M. is then F(17.5) ≈ 5.12 ft.
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(a)
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Find the positions of the point on the rim
of the wheel when t = 2 seconds, 6 seconds, and 10 seconds. Where was this
point when the wheel started to rotate at t = 0?
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(b)
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How will the formulas giving the motion of
the point change if the wheel is rotating counterclockwise.
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(c)
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At what value of t is the point at
(2, 0) for the first time?
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💬 SOLUTION:
As t increases, the point on the rim of the wheel will move around the circle of radius 2.
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(a)
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x(2) ≈
1.902
y(2) ≈
0.618
x(6) ≈ –1.176
x(6) ≈ –1.618
x(10) =
0
y(10) =
2
x(0) = 0
y(0) = 2
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(b)
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x(t)
= –2
sin [(π/5)t], x(t) = 2
cos [(π/5)t]
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(c)
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The
point is at (2, 0) when (π/5)t = (π/2) ; that is, when t = (5/2).
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(a)
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y(t) = 3 sin(πt/5) - 5cos(πt/5) + 2sin((πt/5) - 3)
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(b)
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y(t) = 3 cos(πt/5 - 2) + cos(πt/5) + cos((πt/5) - 3)
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💬 SOLUTION:
Both functions have frequency 2π/10. When you add functions that have the same frequency, the sum has the same frequency.
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(a)
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y(t)
= 3
sin (πt/5) – 5 cos (πt/5) + 2 sin ((πt/5) – 3)
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(b)
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y(t)
= 3
cos ((πt/5) – 2) + cos (πt/5) + cos ((πt/5) – 3)
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(a)
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By expanding sin(ωt + Φ) using the sum of the angles formula, show that the two expressions are equivalent if A = C cos Φ and B = C sin Φ.
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(b)
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Consequently, show that A2 + B2 = C2 and that Φ then satisfies the equation tan Φ = B/A.
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(c)
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Generalize your result to state a proposition about A1 sin(ωt + Φ1) + A2 sin(ωt + Φ2) + A3 sin(ωt + Φ3).
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(d)
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Write an essay, in your own words, that expresses the impor- tance of the identity between A sin(ωt) + B cos(ωt) and C sin(ωt + Φ). Be sure to note that |C| ≥ max(|A|, |B|) and that the identity holds only when you are forming a linear combination (adding and/or subtracting multiples of single powers) of sine and cosine of the same frequency.
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💬 SOLUTION:
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(a)
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C sin (ωt
+ Φ) = (C cos Φ) sin ωt + (C sin Φ)
cos ωt.
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Thus A = C ⋅ cos Φ and B = C ⋅ sin Φ.
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(b)
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A2 + B2 = (C ⋅ cos Φ)2 + (C ⋅ sin Φ)2
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A2 + B2 = C2 (cos2
Φ) + C2 (sin2 Φ)
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A2 + B2 = C2
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Also, (B/A) = (C ⋅ sin Φ)/(C ⋅ cos Φ)
= tan Φ
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(c)
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A1 sin (ωt + Φ1) + A2
sin (ωt + Φ2) + A3 sin (ωt
+ Φ3)
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= A1 (sin ωt cos Φ1
+ cos ωt sin Φ1) + A2 (sin ωt
cos Φ2 + cos ωt sin Φ2) + A3
(sin ωt cos Φ3 + cos ωt sin Φ3)
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= (A1 cos Φ1
+ A2 cos Φ2 + A3 cos Φ3)
sin ωt + (A1 sin Φ1 + A2
sin Φ2 + A3 sin Φ3) cos
ωt
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= C sin (ωt + Φ)
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where C and Φ can be computed
from
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A = A1
cos Φ1 + A2 cos Φ2 + A3
cos Φ3
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B = A1
sin Φ1 + A2 sin Φ2 + A3
sin Φ3
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as in part (b).
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(d)
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Written response. Answers will vary.
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Trigonometric functions that have high frequencies pose special problems for graphing. We now explore how to plot such functions.
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(a)
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[-15, 15]
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(b)
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[-10, 10]
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(c)
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[-8, 8]
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(d)
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[-1, 1]
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(e)
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[-0.25, 0.25]
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Indicate briefly which x-window shows the true behavior of the function, and discuss reasons why the other x-windows give results that look different.
💬 SOLUTION:
(a), (b), and (c) all look similar to this:
(d)
(e)
The windows in (a)-(c) are not helpful because the function oscillates too much over the domain plotted. Plots in (d) or (e) show the behavior of the function.
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(a)
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-5 ≤ x ≤ 5, -1 ≤ y ≤ 1
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(b)
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-1 ≤ x ≤ 1, 0.5 ≤ y ≤ 1.5
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(c)
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-0.1 ≤ x ≤ 0.1, 0.9 ≤ y ≤ 1.1
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Indicate briefly which (x, y)-window shows the true behavior of the function, and discuss reasons why the other (x, y)-windows give results that look different. In this case, is it true that only one window gives the important behavior, or do we need more than one window to graphically communicate the behavior of this function?
💬 SOLUTION:
(a)
(b)
(c)
The plot in (a) shows the long term behavior of the function, but not the short term behavior, whereas the plot in (c) shows the short term behavior, but not the long term behavior. The plot in (b) shows a little of each.
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(a)
|
Use functional composition to form h(x) = (f ○ g)(x) as well as j(x) =(g ○ f)(x).
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(b)
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Find the appropriate window or windows that give a clear picture of h(x)
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(c)
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Find the appropriate window or windows that give a clear picture of j(x)
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💬 SOLUTION:
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(a)
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h(x)
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=
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(f ○ g)(x)
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=
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(3/100) cos (100x) + 2
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(1/100)2 cos2 (100x)
+ 1
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j(x)
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=
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(g ○ f)(x)
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(1/100) cos [100 ⋅ ((3x + 2)/(x2 + 1))]
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(b)
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(c)
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|
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💬 SOLUTION:
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f(x)
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=
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{
|
4(x –〚x〛) + 1 :
x ∈ [n, n + (1/4))
|
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–(4/3)(x –〚x〛) + (7/3)
: x ∈ [n + (1/4), n + 1)
|
where n is an integer.
💬 SOLUTION:
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f(x)
|
=
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{
|
(x – 2n)2 , x ∈ [2n – (1/4), 2n + (1/4)]
|
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0.0625, otherwise
|
where n is an integer.
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BACA JUGA:
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