`Math 150 Exam II Practice Test Solution Key
1)
a)First observe that the new y axis is since there is a vertical transformation that shift the graph 1 down.
Transform the basic
amplitude 3, period , shift to the left
Range: 5 key points
/ / (divide by 2) // / (subtract) /
Original x-axis
(Transformations (optional)
b)
First observe that the new x axis is since there is a vertical transformation that shift the graph 1 up.
Transform the basic graph of y=-cosx
Since the function is multiplied by negative, begin with the reflected graph.
amplitude 1, period , shift to the right
Range: 5 key points
(it is a reflection not affecting the range) / / (divide by 3) // / (add) /
Original x-axis
(it is a reflection not affecting the range) /2)
a)
amplitude undefined , period , shift none
Range / 5 key points(subtract 1) / / (divide by 2) /
The original zeros of cos2x become asymptotes.
b )
amplitude undefined , period , shift none
Range / 5 key points(add 1) / / (divide by 3) /
The original zeros of sin3x+1 become asymptotes.
Zeros(divide by 3) /
3)
a)
Amplitude
Period
Shift to the right.
X-intercepts / Vertical Asymptotes(divide by 2) / /
add / /
b)
Amplitude
Period
Shift to the right.
X-intercepts / Vertical Asymptotes(sub ) / /
4) a) Since the cosine is negative, the triangle should be in the second quadrant , and using the Pythagorean, . The tangent of this triangle is b) Since the sine is negative, the triangle should be in the 4th quadrant , and using the Pythagorean, . The secant of this triangle is
5)a) Use the half angle formula for sin2x:
Apply the half-angle to .
b) The proof is almost the same as part a), except use
6)Since A is in QIV, in the second quadrant. a) is negative, is positive.
7a)
b)
c) d)
8) The first step is to find the quadrant of . Since A is in QIII, So is in Q II, so the sine is positive, cosine and tangent are negative. , ,
9) a) using the Pythagorean, . b) using the Pythagorean, .
10)
11)
Domain / Range/ 1)[-1,1] / 2)
/ 3)[-1.1] / 4)
/ 5)All real numbers / 6)
12) (make a triangle with hyp = 1, adj = x. Then find the y-value by the Pythagorean) B)
C) since the cosine is, r = x, x = 1. Using the Pythagorean, . The cosecant of this triangle is
13) a)
b)
14a)
The new x-axis is y=2.
Transform the basic cosine graph
Period , amplitude 1, shift to the right
Range: 5 key points
/ / (divide by 2) // / (add) /
B )
Transform the basic graph of y=sinx:
The new x-axis is y=1 since the graph is shifted up 1.
Period , amplitude 2, no phase shift
Range: 5 key points
/ / (multiply by 3) /16) a)make a triangle in the second quadrant with x = -3 and r = 5. Then compute the y-value using the Pythagorean (y = 4) (or remember the famous 3-4-5 triangle) b) make a triangle in the first quadrant with y = 5, r = 7.
17) and so . and , so
,
18)
19) This is a complex fraction. Multiply each by cosx.
Now multiply by the conjugate.
20)a)
b)
c) (simplify the complex fraction by multiplying by sinx ) =
21)
, is impossible.
22)
, is impossible.
23)
, k
24) Use double angle to write . The equation simplifies to , which is .
25) . Simplify this and factor the expression to get . Therefore,
26) Use the quadratic formula to solve for using a = 1, b= -2, c = -1 . cosx=2.41 is impossible. To solve cosx= -0.41, use a calculator to find the reference angle . Since cosx is negative, make two triangles, one in QII, the other in QIII.
27) : Using the half-angle , we get
28) ) Use the double angle for cosine to rewrite the equation as . This simplifies as . since the cosine is positive in QI and QIV, is impossible
29) Use the quadratic formula to solve for a) First write the equation using only the sine function: . sinx=1.62 is impossible. To solve sinx= -0.62, use a calculator to find the reference angle . Since sinx is negative, make two triangles, one in QIII, the other in QIV. b) . Cos x=1.62 is impossible. To solve cos x= -0.62, use a calculator to find the reference angle . Since cos x is negative, make two triangles, one in QII, the other in QIII .
30) (impossible) and
31) a) Since and , adding the equations and dividing by 2, we obtain b) this is same as part a) since multiplication is commutative. C) and D) Since and , adding the equations and dividing by 2, we obtain c) and subtracting and dividing by - 2 give d) : c) D)
32) )
33)
34)
35) A) .
Divide both sides by 2 to get . Let k = 0 and k =1,
B)
Divide both sides by 2 to get .
Make triangles, one in QIII, the other in QIV, to get
, . Divide by 2, we get . Pick k = 0 and k = 1,
c)
Make two triangles, one in QII, the other in QIII, , . Divide by 3, we get .
d) :
First find the angle whose cosine is -1: this angle is (x is -1 on the negative x-axis). Solve . . Pick and , we get
36(
a)The values of inverse sine lie in QI or QIV. Since is inverse sine of ½, which is a positive number, its value is in QI. Sketch a short in QI,
1
b)The values of inverse cosine lie in QI or QII. Since is inverse cosine of -½, which is a negative number, its value is in QII. Sketch a tall in QI,
c)The values of inverse tangent lie in QI or QIV. Since is inverse tangent of -1, which is a negative number, its value is in QIV. Sketch a in QIV,
d) The values of inverse sine lie in QI or QIV. Since is inverse sine of a negative number, its value is in QIV. Sketch a tall in QIV,
e) (-1 is on the x-axis)
f) (-1 is on the y-axis)
g) (it is not )
37) a)
5 key points
/ Add :The zeros of become the vertical asymptotes
Zeros
/ Add :b)
/ :The zeros of become the vertical asymptotes. The vertical shift does not affect the vertical asymptotes.
Zeros
/ Divide by 20