## nphuongsun93 Group Title $\cos(x) + \cos(y) - \cos (x+y) = \frac{3}{2}$ one year ago one year ago

1. Zekarias Group Title

Then what?

2. nphuongsun93 Group Title

solve for y, sorry

3. nphuongsun93 Group Title

gonna read this later, bbl 2 hours *-*

4. lgbasallote Group Title

i think not enough information for solve it

6. mathslover Group Title

@nphuongsun93 What you tried? (just asking)

7. mathslover Group Title

$$\cos x + \cos y - \cos x \cos y + \sin x \sin y=\frac{3}{2}$$

8. mathslover Group Title

@asnaseer sir any ideas?

9. mathslover Group Title

@hartnn

10. Coolsector Group Title

i thought of maybe using cos(a) = [e^(ia) + e^(-ia)] / 2 but i dont think it works either

11. asnaseer Group Title

Maybe try this:$\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$and:$\cos(x+y)=2\cos^2(\frac{x+y}{2})-1$That will lead to:$\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})=\frac{1}{4}$

12. asnaseer Group Title

which is a quadratic in $$\cos(\frac{x+y}{2})$$ - although it also contains a $$\cos(\frac{x-y}{2})$$

13. asnaseer Group Title

Solving the quadratic leads to:$\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sqrt{\cos^2(\frac{x-y}{2})-1}}{2}$

14. asnaseer Group Title

Which simplifies to:$\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sin(\frac{x-y}{2})}{2}$

15. asnaseer Group Title

Don't know if this is helpful or not but I have to go now - good luck!

16. nphuongsun93 Group Title

Thanks a lot @asnaseer i get it, that's just what i needed

17. mathslover Group Title

@nphuongsun93 , what and how did you get the answer?

18. nphuongsun93 Group Title

$\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})-0.25 = 0$ then let $a = \cos(\frac{x+y}{2}) \space and \space b = \cos ( \frac{x-y}{2})$ $ab - a^2 - 0.25=0$ taking this as a quadratic in p, b^2 -1 has to be positive to get real solution for a, but the highest value of b = cos((x-y)/2) is 1 (angles), so the only real solution is b =1 or b^2 -1 = 0 $b= \cos(\frac{x-y}{2})=1 \space or \space \frac{x-y}{2}=0 \space or \space x=y$