Math 125 - HW 3 Solutions

8/22/2019 Math 125 - HW 3 Solutions

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HW 3 Problems Solutions (Page 148)

66 y = Step 1: The amplitude = |A| = |

|

[

|

|, |

|]

[ , ]Step 2: For this function, =

The period = =

= 2 . = 16Divide [0,

] = [0, 16] into 4 equal lengths

|___________|___________|___________|____________|

0 4 8 12 16

= 8 = 4 = 12

Step 3: Make the table to find five key points

x 0 4 8 12 16

x 0

sin(x) 0 1 0 1 0

sin(x) 0 0 0

y = sin(x) + 1 2


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Step 4: Connect the points: (0,), (4, 1), (8,

),

(12, 2), (16,)

y = HW 3 Problems Solutions (Page 157)

38 y = 3cot 2Step 1: The period =

=

=

Step 2: Adjacent vertical asymptotes: x = 0 and x = x = 0 and

x =

===> x = 0 and x = 2Step 3: Divide (0, ) into 4 equal lengths

|__________|___________|___________|___________|

0


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=

=

=

Step 4: Make the table for:

first-quarter point ,midpoint ,third-quarter point

Step 5: Connect the points: (, 1), (, ), ( , )

y = 3cot 2

x

x

cot 1 0 1

3 cot 3 0 3

y = 3cot 1


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40 y = 3 sec + 1Were going to draw y = 3 cos + 1 first. Then we connect the maxand the min points with a U-shape.

Step 1: The amplitude = |A| = |3| [ |3|, |3|] [ 3, 3]

Step 2: For this function, = The period =

=

= 2 . 4 = 8

Divide [0, ] = [0, 8] into 4 equal lengths|__________|____________|____________|____________|

0 2 4 6 8

= 4

= 2 = 6

Step 3: Make the table to fine five key points

x 0 2 4 6 8x 0

2

cos(

x) 1 0

1 0 1

3cos(x) 3 0 3 0 3

y = 3cos(x) + 1 4 1 1 4


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Step 4: Connect the points: (0, 4), (2, 1), (4, ),(6, 1), (8, 4)

y = 3 sec + 1

HW 3 Problems Solutions (Page 188)

16 Let = ==> cos = , and The angle is in quadrant I and II based on the restriction. Sincecos < 0, then the only quadrant that fits the cosine is negative isquadrant II, because cosine is positive in quadrant I and we cant use

it. So we find that when well have cos = .Thus,

=

=

22 sin-1 Let = sin-1 ==> sin = , and

We know is in quadrants I and IV, but we also know sin < 0.


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Thus we narrow it down the quadrant is in, which is in quadrant IV.The only angle that gives sin = is ==> = sin-1 =

24 sin-1

Let = sin-1 ==> sin = , We know is in quadrants I and IV, but we also know sin < 0. Thus wenarrow it down the quadrant is in, which is in quadrant IV. The onlyangle that gives sin = is ==> = sin-1 =

40 sin-1[ ]Based on 3a, since is in [ ], then sin-1[ ] =

44 tan-1[ ]We cant use 3e since is not in . Since is in quadrantIII, then tan > 0. We want to find an angle that is in quadrant Ior IV that has tan> 0. We cant use from quadrant IV becausetan < 0 there. Then, the only quadrant left that has withtan = tan

is quadrant I. Finally, we see that tan

= tan

.Then tan-1[ ] = tan-1 =

46 cos[cos-1( ]Based on 3d, since is in [-1, 1], then cos[cos-1( )] =

52 sin[sin-1(1.5)]

Based on 3b, since 1.5 is not in [-1, 1], then sin-1(1.5) is not defined.

Thus sin[sin-1(1.5)] is also not defined.


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HW 3 Problems Solutions (Pages 194)

14 cot ]Let

=

===> sin

=

,

and

So is in Quadrants I and IV, based on the restriction. Since sin ,then we eliminate quadrant I because sin is positive there. Then isin quadrant IV. So we find that when we have sin = .

Thus, cot[ = cot =

=

=

24 cos-1[ ]We cant use 3c since is not in [0, ]. Since is in quadrantIV, then cos > 0. We want to find an angle that is in quadrant Ior II that has cos> 0. We cant use from quadrant II becausecos < 0 there. Then, the only quadrant left that has withcos = cos is quadrant I. Then by trial error, we see thatcos

= cos

==> cos-1[

] = cos-1[

] =

30 csc[tan-1( 2)]Let = tan-1( 2) ==> tan = 2, < < and Since tan < 0 and < < , then is in quadrant IV.Since tan = 2 = =

, then y = 2 and x = 1.

So, r = = = Thus,

csc[tan-1( 2)] = csc = = = 32 cot[ ]

Let = cos-1 ==> cos = 0 and 1


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Since < 0 and , then is in quadrant II.Since cos = = , then x = and r = 3.So, r2 = x2 + y2 ==> y2 = r2 x2 ==> y =

= ()= =

Thus, cot[ ]= cot = = =

. =

=

= =

34 csc Let = ===> tan = , < < and The restriction gives us is in quadrants I and IV. Since tan > 0, thenwe have is in quadrant I.Since tan = , then y = 1 and x = 2So r = = = Thus, csc = csc = = =

36 We have = .

Let = ===> cos = , 0 and 1

We eliminate quadrant I since cos > 0 there.Hence, we find when we have cos =

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Math 125 - HW 3 Solutions