Question

prove that

sinx + siny =2sin(x+y/2)cos(x+y/2)

Answer 1

use identities

Answer 2

Focus first on sin (x+y). There's a trig identity for that. What is that identity? Similarly, there's an identity for cos (x+y). What does it say?

After applying these identities to expand sin (x+y/2) and cos (x+y/2), you'll end up with sin (y/2) and cos (y/2). There are "half angle" formulas that apply here.

Looks like you have a lot of writing to do here.

Answer 3

except this isn't one

Answer 4

all the tricks, man
it will show how much you're reading

Answer 5

you can even prove it wrong laughing out loud

Answer 6

\[\sin(a) + \sin(b) = 2\sin\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x\color{#C00}- y}{2}\right)\]

Answer 7

well, lol - I meant \(\sin(x ) + \sin(y)\) on the LHS

Answer 8

sum-to-product @ParthKohli
I think he just typed it incorrectly, that cos(x-)/2

Answer 9

damn I mistyped laughing out loud

Answer 10

think he just typed it incorrectly, that cos(x-y)/2

Answer 11

Nope, it's \(\cos(\frac{x - y}{2})\). The question is incorrect.

Answer 12

I think it's typo
the dude knows nothing about math to begin with

Answer 13

one can always begin with six-identities and reveal all other identities and properties. it takes time specially in the middle of an exam, but it is not a difficult task

Answer 14

A million dollar question :P

Answer 15

if one has a clue about trig at all, one can reverse back the process
sum-to-product can be derived from product-to-sum, which can be derived from angle addition-substraction

if you can't do that, learn a little geometry
http://math.ucsd.edu/~wgarner/math4c/textbook/chapter6/addsubformulas.htm

Answer 16

OK, here's a proof I like:\[\sin(a + b) = \sin(a)\cos(b) +\cos(a)\sin(b) \]\[\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)\]
\[\sin(a - b) = \sin(a)\cos(b) - \sin(b)\cos(a)\]\[\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)\]

From i and iii,
\[\sin(a + b) + \sin(a-b) = 2\sin(a)\cos(b)\]From ii and iv,\[\cos(a+b) + \cos(a-b) = 2\cos(a)\cos(b)\]
If you let \(x = a + b\) and \(y = a - b\), then \(a = \frac{x + y}{2}\) and \(b = \frac{x- y}{2 }\). Therefore...