Derive the trigonometric addition formula for sine :

- anonymous

Derive the trigonometric addition formula for sine :

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- anonymous

\[\sin(a+b)=sinacosb+cosasinb\]

- 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

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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## More answers

- IrishBoy123

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

- anonymous

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

- 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

what are cofunction identities ?

- anonymous

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

- freckles

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

- anonymous

yes

- 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

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

- 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

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

- IrishBoy123

yeah, go for it!

- anonymous

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

- IrishBoy123

take your time.

- 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

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

- anonymous

these are my notes

##### 4 Attachments

- anonymous

|dw:1440346974875:dw|

- anonymous

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

- anonymous

Yeah , apex doesn't teach that

- anonymous

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

- IrishBoy123

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

- anonymous

https://archive.org/details/TeachYourselfTrigonometry_94

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you learned trig in the 6th grade ?

- anonymous

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

- 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

hold on

- freckles

also i did notice an error in your notes

- anonymous

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

- 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

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

- anonymous

these are the questions

##### 1 Attachment

- freckles

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

- freckles

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

- freckles

I wonder if you can use the cos difference identity

- 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

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

- freckles

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

- freckles

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

- anonymous

the sinacosb-cosasinb ?

- 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

I'm ready to give up :/

- 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

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

- anonymous

where's x ?

- anonymous

jk i see it

- anonymous

is that all or ?

- freckles

don't forget to use the cofunction identities

- freckles

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

- anonymous

what are the cofunction identities

- freckles

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

- freckles

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

- anonymous

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

- anonymous

I'm still confused :/

- freckles

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

- anonymous

no ..

- 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

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

- freckles

right now use the cofunction identities

- anonymous

but you just said to substitute x with that ..

- 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

@jxaf You can do it! :)

- 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

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

- anonymous

@yamyam70 thank you lol
@freckles yeah , I do

- 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

yes

- anonymous

what about the next problem

- freckles

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

- anonymous

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

- anonymous

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

- 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

it becomes 0 ?

- 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

jk , the subtraction distributes

- anonymous

so it'd be 2sinasinb ?

- freckles

yes then what is 1/2*2?

- anonymous

1

- 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

and yes 1/2*2=1

- anonymous

wait why'd you use the half identity thing ?

- anonymous

jk

- anonymous

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

- 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

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

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

- 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

@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

Thank you ! @IrishBoy123

- freckles

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

- freckles

it isn't too long

- anonymous

yeah

- freckles

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

- freckles

the sum of the angles in a triangle equal 180 deg

- freckles

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

- freckles

I can write B in terms of A

- freckles

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

- freckles

|dw:1440350884800:dw|

- freckles

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

- anonymous

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

- 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

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

- 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

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

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

- anonymous

could you help me with another problem ?

- 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

Okay, thank you . Ima make it a different question

- anonymous

*

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