Derive the trigonometric addition formula for sine :
\[\sin(a+b)=sinacosb+cosasinb\]
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).
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|
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