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

A. sin a

B. sin b

C. cos a

D. cos b

***my answer; A. sin a

is that right? @terenzreignz :)

Question

_____ cos b = 1/2[sin(a+b) + sin(a-b)]

A.  sin a 

B.  sin b

C.  cos a

D.  cos b

***my answer; A.  sin a

is that right? @terenzreignz :)

terenzreignz · Answer

hmm... I hate these identities.
I don't recall when I ever had to use them...

terenzreignz · Answer

Let's see...
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

thus
sin(a+b) + sin(a-b) = 2sin(a)cos(b)
halving that, you get
sin(a)cos(b)

how about that, eh? :D

anonymous · Answer

ooh!! yay!! :D 

thanks!! :D haha woohoo! :D