Math 150 Sample Final Solutions

1) First write in the standard form by factoring out the coefficient of x:

amplitude 3, period , shift to the right

:

Range / y-axis / 5 key points
/ / / (divide by 3) /
/ / / (add) /

.

2)

X / Y=tanx / Description
/ 0 / x-intercept
/ Undefined / Vertical Asymptote
/ 0 / x-intercept
/ Undefined / Vertical Asymptote
/ 0 / x-intercept
X-intercepts / Vertical Asymptote
Range / y-axis
/ / Y=0
/ / Y=1

3) : Since the equation contains sine and cosine, use to convert cosine to sine: . This factors as . is impossible.

4)

5) See the picture: First find x: .

6)

7) and so . and , so

,

8)

a)

(observe that )

b) Add the fractions using the cross multiplication formula ,

c) Multiply by the conjugate:

9) see the picture: First find x: . ft.

10) a) b)

11) Sketch its Cartesian graph by first finding the x-intercept: ; Make two 30-60-90 triangles; one in QI and the other in QII to solve this: The x-intercepts are . First reflect the graph of about the x-axis. Then stretch the graph vertically between -2 and 2. The final transformation is to shift the graph up 1units.

12) )First make two triangles, one in QII, the other in QIII with the reference angle 60. . Divide by 4 to get . Pick k = 0, 1, 2, 3

13)

a) : Sine negative in QIII and QIV. But you take only the one ion QIV using a negative angle. Making a short triangle, b) : Cosine is 0 when x is 0 on the unit circle: or. But take only the one in the range of : So c) Tangent is negative in QII and QIV. But you take only the one ion QIV using a negative angle. Making a triangle, .

14) ): Use sin2A=2sinAcosA. Note that , not since inverse cosine of a negative number would be an angle in QII, where sine is positive.

B): use cos(A+B)=cosAcosB-sinAsinBc) Use :

15)) a) ) First multiply both sides by r to get To complete the square, . It is a circle of radius 2 centered at (0,2)

b) . Divide both sides by we get

16) a) This is a 30-60-80 triangle with the reference angle 60. To get the radius, multiply the short side by 2.

b)This is a 30-60-80 triangle with the reference angle 630.

.

17) Sine is -1 when the angle is on the negative y-axis.. By choosing k = 0 and k =1, we get

18)

19) )Since the equation contains x and 2x, first use double angle. Factor this expression and obtain (look at the graph or use the unit circle), (making triangles in QIII, QIV)

20).

This is a circle of radius 1, centered at (3,2) moving counterclockwise

t / /
/ 4 / 2
/ 3 / 3
/ 2 / 2

b) : Take the second equation and solve for t. Then substitute into the first. . . This is a line.

t / /
/ 1 / 1
1 / -2 / 2
2 / -5 / 3

21) Recall that plane + wind = actual. Angle = 450-bearing Plane = Wind = Plane + wind = . The actual speed is mph. For the true course, the reference angle is . Notice that the vector is in the 2ndquad since x-comp is neg and the y-comp is pos. Therefore, thebearingis 270+48.4 = 318.4 degrees.

22) . First solve :

make two triangles: one in QI, the other in QIII. The reference angle is . . Divide by 2 to get Next solve : Make two triangles: one in QII, the other in QIV with the reference angle .

. Divide by 2 to get Picking k=0 and k=1,

23) Make a triangle in QI since the sine is positive : r = 1 and y = x. Then the x-value is . Use double angle

24)Since the equation contains and , first use double angle to write. Then set one side =0.. Factor out , the equation becomes . Set each factor = 0, . .

25)

26) a) <1,3> and <-1,1> b) : . For , first find the reference angle: . Then since is in QII, Therefore, c) Using the conversion formulas, , . (-1.7, 2.5)