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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(a+b)=sinacosb+cosasinb\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also, derive these identities using the addition or subtraction formulas for sine or cosine : \[sinasinb=\frac{ 1 }{ 2 }(\cos(ab)\cos(a+b))\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4easiest 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 (ab).

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4if you post some stuff, i'm sure you will get all the help you need :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've never seen that first formula before . I'm even more confused

freckles
 one year ago
Best ResponseYou've already chosen the best response.6you can derive cos(ab)=cos(a)cos(b)+sin(a)sin(b) using the distance formula then you can use cos(ab)=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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are cofunction identities ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is an algebra 2 class and they threw in a trig section

freckles
 one year ago
Best ResponseYou've already chosen the best response.6you are asking to derive the formula sin(a+b)=sin(a)cos(b)+sin(b)cos(a) right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6hmmm maybe you don't remember the name of the identity but you remember that cos(90A)=sin(A) and sin(90A=cos(A)? If not you easily show these by drawing right triangles and comparing the above ratios.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the notes didn't name the identities and I'm sure I learned that . Trigonometry is just really hard for me to understand

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4hi @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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um, that looks really complex. Do you want a picture of my notes to see what I learned ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay it might take a bit cause my computer is really slow

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4well 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

yamyam70
 one year ago
Best ResponseYou've already chosen the best response.1Try searching , sum and difference trigonometric identities, that's the name of the first one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440346974875:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you guys keep doing the same triangle and I haven't seen that :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah , apex doesn't teach that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they probably want you to just memorize and expect to learn.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4@jxaf i'll happily have a read of your notes once i get back, am popping out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk why they would when I'm taking an algebra 2 class and @IrishBoy123 okay (:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your identities are covered in that really old teach yourself trig book beginning from pp 98

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is the same supplementary textbook I used in 6th grade

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you learned trig in the 6th grade ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they teach this early as 6th grade in some parts of Asia

freckles
 one year ago
Best ResponseYou've already chosen the best response.6after 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6also i did notice an error in your notes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sure there is . Apex doesn't make sense when it comes to the note taking

freckles
 one year ago
Best ResponseYou've already chosen the best response.6"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)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6sin^2(theta)+cos^2(theta)=1 is true but cos(a+b) doesn't equal sin^2(theta)+cos^2(theta)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these are the questions

freckles
 one year ago
Best ResponseYou've already chosen the best response.6oh and page 1 ... cos(a+b)=cos(a)cos(b)sin(a)sin(b)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6i guess i'm wrong it is to derive sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6I wonder if you can use the cos difference identity

freckles
 one year ago
Best ResponseYou've already chosen the best response.6\[\cos(ab)=\cos(a)\cos(b)+\sin(a)\sin(b) \\ \sin(a+b)=\cos(\frac{\pi}{2}[a+b])= \cos(\frac{\pi}{2}ab) \\ =\cos([\frac{\pi}{2}a][b])\] then plug into the cos difference identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea lol . I need this packet done by tomorrow and it's stressing me out

freckles
 one year ago
Best ResponseYou've already chosen the best response.6well I don't know what you can use to derive I don't know your tool set for the problem

freckles
 one year ago
Best ResponseYou've already chosen the best response.6but if you can use the cos difference identity then that way I mentioned above is one way to do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the sinacosbcosasinb ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6if you aren't allowed to use that then you have to derive also the identity cos(ab)=cos(a)cos(b)+sin(a)sin(b) and as mentioned earlier one way to do that is using the distance formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm ready to give up :/

freckles
 one year ago
Best ResponseYou've already chosen the best response.6just use this then: \[\cos(xb)=\cos(x)\cos(b)+\sin(x)\sin(b) \\ \sin(a+b)=\cos([\frac{\pi}{2}a]b)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.6replace x with pi/2a and replace b well with b

freckles
 one year ago
Best ResponseYou've already chosen the best response.6don't forget to use the cofunction identities

freckles
 one year ago
Best ResponseYou've already chosen the best response.6and then you will have sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are the cofunction identities

freckles
 one year ago
Best ResponseYou've already chosen the best response.6hmmm... I thought I mentioned two of them earlier...

freckles
 one year ago
Best ResponseYou've already chosen the best response.6cos(pi/2a)=sin(a) sin(pi/2a)=cos(a)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you probably did , I just forgot trying to understand everything else

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still confused :/

freckles
 one year ago
Best ResponseYou've already chosen the best response.6ok have you expanded \[\cos((\frac{\pi}{2}a)b) \text{ yet using the cosine difference identity ?}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.6\[\cos(xb)=\cos(x)\cos(b)+\sin(x)\sin(b)\] I was asking you to replace x with pi/2a and b can just stay b so you haven't that yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it'd be cos(pi/2a)cos(b)+sin(pi/2a)sin(b) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6right now use the cofunction identities

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but you just said to substitute x with that ..

freckles
 one year ago
Best ResponseYou've already chosen the best response.6\[\sin(a+b) \\ =\cos(\frac{\pi}{2}(a+b)) \\ =\cos(\frac{\pi}{2}ab) \\ =\cos([\frac{\pi}{2}a][b]) \\ =\cos(\frac{\pi}{2}a)\cos(b)+\sin(\frac{\pi}{2}a)\sin(b) \] one last step and you are done the last step is to apply cofunction identities

yamyam70
 one year ago
Best ResponseYou've already chosen the best response.1@jxaf You can do it! :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6"but you just said to substitute x with that .." I was saying you did that part right... but you have one last step

freckles
 one year ago
Best ResponseYou've already chosen the best response.6remember I just say above cos(pi/2a)=? and sin(pi/2a)=?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@yamyam70 thank you lol @freckles yeah , I do

freckles
 one year ago
Best ResponseYou've already chosen the best response.6so you have completed the problem right? because you replaced cos(pi/2a) with sin(a) and sin(pi/2a) with cos(a)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the next problem

freckles
 one year ago
Best ResponseYou've already chosen the best response.6try expanding cos(ab) and cos(a+b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the cos(ab) would be cosacosb+sinasinb ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and cos(a+b) would be cosacosbsinasinb ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6what 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)]

freckles
 one year ago
Best ResponseYou've already chosen the best response.6well sin(a)sin(b)+sin(a)sin(b) doesn't equal 0 but cos(a)cos(b)cos(a)cos(b) is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0jk , the subtraction distributes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it'd be 2sinasinb ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6yes then what is 1/2*2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6\[\frac{1}{2}(\cos(ab)\cos(a+b)) \\ \frac{1}{2}(\cos(a)\cos(b)+\sin(a)\sin(b)\cos(a)\cos(b)+\sin(a)\sin(b)) \\ \frac{1}{2}(2 \sin(a)\sin(b)) \\ (\frac{1}{2} \cdot 2) \sin(a)\sin(b)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait why'd you use the half identity thing ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see why . I don't think straight when I'm stressed as you can see

freckles
 one year ago
Best ResponseYou've already chosen the best response.6don'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.6awww yeah i guess life can get in the way of putting aside time for math hope she is doing better

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And plus the thing I have to use to get the notes makes no sense to me at all and she is , thank you

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4@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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you ! @IrishBoy123

freckles
 one year ago
Best ResponseYou've already chosen the best response.6did you want me to show you why the cofunction identities are true @jxaf

freckles
 one year ago
Best ResponseYou've already chosen the best response.6okay this is a right triangle (which I have labeled) dw:1440350795124:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.6the sum of the angles in a triangle equal 180 deg

freckles
 one year ago
Best ResponseYou've already chosen the best response.6I can write B in terms of A

freckles
 one year ago
Best ResponseYou've already chosen the best response.6B+A=90 subtract A on both sides B=90A

freckles
 one year ago
Best ResponseYou've already chosen the best response.6dw:1440350884800:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.6so do you know how to find cos(B) and sin(A) using this right triangle?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos(b) would be a/c sin(a) would be a/c also ?????????????

freckles
 one year ago
Best ResponseYou've already chosen the best response.6and 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=90A so that means cos(90A)=sin(A)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6this is one of the cofunction identities and you can find 5 other cofunction identities using this right triangle

freckles
 one year ago
Best ResponseYou've already chosen the best response.6\[\cos(B)=\sin(A)=\frac{a}{c} \\ \text{ so } \cos(90A)=\sin(A) \\ \sin(B)=\cos(A)=\frac{b}{c} \\ \text{ so } \sin(90A)=\cos(A) \\ \tan(B)=\cot(A)=\frac{b}{a} \\ \text{ so } \tan(90A)=\cot(A) \\ \csc(B)=\sec(A) =\frac{c}{b}\\ \text{ so } \csc(90A)=\sec(A) \\ \] you can also from the triangle see that cot(90A)=tan(A) and sec(90A)=csc(A)

freckles
 one year ago
Best ResponseYou've already chosen the best response.6by 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No it makes more sense with degrees lol . the whole pi stuff gets me confused . I need the unit circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you help me with another problem ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.6i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, thank you . Ima make it a different question
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