Derive the trigonometric addition formula for sine :

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Derive the trigonometric addition formula for sine :

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

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if you post some stuff, i'm sure you will get all the help you need :p
I've never seen that first formula before . I'm even more confused
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)
what are cofunction identities ?
This is an algebra 2 class and they threw in a trig section
you are asking to derive the formula sin(a+b)=sin(a)cos(b)+sin(b)cos(a) right?
yes
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.
the notes didn't name the identities and I'm sure I learned that . Trigonometry is just really hard for me to understand
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
Um, that looks really complex. Do you want a picture of my notes to see what I learned ?
yeah, go for it!
okay it might take a bit cause my computer is really slow
take your time.
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|
Try searching , sum and difference trigonometric identities, that's the name of the first one
|dw:1440346974875:dw|
you guys keep doing the same triangle and I haven't seen that :/
Yeah , apex doesn't teach that
they probably want you to just memorize and expect to learn.
@jxaf i'll happily have a read of your notes once i get back, am popping out
https://archive.org/details/TeachYourselfTrigonometry_94
idk why they would when I'm taking an algebra 2 class and @IrishBoy123 okay (:
your identities are covered in that really old teach yourself trig book beginning from pp 98
it is the same supplementary textbook I used in 6th grade
you learned trig in the 6th grade ?
they teach this early as 6th grade in some parts of Asia
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?
hold on
also i did notice an error in your notes
I'm sure there is . Apex doesn't make sense when it comes to the note taking
"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)
sin^2(theta)+cos^2(theta)=1 is true but cos(a+b) doesn't equal sin^2(theta)+cos^2(theta)
these are the questions
1 Attachment
oh and page 1 ... cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
i guess i'm wrong it is to derive sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
I wonder if you can use the cos difference identity
\[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \\ \sin(a+b)=\cos(\frac{\pi}{2}-[a+b])= \cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b])\] then plug into the cos difference identity
I have no idea lol . I need this packet done by tomorrow and it's stressing me out
well I don't know what you can use to derive I don't know your tool set for the problem
but if you can use the cos difference identity then that way I mentioned above is one way to do it
the sinacosb-cosasinb ?
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
I'm ready to give up :/
just use this then: \[\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b) \\ \sin(a+b)=\cos([\frac{\pi}{2}-a]-b)\]
replace x with pi/2-a and replace b well with b
where's x ?
jk i see it
is that all or ?
don't forget to use the cofunction identities
and then you will have sin(a+b)=sin(a)cos(b)+sin(b)cos(a)
what are the cofunction identities
hmmm... I thought I mentioned two of them earlier...
cos(pi/2-a)=sin(a) sin(pi/2-a)=cos(a)
you probably did , I just forgot trying to understand everything else
I'm still confused :/
ok have you expanded \[\cos((\frac{\pi}{2}-a)-b) \text{ yet using the cosine difference identity ?}\]
no ..
\[\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b)\] I was asking you to replace x with pi/2-a and b can just stay b so you haven't that yet?
so it'd be cos(pi/2-a)cos(b)+sin(pi/2-a)sin(b) ?
right now use the cofunction identities
but you just said to substitute x with that ..
\[\sin(a+b) \\ =\cos(\frac{\pi}{2}-(a+b)) \\ =\cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b]) \\ =\cos(\frac{\pi}{2}-a)\cos(b)+\sin(\frac{\pi}{2}-a)\sin(b) \] one last step and you are done the last step is to apply cofunction identities
@jxaf You can do it! :)
"but you just said to substitute x with that .." I was saying you did that part right... but you have one last step
remember I just say above cos(pi/2-a)=? and sin(pi/2-a)=?
@yamyam70 thank you lol @freckles yeah , I do
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)?
yes
what about the next problem
try expanding cos(a-b) and cos(a+b)
the cos(a-b) would be cosacosb+sinasinb ?
and cos(a+b) would be cosacosb-sinasinb ?
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)]
it becomes 0 ?
well sin(a)sin(b)+sin(a)sin(b) doesn't equal 0 but cos(a)cos(b)-cos(a)cos(b) is 0
jk , the subtraction distributes
so it'd be 2sinasinb ?
yes then what is 1/2*2?
1
\[\frac{1}{2}(\cos(a-b)-\cos(a+b)) \\ \frac{1}{2}(\cos(a)\cos(b)+\sin(a)\sin(b)-\cos(a)\cos(b)+\sin(a)\sin(b)) \\ \frac{1}{2}(2 \sin(a)\sin(b)) \\ (\frac{1}{2} \cdot 2) \sin(a)\sin(b)\]
and yes 1/2*2=1
wait why'd you use the half identity thing ?
jk
I see why . I don't think straight when I'm stressed as you can see
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
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
awww yeah i guess life can get in the way of putting aside time for math hope she is doing better
And plus the thing I have to use to get the notes makes no sense to me at all and she is , thank you
@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!
Thank you ! @IrishBoy123
did you want me to show you why the cofunction identities are true @jxaf
it isn't too long
yeah
okay this is a right triangle (which I have labeled) |dw:1440350795124:dw|
the sum of the angles in a triangle equal 180 deg
B+A+90=180 B+A=90
I can write B in terms of A
B+A=90 subtract A on both sides B=90-A
|dw:1440350884800:dw|
so do you know how to find cos(B) and sin(A) using this right triangle?
cos(b) would be a/c sin(a) would be a/c also ?????????????
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)
this is one of the cofunction identities and you can find 5 other cofunction identities using this right triangle
\[\cos(B)=\sin(A)=\frac{a}{c} \\ \text{ so } \cos(90-A)=\sin(A) \\ \sin(B)=\cos(A)=\frac{b}{c} \\ \text{ so } \sin(90-A)=\cos(A) \\ \tan(B)=\cot(A)=\frac{b}{a} \\ \text{ so } \tan(90-A)=\cot(A) \\ \csc(B)=\sec(A) =\frac{c}{b}\\ \text{ so } \csc(90-A)=\sec(A) \\ \] you can also from the triangle see that cot(90-A)=tan(A) and sec(90-A)=csc(A)
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
No it makes more sense with degrees lol . the whole pi stuff gets me confused . I need the unit circle
could you help me with another problem ?
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
Okay, thank you . Ima make it a different question
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