## brazellnicole25 Group Title Can anyone help me with a trig problem. I need to turn y1=-4sin(t/2)+3cos(t/2) into the form y2=Asin(Bt+C) one year ago one year ago

1. brazellnicole25 Group Title

I would like to know how to obtain the answer step by step. I don't want to just know the answer. I would like to understand how it was obtained.

2. asnaseer Group Title

do you know how to expand sin(Bt+C)?

3. asnaseer Group Title

using the angle sum formulae?

4. asnaseer Group Title

@brazellnicole25 I can only help you if you respond to my questions.

5. brazellnicole25 Group Title

I will repond. i know i'm suppost to use the transformation identity to make the first equation the other one. and c is suppost to be the smallest positive ange.

6. asnaseer Group Title

first expand sin(Bt+C) using the angle sum formulae - do you know how to do that?

7. asnaseer Group Title

i.e do you know how to express sin(x+y) in terms of sin(x) and cos(x)?

8. asnaseer Group Title

and sin(y) and cos(y)

9. brazellnicole25 Group Title

yes i have done that before

10. asnaseer Group Title

ok, good, so please first write sin(Bt+C) in its expanded form

11. brazellnicole25 Group Title

okay is it sin(a+-B)=sinacosb+-cos(a)sin(b)

12. asnaseer Group Title

yes - that is the correct formulae to use here

13. brazellnicole25 Group Title

okay.

14. asnaseer Group Title

so sin(Bt+C)=?

15. brazellnicole25 Group Title

does it equal the equation that I wrote before? I am stuck here for some reason

16. asnaseer Group Title

ok, let me take you through this one step-by-step and then hopefully you will be able to do others that are similar to this one

17. brazellnicole25 Group Title

This is where I'm having my block for this problem for some reason. I've used the above before but I dont' know why i'm having so much trouble. I'm sorry I'm really trying

18. asnaseer Group Title

no problem - lets work through it together. we start by first expanding sin(Bt+C) to get: sin(Bt+C) = sin(Bt)cos(C) + cos(Bt)sin(C) agreed?

19. brazellnicole25 Group Title

yes

20. asnaseer Group Title

good, next we multiply this by A to get: Asin(Bt+C) = Asin(Bt)cos(C) + Acos(Bt)sin(C) agreed?

21. brazellnicole25 Group Title

yes because we have to do the same to all of them (distribute) okay i'm following

22. asnaseer Group Title

great, now lets rearrange this to get it into a similar form to the original equation. we therefore get: Asin(Bt+C) = Acos(C) * sin(Bt) + Asin(C) * cos(Bt)

23. asnaseer Group Title

compare this to the original equation which is: -4*sin(t/2) + 3*cos(t/2)

24. asnaseer Group Title

can you now infer what the value of B must be?

25. asnaseer Group Title

compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)

26. brazellnicole25 Group Title

b must be t is that right?

27. brazellnicole25 Group Title

t/2

28. asnaseer Group Title

not quite - if you compare these terms then we see that: Bt = t/2 therefore B = ?

29. asnaseer Group Title

divide both sides by 't'

30. sirm3d Group Title

if you write $$t/2$$ as $$(1/2)t$$ you should be able to infer the value of $$B$$

31. brazellnicole25 Group Title

2

32. asnaseer Group Title

we have B times 't' is equal to (1/2) times 't' so what must B equal?

33. brazellnicole25 Group Title

1/2

34. asnaseer Group Title

perfect!

35. asnaseer Group Title

ok, so now we can use this to get: -4*sin(t/2) + 3*cos(t/2) = Acos(C) * sin(t/2) + Asin(C) * cos(t/2)

36. asnaseer Group Title

now compare the terms that are multiplying the sin(t/2) terms and the cos(t/2) terms on both sides of the equals sign

37. asnaseer Group Title

can you see that this means that: -4 = Acos(C) 3 = Asin(C)

38. brazellnicole25 Group Title

yes I can see that. The right side of the equal sign it just streched out. yeah that makes sense

39. asnaseer Group Title

good, next are you aware of the identity:$\sin^2(x)+\cos^2(x)=1$

40. brazellnicole25 Group Title

yes and i know that can me manipulated in a couple of ways. that is one of the pythagorean identies

41. asnaseer Group Title

correct, so if we use this identity, we can square both sides of: -4 = Acos(C) 3 = Asin(C) to get:$16 = A^2\cos^2(C)$$9=A^2\sin^2(C)$and then add them together to get:$16+9=A^2\cos^2C)+A^2\sin^2(C)$

42. asnaseer Group Title

therefore:$25=A^2(\cos^2(C)+\sin^2(C))=A^2(1)=A^2$does that make sense?

43. brazellnicole25 Group Title

yes it does your just combining the two to make one equation.

44. asnaseer Group Title

correct - now we take square roots of both sides to get:$A=\sqrt{25}=5$so now we have found A=5 and B=1/2 which just leaves C to find

45. brazellnicole25 Group Title

yeah that's right

46. asnaseer Group Title

we can find C using either one of the two earlier equations we got to: -4 = Acos(C) 3 = Asin(C) lets use the second one to get: 3 = Asin(C) therefore: sin(C) = 3/A = 3/5 therefore: C = arcsin(3/5) = 36.87 degrees (approximately)

47. asnaseer Group Title

make sense?

48. brazellnicole25 Group Title

yes that does. you solved for c.

49. brazellnicole25 Group Title

wow i made that harder than it had to be for sure.

50. asnaseer Group Title

so we can finally write: y2 = 5sin(t/2 + 36.87) or, if you prefer: y2 = 5sin(t/2 + arcsin(3/5))

51. asnaseer Group Title

I hope you understood all the steps

52. brazellnicole25 Group Title

yes I did. Thank you so much. I really appriciate it. I was looking back at some of these problems that I missed on a previous test and it was bothering me so I wanted to figure them out. Just so I would know. I don't want to be behind in my calculus class next semester.

53. sirm3d Group Title

hmmm, i think C should be 180-36.87.

54. brazellnicole25 Group Title

I thank you so much for your time.

55. asnaseer Group Title

yw :) @sirm3d - let me check what you are saying...

56. asnaseer Group Title

both are correct @sirm3d since:$\sin(x)=\sin(180-x)$

57. asnaseer Group Title

actually no

58. sirm3d Group Title

$y_2=5 \sin (\frac{1}{2}t +143.17)$ because $$-4=5 \cos C$$ abd there's no way to reconcile the sign if C is in quadrant I

59. asnaseer Group Title

I believe you are right @sirm3d since: -4 = Acos(C)

60. asnaseer Group Title

which implies cos(C) must be negative thank you for correcting me. :)

61. asnaseer Group Title

@brazellnicole25 - to avoid this type of mistake, what we should have done is gone from these two equations: -4 = Acos(C) 3 = Asin(C) to this one by diving the two: tan(C) = -3/4 that guarantees that we get the correct quadrant.

62. sirm3d Group Title

dividing the two equations should yield an equation in tangent.$\frac{3}{-4}=\frac{\sin C}{\cos C}=\tan C$

63. asnaseer Group Title

exactly

64. asnaseer Group Title

I hope that is also clear to you @brazellnicole25

65. brazellnicole25 Group Title

okay yeah so you combined the two right? and then solved for C. is that correct

66. asnaseer Group Title

yes because the angle C has to satisfy BOTH equations: -4 = Acos(C) 3 = Asin(C)

67. asnaseer Group Title

so we just divided the second one by the first one to get:$\frac{3}{-4}=\frac{A\sin(C)}{A\cos(C)}$which lead to: tan(C) = -3/4

68. asnaseer Group Title

so the correct solution is: y2 = 5sin(t/2 - arctan(3/4))

69. asnaseer Group Title

sorry: y2 = 5sin(t/2 + arctan(-3/4))

70. brazellnicole25 Group Title

71. asnaseer Group Title

if its related to this question then sure

72. brazellnicole25 Group Title

y1 and y2 would be the same equations right? I mean if I was to graph them?

73. asnaseer Group Title

yes

74. brazellnicole25 Group Title

75. brazellnicole25 Group Title

okay thank you so much. I really appriciate it. and your time.

76. brazellnicole25 Group Title

I haven't ever used this site is there anything I can do to help you get a beter rating or anything? I don't know alot about this site.

77. asnaseer Group Title

I am glad I was able to help - I do this because I enjoy teaching those who are willing to learn, not for any rating. But, in general, the usual workflow is: 1. You post a question. 2. One or more people come to your question to help you out. 3. You select the one that best answered your question by clicking on the blue Best Response button next to any one of their replies. 4. You close the question by clicking on the Close button then the whole process starts again.

78. brazellnicole25 Group Title

79. asnaseer Group Title

BTW: You can use wolframalpha to see the graphs as follows: y1: http://www.wolframalpha.com/input/?i=y%3D-4sin%28t%2F2%29%2B3cos%28t%2F2%29 y2: http://www.wolframalpha.com/input/?i=y+%3D+5sin%28t%2F2+%2B+2.4981%29

80. brazellnicole25 Group Title

would the amplitude of this be 5 , the period be 2pie (symbol i mean) and the phanse shift be .... sorry i'm thinking about that for a min.

81. brazellnicole25 Group Title

okay thanks for showing me the graphs. I got the same graphs on my calculator at on these you gave me. Thank you so much for that.

82. asnaseer Group Title

83. asnaseer Group Title

in general:$Y=A\sin(wt+\phi)$means: A=amplitude $$w=2*\pi/P$$ P= period $$\phi$$=phase shift

84. brazellnicole25 Group Title

okay sorry to be so full of questions. I'm just rying to fully understand it. I hope i'm not being to much trouble. I really do appriciate it though.

85. asnaseer Group Title

not at all - I am glad you find the subject interesting :)

86. brazellnicole25 Group Title

and apply things I've already learned in class. That's why i'm asking so much.

87. brazellnicole25 Group Title

I do love math and I like to do it more when I understand it more.

88. asnaseer Group Title

it is the same with any subject - if you really enjoy a subject then you will naturally get better at it.

89. asnaseer Group Title

a thirst for knowledge is always a good sign - I am sure you will do really well and achieve all that you want to achieve. :)

90. brazellnicole25 Group Title

Thank you so much. I really appriciate it. Your very smart though. do you teach math somewhere? Not that I mean to be to nosey. Your good at it though.

91. asnaseer Group Title

again, you are more than welcome. And no I am not an actual maths teacher, I work as a fulltime software engineer but always enjoyed maths since high school, so I try and keep up with the subject by /trying/ to teach others. :)

92. asnaseer Group Title

I have also learnt a LOT of things from other on this site - @sirm3d being one of them. :)

93. asnaseer Group Title

*others

94. sirm3d Group Title

^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.

95. asnaseer Group Title

what age group do you teach @sirm3d ?

96. brazellnicole25 Group Title

well you are good I mean great at it. I like how you take the time to help others and make sure the person your helping will interact with you. if they really want to learn then they would have no problem answering you.

97. sirm3d Group Title

college math. up to engineering mathematics, @asnaseer

98. brazellnicole25 Group Title

I like sirm3d how you helped us to.

99. asnaseer Group Title

thats good to know @sirm3d - I'll keep you in mind if I run into any problems that I get stuck in. ok folks - I have to go now as it is 1:40am where I am and I need to catch some Zzzz's.... :)

100. sirm3d Group Title

i love math, and i find joy seeing students learn and appreciate math.

101. brazellnicole25 Group Title

you were very polite in giving your input. I have gone to get help before and some people are just so rude. but your so nice about helping.

102. brazellnicole25 Group Title

okay asnaseer I hope you have a good night. thanks alot.

103. asnaseer Group Title

night night... :)