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Math 110 Sample Test 4 Answers
1. Do p. 809 #12.
Plot the graph of y = 4sin(−2πx + π).
Use the odd identity that sin( -w ) = -sin( w ).
y = −4sin(2πx − π) Use the sine of a difference formula.
y = −4sin(2πx − π) = −4(sin(2πx)cos(π) − cos(2πx)sin(π) But cos(π) = −1 and sin(π) = 0
So y = 4sin(2πx)
2π
= 1 The phase shift is 0.
The amplitude is 4. The period is 2π
So the points to plot are ( 0, 0 ), ( 1/4, 4 ), ( 1/2, 0 ), (3/4, -4 ) and ( 1, 0 ).
4 y
3
2
1
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
−1
−2
−3
−4
2. Do p. 809 #21.
Plot the graph of y = csc(−x − π/4) − 2on graph paper.
Use the odd identity that csc( -w ) = -csc( w ).
y = −csc(x + π/4) − 2
The period is 2π. The phase shift is −π/4.
The graph has a vertical asymtotes at x = −pi/4, 3π/4 and 7π/4
Points to plot are ( 0, -sqrt(2)-2 ), ( pi/4, -3 ), (pi/2, -sqrt(2) - 2 ), (5π/4, -1 ), ( 3π/2, sqrt(2) − 2)
8
p(x)
6
4
2
x
1
2
3
4
5
−2
−4
−6
−8
−10
−12
3. Do p. 841 #11 & #35.
Evaluate arccos(-sqrt(3)/2).
1
arccos(-sqrt(3)/2) = t means cos( t ) = -sqrt( 3 ) / 2 and 0 ≤ t ≤ π
So t = 5π/6. Evaluate arcsec( sqrt( 2 ) ).
arcsec( sqrt( 2 ) ) = t means sec( t ) = sqrt( 2 ) and 0 ≤ x ≤ pi and x 6= π/2
Invert. cos( t ) = 1/sqrt( 2 ) and t = π/4.
4. Do p. 842 #159.
Evaluate sin( 2 arcsec(13/5) ).
arcsec(13/5) = t means sec( t ) = 13/5
The hypotenuse of the right triangle with sec( t ) = 13/5 is 13 and one leg is 5.
By the Pythagorean Theorem the other leg b satisfies 52 + b2 = 132 .
So b2 = 132 − 52 = 169 − 25 = 144
So b = +12 since t is in the first quadrant.
sin( 2 arcsec(13/5) ) = sin( 2t ) = 2 sin( t ) cos( t ) by the double angle identity for sine.
So sin( 2t ) = 2 · 12/13 · 5/13 = 120/169.
5. Do p. 845 #184.
Evaluate sin( arctan( x ) / 2 ).
Let t = arctan( x ). Then tan( t ) = x = x / 1.
The
√ hypotenuse of the triangle with angle t and legs 1 and x is by the Pythagorean theorem
1 + x2
Now sin( arctan( x ) / 2 ) = sin(
r t / 2 ). Use a half angle identity for sine,
q
1− √ 1 2
1+x
sin( t / 2 )= ± 1−cos(t)
=
±
2
2
√
Multiply top q
and bottom inside the outer sqrt by 1 + x2
√
2 −1
√
sin(t/2) = ± 21+x
Choose plus if x ≥ 0 or minus if x <0.
1+x2
6. Do p. 846 #215.
A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National
Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the
wilderness directly towards the tower. Let t denote the angle of depression from the top of the tower
to a point on the ground. If the range of the rifle with a tranquilizer dart is 300 feet, find the
smallest value of t for which the corresponding point on the ground is in range of the rifle.
The angle TSB equals the angle of depression.
So sin( t ) = sin( TSB ) = 200 / 300 and t = arcsin( 2/3 ).
7. Do p. 847 #223.
Find the domain of f (x) = arccos( 3x−1
2 ).
The domain of arccos( w ) is −1 ≤ w ≤ 1
So the domain of f is −1 ≤ x−1
Solve for x.
2 ≤1
−2 ≤ 3x − 1 ≤ 2
Add 1.
−1 ≤ 3x ≤ 3
Divide by 3.
−1
≤
x
≤
1.
3
Multiply by 2.
8. Plot a 5 point graph for f( x ) = 2 arccos( x - 1/2 ).
The domain of f is −1 ≤ x − 1/2 ≤ 1.
Solve for x. Add 1/2. −1 + 1/2 ≤ x ≤ 1 +√1/2 or −1/2 <√x < 3/2.
Five points to plot are x = -1/2, 1/2 - 1/ 2, 0, 1/2 + 1/ 2,1/2.
Now f( -1/2
√ ) = 2 arccos( -1 ) =√2π.
f( 1/2 - 1/ 2 ) = 2 arccos( - 1/ 2 ) = 2 · 3π/4 = 3π/2
2
f( 0 ) = 2 arccos(
-1/2 ) = 2 · 2π/3
√ = 4π/3.
√
f( 1/2 + 1/ 2 ) = 2 arccos( 1/ 2 ) = 2 · π/4 = π/2
f( 3/2 ) = 2 arccos( 1 ) = 0.
6 r(x)
5
4
3
2
1
x
−0.4 −0.2
0.2
0.4
0.6
0.8
1
1.2
9. Do p. 874 #9.
√
Solve sin(x/3) = 2/2
√ for all x.
Solve sin(x/3) = 1/ 2
So x/3 = π/4 + 2kπ or 3π/4 + 2kπ for an integer k
So x = 3π/4 + 6kπ or 9π/4 + 6kπ for an integer k
The only solution between 0 and 2π is 3π/4
10. Do p. 874 #28.
Solve 2sec(x)2 = 3 − tan(x) for 0 ≤ x < 2π.
Use the Pythagorean identity sec(x)2 = tan(x)2 + 1
2(tan(x)2 + 1) = 3 − tan(x).
2tan(x)2 + 2 = 3 − tan(x).
Add tan( x ) - 3.
2tan(x)2 + tan(x) − 1 = 0.
Factor.
( 2tan( x ) - 1 )( tan( x ) + 1 ) = 0
2tan( x ) - 1 = 0 or tan( x ) + 1 = 0.
tan( x ) = 1/2 or tan( x ) = -1.
So x = arctan( 1/2 ) or π + arctan(1/2) or π/4 or 3π/4
11. Do p. 875 #59.
Solve arccos( 2x ) = π
arccos( 2x ) = π means cos(π) = 2x. So -1 = 2x and x = -1/2.
3
1.4