## alysa111 one year ago For an angle Θ with the point (−20, −21) on its terminating side, what is the value of cosine?

1. alysa111

@kelliegirl33 can you help me

2. myininaya

If it helps you graph the point. Can you do that for me first?

3. alysa111

@myininaya just plot -20,-21 ?

4. myininaya

Yep plot the point as a first step.

5. myininaya

You can use the draw tool.

6. alysa111

ok

7. alysa111

|dw:1403474666905:dw|

8. alysa111

@myininaya

9. myininaya

|dw:1403474755600:dw| Now I will simply make a right triangle using the center of the graph for my central angle.

10. myininaya

what is the ratio definition of cosine of an angle?

11. alysa111

legnth of adjacent / legnth of hyp

12. myininaya

right you already have the adjacent side given in the picture the only thing you need now is the hyp of that triangle, right?

13. alysa111

yes

14. myininaya

Find the hyp of that triangle using Pythagorean Theorem.

15. alysa111

16. myininaya

You are correct to say that.

17. myininaya

Can you find the length of the hyp?

18. alysa111

yes hold on wouldnt it be -20^2 + 21^2 = c^2

19. myininaya

20^2+21^2=c^2 and then take the square root of both sides to give you c.

20. alysa111

29

21. myininaya

so that is the length of the hyp so what is cos(theta)=?

22. alysa111

29 ?

23. myininaya

you said earlier cos(theta) is adj/hyp you never said cos(theta) is hyp

24. alysa111

20/29

25. myininaya

don't forget the negative sign

26. alysa111

thank you so much i have a few more cn yu help me

27. myininaya

if the angle was on the side of the graph like on the right hand side of the y-axis cos would have been indeed a positive number but since the angle was to the left of the y-axis cos will be negative

28. myininaya

|dw:1403475467323:dw|

29. alysa111

How does changing the function from f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function?

30. myininaya

we could even talk about sine sin is positive when above the x-axis and sine is negative when below the x-axis |dw:1403475522177:dw|

31. myininaya

Well you are talking about shifting the graph up when you have that +5

32. alysa111

so the function shifts up 5 units so the range changes from -1 to 1

33. alysa111

in f(x) to 4 to 6 in g(x)

34. myininaya

For example, pretend we have y=sin(x) the range of sin(x) is from -1 to 1 (inclusive). but when we look at y=sin(x)+1 that means you are moving every point of y=sin(x) up one unit so the range of sin(x)+1 is from -1+1..1+1( inclusive) doing the addition gives 0..2 (inclusive)

35. alysa111

am i right

36. myininaya

well first what is the range of y=3sin(x)?

37. alysa111

f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function

38. myininaya

yes the range of the graph of f would be affected if you need f+5

39. myininaya

did f+5 (not need)

40. myininaya

but the range of y=3sin(x) is not -1 to 1.

41. alysa111

Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|

42. alysa111

its 2 to 8

43. myininaya

Let y=sin(x). Let f(x)=3sin(x). Therefore you could say f(x)=3y where y=sin(x). If you have 3 times every y per x wouldn't the range be stretched more?

44. alysa111

yes

45. myininaya

like the range for y=sin(x) is -1 to 1 (when we talk about range we are talking about the y values) the lowest y value for y=3sin(x) would be what then?

46. alysa111

so f(x) has greater y-intercerpt

47. myininaya

the lowest y value for y=sin(x) is -1 so the lowest y value for f(x)=3sin(x) is ?

48. alysa111

-3

49. myininaya

great so it shouldn't be hard for you to see the highest y-value for f(x)=3sin(x) would be 3(1)=3 so the range for y=3sin(x) is -3 to 3 (inclusive)

50. myininaya

now what is the range for y=3sin(x)+5

51. alysa111

2 to 8

52. myininaya

omg you are great so now you can answer that one question about how the range is affected and actually state the ranges of both functions

53. alysa111

now for Which of the following functions has the greatest y-intercept? wouldnt it be f(x) since the y is 9

54. myininaya

Also when answering the question it might be better to choose to answer in interval notation or inequality notation. 2 to 8 isn't really something people say to state an interval (not that I have ever seen anyways)

55. alysa111

Okay , so it would be stated has 2 to 8 in g(x)

56. myininaya

well like i said you might want to choose to answer in interval notation (or inequality) y is an element of the interval [2,8] <--interval (i put brackets to mean to included endpoints inequality notation: $2 \le y \le 8$ and to your next question.... |dw:1403476409004:dw| do you what the y-intercept is for a curve? It is the part of the curve that goes through the y-axis. When I say part, I'm talking about the specific point in which the curve does that. What do you know about all the points on the y-axis? What do they all have in common?

57. myininaya

I will give you a hint. All the x values of any point on the y-axis are _____?___

58. alysa111

Im not sure

59. alysa111

4-9 = -5 which is the same for g(x)

60. alysa111

the same y-intercept

61. myininaya

Ok since you can't answer that, I will ask you a series of questions now? Where does (0,10) lie? Where does (0,5) lie? Where does (0,2) lie? Where does (0,0) lie? Where does (0,-1) lie? Where does (0,-6) lie? where does (0,-1000) lie?

62. alysa111

on the x axis

63. myininaya

That is incorrect. They all lie on the y-axis. The answer to my question: All the x-values of any point on the y-axis is ____?__ All the x-values of any point on the y-axis is zero. So to find the y-intercept for a curve you just need to find y when x is zero. You just need to replace x with zero and solve for y.

64. alysa111

so 0-9 ? 0-4 ?

65. myininaya

For the one function, it should be clear (0,0) is the y-intercept (that is the origin of the graph after all; is is all an x-intercept) The other function you did right even though it doesn't seem you know why you were doing it since you were not able to answer the question.

66. alysa111

Yeah im confused now

67. myininaya

|dw:1403477105487:dw| This is the information given. You are asking to find the y-intercept for both graphs right? Can you tell me which point from f have the x is 0?

68. alysa111

functions has the greatest y-intercept

69. alysa111

1,1 ?

70. myininaya

I'm asking you to tell which of the points from f have the x value is 0?

71. myininaya

|dw:1403477214481:dw|

72. alysa111

its 0,0 ?

73. myininaya

|dw:1403477221682:dw| Do you see any of these x's being 0?

74. alysa111

yes

75. myininaya

yes the point (0,0) is a y-intercept because the x is 0

76. myininaya

now we need to look at g(x)=5cos(3x)-5

77. myininaya

what point on g gives us the x is 0?

78. alysa111

-5

79. myininaya

how did you get that?

80. myininaya

Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.

81. myininaya

82. myininaya

Replace the x with 0 like so y-intercept of g is g(0)=5cos(3*0)-5

83. myininaya

evaluate 5cos(3*0)-5

84. alysa111

5cos(0)-5 5cos

85. myininaya

you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5 but where do we go from there?

86. alysa111

1.41

87. myininaya

:( 5cos(0)-5 what is cos(0)?

88. alysa111

1

89. myininaya

5(1)-5 can you simplify from here now?

90. alysa111

0

91. alysa111

same y

92. myininaya

right so the y-intercept for g is (0,0) as well they have the same y-intercept

93. alysa111

1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.

94. myininaya

draw a right triangle using this information sin(theta)=15/17

95. alysa111

$\sin^2 \theta + \cos ^2\theta = 1$ will be what we use ?

96. myininaya

you can use that

97. alysa111

|dw:1403477725033:dw|

98. alysa111

|dw:1403477767271:dw|

99. myininaya

if you want to use that specific identity you can replace sin(theta) with 15/17 and solve for cos(theta)

100. alysa111

sin(15) = 0.650

101. alysa111

sin*17) = -0.96

102. myininaya

can you replace sin(theta) with 15/17 in the following equation $\sin^2(\theta)+\cos^2(\theta)=1$

103. alysa111

no

104. myininaya

why not?

105. alysa111

well wouldnt you put them in front of sin and cos

106. myininaya

just so you know $\sin^2(\theta)$ and $(\sin(\theta))^2$ have exactly the same meaning

107. myininaya

you are given sin(theta) is 15/17 so you replace the whole sin(theta) thing with 15/17

108. myininaya

$\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1$ So we have $(\sin(\theta))^2+(\cos(\theta))^2=1 \\ (\frac{15}{17})^2+(\cos(\theta))^2 =1 \text{ I did this because I was given that } \sin(\theta)=15/17$

109. myininaya

now solve for cos(theta)

110. myininaya

if that confuses you solve the following equation for u $(\frac{15}{17})^2+u^2=1$

111. myininaya

whatever value(s) you get for u is your answer since u is cos(theta)

112. myininaya

and cos(theta) is the objective

113. alysa111

225/285

114. alysa111

|dw:1403478332917:dw|

115. myininaya

so what did you do to get that?

116. alysa111

found cos of theta

117. myininaya

you solve the equation (15/17)^2+u^2=1 for u (which is cos(theta) )?

118. alysa111

yes

119. myininaya

you might want to do that one more time

120. myininaya

the denominator is right but your numerator is off

121. alysa111

8 / 17

122. myininaya

ok so cos could either be 8/17 or -8/17 when sin is 15/17

123. alysa111

|dw:1403478613956:dw| my only option

124. myininaya

Agreed

125. alysa111

In Shadow Bay in July, high tide is at 1:30 pm. The water level is 4 feet at high tide and 0 feet at low tide. Assuming the next high tide is exactly 12 hours later and the height of the water can be modeled by a cosine curve, find an equation for Shadow Bay's water level in July as a function of time (t).

126. alysa111

okay so i got f(t) = 4cos pie/2t +1

127. alysa111

@myininaya