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I have tried all kinds of different method, and have spent more than an hour on this problem, also looked upon internet looking for similar problems but none worked. All I got in the end, I can get my answers to $\dfrac{\csc y+\csc x}{\tan x+\tan y}$, but pretty sure that doesn't answer the problem, and tried using identities but in the end they always become much more complicated than I couldn't clear up, so I pretty much did the opposite of simplify.

thank u for your comment
how to u use the sum to product rule on sin(x+y) i only found identity that sin(x+y)= sinxcosy+cosxsiny? does sin(x+y)= sinx+siny? or u used the sum rule and then the product rule of sinx*cosy identity which didn't work for me

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$\frac{\cos(x)+\cos(y)}{\sin(x+y)}=\frac{2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}{\sin(x+y)}$ by sum-to-product. In turn, this expression

$$=\frac{2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}{2\sin(\frac{x+y}{2})\cos(\frac{x+y}{2})}$$ by the double angle fromula for sin.

Hence the simplified expression is $$=\frac{\cos(\frac{x-y}{2})}{\sin(\frac{x+y}{2})}$$

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  • $\begingroup$ how to u use the sum to product rule on sin(x+y) i only found identity that sin(x+y)= sinxcosy+cosxsiny $\endgroup$ – Luna Oct 26 '16 at 21:58
  • $\begingroup$ I really dont get how you use the double angle formula for sine to get that. $\endgroup$ – Luna Oct 26 '16 at 22:48
  • $\begingroup$ Sum-to-product was only used in the very first step, where $\cos(x)+\cos(y)$ was turned into $2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$. The second step involved using the double angle formula for sine: $\sin(2a)=2\sin(a)\cos(a)$, just replace $a$ with $\frac{x+y}{2}$ to obtain the fact I used. $\endgroup$ – Jed Oct 26 '16 at 23:48
  • $\begingroup$ AHHHHHHHHHHHHHHH now i see omg thank u so much $\endgroup$ – Luna Oct 27 '16 at 1:06
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Here is what I think you want: $$cosx + cosy = 2.cos(\frac{x+y}{2}).cos(\frac{x-y}{2}).$$ Also: $$sin(x+y) = 2.sin(\frac{x+y}{2}).cos(\frac{x+y}{2}).$$ Therefore: $$\frac{cosx + cosy}{sin(x+y)} = \frac{cos(\frac{x-y}{2})}{sin(\frac{x+y}{2})}$$ That's as simplified as I could get it. At least there are only products, right?

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  • $\begingroup$ Jed came earlier. $\endgroup$ – Matheus Rotta Oct 26 '16 at 20:40
  • $\begingroup$ how to u use the sum to product rule on sin(x+y) i only found identity that sin(x+y)= sinxcosy+cosxsiny $\endgroup$ – Luna Oct 26 '16 at 21:58

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