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

1. alysa111 Group Title

@kelliegirl33 can you help me

2. myininaya Group Title

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

3. alysa111 Group Title

@myininaya just plot -20,-21 ?

4. myininaya Group Title

Yep plot the point as a first step.

5. myininaya Group Title

You can use the draw tool.

6. alysa111 Group Title

ok

7. alysa111 Group Title

|dw:1403474666905:dw|

8. alysa111 Group Title

@myininaya

9. myininaya Group Title

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

10. myininaya Group Title

what is the ratio definition of cosine of an angle?

11. alysa111 Group Title

legnth of adjacent / legnth of hyp

12. myininaya Group Title

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 Group Title

yes

14. myininaya Group Title

Find the hyp of that triangle using Pythagorean Theorem.

15. alysa111 Group Title

16. myininaya Group Title

You are correct to say that.

17. myininaya Group Title

Can you find the length of the hyp?

18. alysa111 Group Title

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

19. myininaya Group Title

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

20. alysa111 Group Title

29

21. myininaya Group Title

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

22. alysa111 Group Title

29 ?

23. myininaya Group Title

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

24. alysa111 Group Title

20/29

25. myininaya Group Title

don't forget the negative sign

26. alysa111 Group Title

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

27. myininaya Group Title

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 Group Title

|dw:1403475467323:dw|

29. alysa111 Group Title

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 Group Title

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 Group Title

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

32. alysa111 Group Title

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

33. alysa111 Group Title

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

34. myininaya Group Title

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 Group Title

am i right

36. myininaya Group Title

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

37. alysa111 Group Title

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

38. myininaya Group Title

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

39. myininaya Group Title

did f+5 (not need)

40. myininaya Group Title

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

41. alysa111 Group Title

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

42. alysa111 Group Title

its 2 to 8

43. myininaya Group Title

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 Group Title

yes

45. myininaya Group Title

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 Group Title

so f(x) has greater y-intercerpt

47. myininaya Group Title

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

48. alysa111 Group Title

-3

49. myininaya Group Title

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 Group Title

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

51. alysa111 Group Title

2 to 8

52. myininaya Group Title

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 Group Title

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

54. myininaya Group Title

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 Group Title

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

56. myininaya Group Title

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 Group Title

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

58. alysa111 Group Title

Im not sure

59. alysa111 Group Title

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

60. alysa111 Group Title

the same y-intercept

61. myininaya Group Title

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 Group Title

on the x axis

63. myininaya Group Title

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 Group Title

so 0-9 ? 0-4 ?

65. myininaya Group Title

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 Group Title

Yeah im confused now

67. myininaya Group Title

|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 Group Title

functions has the greatest y-intercept

69. alysa111 Group Title

1,1 ?

70. myininaya Group Title

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

71. myininaya Group Title

|dw:1403477214481:dw|

72. alysa111 Group Title

its 0,0 ?

73. myininaya Group Title

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

74. alysa111 Group Title

yes

75. myininaya Group Title

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

76. myininaya Group Title

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

77. myininaya Group Title

what point on g gives us the x is 0?

78. alysa111 Group Title

-5

79. myininaya Group Title

how did you get that?

80. myininaya Group Title

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

81. myininaya Group Title

82. myininaya Group Title

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

83. myininaya Group Title

evaluate 5cos(3*0)-5

84. alysa111 Group Title

5cos(0)-5 5cos

85. myininaya Group Title

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 Group Title

1.41

87. myininaya Group Title

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

88. alysa111 Group Title

1

89. myininaya Group Title

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

90. alysa111 Group Title

0

91. alysa111 Group Title

same y

92. myininaya Group Title

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

93. alysa111 Group Title

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

94. myininaya Group Title

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

95. alysa111 Group Title

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

96. myininaya Group Title

you can use that

97. alysa111 Group Title

|dw:1403477725033:dw|

98. alysa111 Group Title

|dw:1403477767271:dw|

99. myininaya Group Title

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

100. alysa111 Group Title

sin(15) = 0.650

101. alysa111 Group Title

sin*17) = -0.96

102. myininaya Group Title

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

103. alysa111 Group Title

no

104. myininaya Group Title

why not?

105. alysa111 Group Title

well wouldnt you put them in front of sin and cos

106. myininaya Group Title

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

107. myininaya Group Title

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

108. myininaya Group Title

$\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1$ 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 Group Title

now solve for cos(theta)

110. myininaya Group Title

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

111. myininaya Group Title

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

112. myininaya Group Title

and cos(theta) is the objective

113. alysa111 Group Title

225/285

114. alysa111 Group Title

|dw:1403478332917:dw|

115. myininaya Group Title

so what did you do to get that?

116. alysa111 Group Title

found cos of theta

117. myininaya Group Title

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

118. alysa111 Group Title

yes

119. myininaya Group Title

you might want to do that one more time

120. myininaya Group Title

the denominator is right but your numerator is off

121. alysa111 Group Title

8 / 17

122. myininaya Group Title

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

123. alysa111 Group Title

|dw:1403478613956:dw| my only option

124. myininaya Group Title

Agreed

125. alysa111 Group Title

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 Group Title

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

127. alysa111 Group Title

@myininaya