sin(a) _______ = 1/2 [sin(a+b) + sin(a-b)]

a) sin(b)
b) cos(a)
c) cos(b)
d) sin(a)

Question

sin(a) _______ = 1/2 [sin(a+b) + sin(a-b)]

a) sin(b)
b) cos(a)
c) cos(b)
d) sin(a)

anonymous · Answer

@ganeshie8 @hartnn @JackJordan

hartnn · Answer

can you expand 
sin (a+b)
and 
sin(a-b) 
?

know the formulas?

anonymous · Answer

no..haha

hartnn · Answer

ok, here you go!
$\Large \sin (a+b) = \sin a \cos b +\cos a \sin b$
$\Large \sin (a-b) = \sin a \cos b -\cos a \sin b$

just plug these in, on the right side.

hartnn · Answer

like add them and see what u get...

hartnn · Answer

anything getting cancelld out ?

anonymous · Answer

idk why but i couldn't get anything from it

hartnn · Answer

did you add? 

$\Large \sin (a+b) + \Large \sin (a-b) \ \Large = \sin a \cos b +\cos a \sin b+  \sin a \cos b -\cos a \sin b$

now you see any term getting cancelled out ?

anonymous · Answer

So when you add them, the latter terms cancel.
You will be left with two lots of sin(a)cos(b)
Now dividing by two, what do we get?

anonymous · Answer

so is it cos (b)?

hartnn · Answer

jack buddy! let him/her do something :P

hartnn · Answer

yep,
sin a cos b 
so yeah

anonymous · Answer

hrtnn why'd you give him the answer? He thought it was cos b

hartnn · Answer

we all know who gave out the answer, so cut it out lol