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

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

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

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.

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

these are my notes

|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

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

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

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

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 ?

I'm ready to give up :/

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 :/

no ..

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 ..

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

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?

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

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

Thank you ! @IrishBoy123

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 ?????????????

could you help me with another problem ?

Okay, thank you . Ima make it a different question