## anonymous one year ago Derive the trigonometric addition formula for sine :

1. anonymous

$\sin(a+b)=sinacosb+cosasinb$

2. anonymous

also, derive these identities using the addition or subtraction formulas for sine or cosine : $sinasinb=\frac{ 1 }{ 2 }(\cos(a-b)-\cos(a+b))$

3. IrishBoy123

easiest i know is: $$e^{i \ a} = (cos a + i \ sin a)$$ $$e^{i \ b} = (cosb + i \ sin b)$$ $$e^{i \ a} . e^{i \ b} = e^{i \ (a+b)} =$$ then multiply out and equate the real and complex bits for the second bit, just churn the formula. eg expand the RHS using the cos (a+b) and cos (a-b).

4. IrishBoy123

if you post some stuff, i'm sure you will get all the help you need :p

5. anonymous

I've never seen that first formula before . I'm even more confused

6. freckles

you can derive cos(a-b)=cos(a)cos(b)+sin(a)sin(b) using the distance formula then you can use cos(a-b)=cos(a)cos(b)+sin(a)sin(b) along with cofunction identities to show sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

7. anonymous

what are cofunction identities ?

8. anonymous

This is an algebra 2 class and they threw in a trig section

9. freckles

you are asking to derive the formula sin(a+b)=sin(a)cos(b)+sin(b)cos(a) right?

10. anonymous

yes

11. freckles

hmmm maybe you don't remember the name of the identity but you remember that cos(90-A)=sin(A) and sin(90-A=cos(A)? If not you easily show these by drawing right triangles and comparing the above ratios.

12. anonymous

the notes didn't name the identities and I'm sure I learned that . Trigonometry is just really hard for me to understand

13. IrishBoy123

hi @jxaf sorry for the confusion, i should have asked first what you knew. do you think you could follow this proof, with some guidance. https://gyazo.com/be936132eb2a67f24dbd6841f0cf94a3

14. anonymous

Um, that looks really complex. Do you want a picture of my notes to see what I learned ?

15. IrishBoy123

yeah, go for it!

16. anonymous

okay it might take a bit cause my computer is really slow

17. IrishBoy123

18. IrishBoy123

well i've drawn the shape for the proof anyways just in case it comes in useful. i think it's more straightforward that it might appear. |dw:1440346130714:dw|

19. yamyam70

Try searching , sum and difference trigonometric identities, that's the name of the first one

20. anonymous

these are my notes

21. anonymous

|dw:1440346974875:dw|

22. anonymous

you guys keep doing the same triangle and I haven't seen that :/

23. anonymous

Yeah , apex doesn't teach that

24. anonymous

they probably want you to just memorize and expect to learn.

25. IrishBoy123

@jxaf i'll happily have a read of your notes once i get back, am popping out

26. anonymous
27. anonymous

idk why they would when I'm taking an algebra 2 class and @IrishBoy123 okay (:

28. anonymous

your identities are covered in that really old teach yourself trig book beginning from pp 98

29. anonymous

it is the same supplementary textbook I used in 6th grade

30. anonymous

you learned trig in the 6th grade ?

31. anonymous

they teach this early as 6th grade in some parts of Asia

32. freckles

after seeing your notes...I'm unsure the question is to actually derive the sine sum formula you know derive this formula: sin(a+b)=sin(a)cos(b)+sin(b)cos(a) are you sure it says derive that formula?

33. anonymous

hold on

34. freckles

also i did notice an error in your notes

35. anonymous

I'm sure there is . Apex doesn't make sense when it comes to the note taking

36. freckles

"page 3: a. The trigonometric sum identity for sine is san(a+b)=sin(a)cos(-b)+cos(a)sin(-b) ...." That should be sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

37. freckles

sin^2(theta)+cos^2(theta)=1 is true but cos(a+b) doesn't equal sin^2(theta)+cos^2(theta)

38. anonymous

these are the questions

39. freckles

oh and page 1 ... cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

40. freckles

i guess i'm wrong it is to derive sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

41. freckles

I wonder if you can use the cos difference identity

42. freckles

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \\ \sin(a+b)=\cos(\frac{\pi}{2}-[a+b])= \cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b])$ then plug into the cos difference identity

43. anonymous

I have no idea lol . I need this packet done by tomorrow and it's stressing me out

44. freckles

well I don't know what you can use to derive I don't know your tool set for the problem

45. freckles

but if you can use the cos difference identity then that way I mentioned above is one way to do it

46. anonymous

the sinacosb-cosasinb ?

47. freckles

if you aren't allowed to use that then you have to derive also the identity cos(a-b)=cos(a)cos(b)+sin(a)sin(b) and as mentioned earlier one way to do that is using the distance formula

48. anonymous

I'm ready to give up :/

49. freckles

just use this then: $\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b) \\ \sin(a+b)=\cos([\frac{\pi}{2}-a]-b)$

50. freckles

replace x with pi/2-a and replace b well with b

51. anonymous

where's x ?

52. anonymous

jk i see it

53. anonymous

is that all or ?

54. freckles

don't forget to use the cofunction identities

55. freckles

and then you will have sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

56. anonymous

what are the cofunction identities

57. freckles

hmmm... I thought I mentioned two of them earlier...

58. freckles

cos(pi/2-a)=sin(a) sin(pi/2-a)=cos(a)

59. anonymous

you probably did , I just forgot trying to understand everything else

60. anonymous

I'm still confused :/

61. freckles

ok have you expanded $\cos((\frac{\pi}{2}-a)-b) \text{ yet using the cosine difference identity ?}$

62. anonymous

no ..

63. freckles

$\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b)$ I was asking you to replace x with pi/2-a and b can just stay b so you haven't that yet?

64. anonymous

so it'd be cos(pi/2-a)cos(b)+sin(pi/2-a)sin(b) ?

65. freckles

right now use the cofunction identities

66. anonymous

but you just said to substitute x with that ..

67. freckles

$\sin(a+b) \\ =\cos(\frac{\pi}{2}-(a+b)) \\ =\cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b]) \\ =\cos(\frac{\pi}{2}-a)\cos(b)+\sin(\frac{\pi}{2}-a)\sin(b)$ one last step and you are done the last step is to apply cofunction identities

68. yamyam70

@jxaf You can do it! :)

69. freckles

"but you just said to substitute x with that .." I was saying you did that part right... but you have one last step

70. freckles

remember I just say above cos(pi/2-a)=? and sin(pi/2-a)=?

71. anonymous

@yamyam70 thank you lol @freckles yeah , I do

72. freckles

so you have completed the problem right? because you replaced cos(pi/2-a) with sin(a) and sin(pi/2-a) with cos(a)?

73. anonymous

yes

74. anonymous

75. freckles

try expanding cos(a-b) and cos(a+b)

76. anonymous

the cos(a-b) would be cosacosb+sinasinb ?

77. anonymous

and cos(a+b) would be cosacosb-sinasinb ?

78. freckles

what happens when you subtract that last from the first one you wrote that is can you simplify: [cos(a)cos(b)+sin(a)sin(b)]-[cos(a)cos(b)-sin(a)sin(b)]

79. anonymous

it becomes 0 ?

80. freckles

well sin(a)sin(b)+sin(a)sin(b) doesn't equal 0 but cos(a)cos(b)-cos(a)cos(b) is 0

81. anonymous

jk , the subtraction distributes

82. anonymous

so it'd be 2sinasinb ?

83. freckles

yes then what is 1/2*2?

84. anonymous

1

85. freckles

$\frac{1}{2}(\cos(a-b)-\cos(a+b)) \\ \frac{1}{2}(\cos(a)\cos(b)+\sin(a)\sin(b)-\cos(a)\cos(b)+\sin(a)\sin(b)) \\ \frac{1}{2}(2 \sin(a)\sin(b)) \\ (\frac{1}{2} \cdot 2) \sin(a)\sin(b)$

86. freckles

and yes 1/2*2=1

87. anonymous

wait why'd you use the half identity thing ?

88. anonymous

jk

89. anonymous

I see why . I don't think straight when I'm stressed as you can see

90. freckles

don't be stressed take a deep breath and enjoy math you should always put aside enough time to think about math so I can see why you are stressed if you have put yourself so close to a deadline (if you have this time) just don't do that again and you might enjoy math

91. anonymous

I like math , but I've been introduced to trigonometry since geometry and I have never been able to get it and the packet I'm doing was supposed to be due on Friday and I didn't get it done because my mom had major surgery and I was helping her and I just didn't have time to do it but I got an extension

92. freckles

awww yeah i guess life can get in the way of putting aside time for math hope she is doing better

93. anonymous

And plus the thing I have to use to get the notes makes no sense to me at all and she is , thank you

94. IrishBoy123

@jxaf awesome job, so go back and look again at P1, P3a, P8b & P8d in your worksheet. there may be others but those are signs that you are cutting corners and doing things whilst stressed..... well done!

95. anonymous

Thank you ! @IrishBoy123

96. freckles

did you want me to show you why the cofunction identities are true @jxaf

97. freckles

it isn't too long

98. anonymous

yeah

99. freckles

okay this is a right triangle (which I have labeled) |dw:1440350795124:dw|

100. freckles

the sum of the angles in a triangle equal 180 deg

101. freckles

B+A+90=180 B+A=90

102. freckles

I can write B in terms of A

103. freckles

B+A=90 subtract A on both sides B=90-A

104. freckles

|dw:1440350884800:dw|

105. freckles

so do you know how to find cos(B) and sin(A) using this right triangle?

106. anonymous

cos(b) would be a/c sin(a) would be a/c also ?????????????

107. freckles

and that should be cos(B) and sin(A) respectively right guess what ? that makes cos(B)=sin(A) since they are both equal to a/c but B=90-A so that means cos(90-A)=sin(A)

108. freckles

this is one of the cofunction identities and you can find 5 other cofunction identities using this right triangle

109. freckles

$\cos(B)=\sin(A)=\frac{a}{c} \\ \text{ so } \cos(90-A)=\sin(A) \\ \sin(B)=\cos(A)=\frac{b}{c} \\ \text{ so } \sin(90-A)=\cos(A) \\ \tan(B)=\cot(A)=\frac{b}{a} \\ \text{ so } \tan(90-A)=\cot(A) \\ \csc(B)=\sec(A) =\frac{c}{b}\\ \text{ so } \csc(90-A)=\sec(A) \\$ you can also from the triangle see that cot(90-A)=tan(A) and sec(90-A)=csc(A)

110. freckles

by the way all of the above measurements I used were in degrees I hope that was obvious by me choose 90 instead of pi/2

111. anonymous

No it makes more sense with degrees lol . the whole pi stuff gets me confused . I need the unit circle

112. anonymous

could you help me with another problem ?

113. freckles

i will be back in 5 minutes and if you still need help on it you know if someone else hasn't helped you yet i will try to

114. anonymous

Okay, thank you . Ima make it a different question

115. anonymous

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