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Mathematics 13 Online
OpenStudy (anonymous):

Prove this identity sin(2A)=2sinAcosA.

OpenStudy (anonymous):

Start by considering 2A as A+A ... does that make you think of any particular theorem?

OpenStudy (turingtest):

...of course that makes me want to know how to prove the theorem you are thinking of @Tonks

OpenStudy (anonymous):

sin(a+a)=sinacosb+cosasinb put the value of a and be here.

OpenStudy (anonymous):

We know that sin(2A) = sin(A + A) sin (A + A) = cosA sinA + cosA sinA = 2 sinA cosA QED

OpenStudy (anonymous):

@TuringTest try stacking triangles in the unit circle... or is that for cosine?

OpenStudy (turingtest):

I know it was some geometric proof like that, I don't remember those darn things...

OpenStudy (anonymous):

Is that enough or do you need proof for sin (A + B) = cosA sinB + cosB sinA

OpenStudy (anonymous):

http://youtu.be/BlEtOG4NRsg

OpenStudy (anonymous):

formula you have given is wrong @micahwood50

OpenStudy (turingtest):

it's correct

OpenStudy (anonymous):

For proof of sin (A + B) = cosA sinB + cosB sinA: http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Sine

OpenStudy (turingtest):

yep, that's the one...

OpenStudy (anonymous):

@muhammad9t5 It's the same. Your point is?

OpenStudy (turingtest):

perhaps you are neglecting the fact that multiplication is commutative muhammad?

OpenStudy (anonymous):

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