## anonymous 4 years ago 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)

1. anonymous

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

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

3. asnaseer

using the angle sum formulae?

4. asnaseer

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

5. anonymous

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

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

7. asnaseer

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

8. asnaseer

and sin(y) and cos(y)

9. anonymous

yes i have done that before

10. asnaseer

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

11. anonymous

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

12. asnaseer

yes - that is the correct formulae to use here

13. anonymous

okay.

14. asnaseer

so sin(Bt+C)=?

15. anonymous

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

16. asnaseer

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. anonymous

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

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. anonymous

yes

20. asnaseer

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

21. anonymous

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

22. asnaseer

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

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

24. asnaseer

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

25. asnaseer

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

26. anonymous

b must be t is that right?

27. anonymous

t/2

28. asnaseer

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

29. asnaseer

divide both sides by 't'

30. sirm3d

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

31. anonymous

2

32. asnaseer

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

33. anonymous

1/2

34. asnaseer

perfect!

35. asnaseer

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

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

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

38. anonymous

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

39. asnaseer

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

40. anonymous

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

41. asnaseer

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

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

43. anonymous

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

44. asnaseer

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. anonymous

yeah that's right

46. asnaseer

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

make sense?

48. anonymous

yes that does. you solved for c.

49. anonymous

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

50. asnaseer

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

51. asnaseer

I hope you understood all the steps

52. anonymous

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

hmmm, i think C should be 180-36.87.

54. anonymous

I thank you so much for your time.

55. asnaseer

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

56. asnaseer

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

57. asnaseer

actually no

58. sirm3d

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

59. asnaseer

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

60. asnaseer

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

61. asnaseer

@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

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

63. asnaseer

exactly

64. asnaseer

I hope that is also clear to you @brazellnicole25

65. anonymous

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

66. asnaseer

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

67. asnaseer

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

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

69. asnaseer

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

70. anonymous

71. asnaseer

if its related to this question then sure

72. anonymous

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

73. asnaseer

yes

74. anonymous

75. anonymous

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

76. anonymous

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

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. anonymous

79. asnaseer

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. anonymous

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. anonymous

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

83. asnaseer

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

84. anonymous

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

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

86. anonymous

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

87. anonymous

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

88. asnaseer

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

89. asnaseer

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. anonymous

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

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

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

93. asnaseer

*others

94. sirm3d

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

95. asnaseer

what age group do you teach @sirm3d ?

96. anonymous

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

college math. up to engineering mathematics, @asnaseer

98. anonymous

I like sirm3d how you helped us to.

99. asnaseer

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

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

101. anonymous

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. anonymous

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

103. asnaseer

night night... :)