anonymous
  • anonymous
Derive the trigonometric addition formula for sine :
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sin(a+b)=sinacosb+cosasinb\]
anonymous
  • anonymous
also, derive these identities using the addition or subtraction formulas for sine or cosine : \[sinasinb=\frac{ 1 }{ 2 }(\cos(a-b)-\cos(a+b))\]
IrishBoy123
  • 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).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.