sinxcosy = 1/2 (sin(x+y) + cos(x-y))
True or False?

Question

sinxcosy = 1/2 (sin(x+y) + cos(x-y)) 
True or False?

anonymous · Answer

sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

cos(x-y) = cos(x)cos(y) + sin(x)sin(y)

can do you it now ?

anonymous · Answer

er true?

anonymous · Answer

$$\sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y)$$
$$\cos(x-y) = \cos(x) \cos(y) + \sin(x) \sin(y)$$

anonymous · Answer

Can you add now $sin(x+y)$ and $cos(x-y)$ ??

anonymous · Answer

huh?

anonymous · Answer

its (sin(x+y)) + (cos(x-y))   ._.

anonymous · Answer

I said add expansion that we have given there above..

Add the expansion of sin(x+y) and cos(x-y) and tell me what did you get ??

anonymous · Answer

i just told you what "I" can get

anonymous · Answer

using :
sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

cos(x-y) = cos(x)cos(y) + sin(x)sin(y)
your equation becomes
sinxcosy = 1/2 (sin(x)cos(y) + sin(y)cos(x) + cos(x)cos(y) + sin(x)sin(y))

anonymous · Answer

so ....true then

anonymous · Answer

no false
bc thats sin x cos y = sinx cosy 
...wait ..waht?

anonymous · Answer

show me

calculusfunctions · Answer

Wow! @Mpost knows how to show off but is neither elegant in his writing nor can he solve such simple simple questions. It seems like the only thing he does know is how to be disrespectful towards authority figures. Needs serious counselling. Almost feel sorry for this arrogant sorry excuse for a human.

anonymous · Answer

It is wrong...

anonymous · Answer

$$\rm 2\sin(x)\cos(y) \ne \sin(x)\cos(y) + \sin(y)\cos(x) + \sin(x)\sin(y) + \cos(x)\cos(y)$$