# Trig factor law help

1. Nov 18, 2008

### icystrike

1. The problem statement, all variables and given/known data
i encountered some problems while solving this question.
Hope someone would show me the necessary working and solution, thanks in advance (:

Question:

Given that sin x + sin y = a and cos x + cos y = a. where a$$\neq$$0 , find sin x + cos x in terms of a.

2. Relevant equations

3. The attempt at a solution

2. Nov 18, 2008

### tiny-tim

Welcome to PF!

Hi icystrike! Welcome to PF!

Hint: instead of x and y, write both equations in terms of (x + y)/2 and (x - y)/2.

(or you could just subtract the two equations, and use a formula for sinx - cosx)

3. Nov 18, 2008

### icystrike

Re: Trigonometry

i tried tat method alr..
factor law fellowed by double angle law

i derive with 2(sin x/2)(cos y/2)=0

but nothing can move on..

4. Nov 18, 2008

### tiny-tim

No, you shouldn't have got that

5. Nov 18, 2008

### icystrike

Re: Trigonometry

6. Nov 18, 2008

### tiny-tim

ah … you went a bit doo-lally :uhh: after sin(x+y)/2 = cos(x+y)/2 …

just divide by the RHS, and you get tan(x+y)/2 = 1,

and so x+y = … ?

7. Nov 18, 2008

### icystrike

Re: Trigonometry

yup.
(x+y)/2=45 deg , 225 deg

8. Nov 18, 2008

### tiny-tim

Go on then … so x+y = … ?

and so sinx + cosx = … ?

9. Nov 18, 2008

### icystrike

Re: Trigonometry

eh. x+y=90,450 deg

sinx+cosx
i cant find..

10. Nov 18, 2008

### tiny-tim

i] for a start you can forget 450 … it's more than 360

ii] so x + y = 90º … so siny = … ?

11. Nov 18, 2008

### icystrike

Re: Trigonometry

i think cant drop 450 deg cos they did not mention tat it must be larger or equal to 0 and smaller or equal to 360. cos it produces different answers in the end..

siny=cosx or 2(siny/2)(cosy/2)
theefore sinx+cosx=sinx+siny=a

12. Nov 18, 2008

### tiny-tim

Yup!
But if x + y = 450, then y = 450 - x, so siny = sin(450-x) = sin(90-x) = cosx …

multiples of 360 make no difference.
uhh?

2(siny/2)(cosy/2) = siny whatever x is … how does that help?

13. Nov 18, 2008

### Дьявол

Re: Trigonometry

Rather doing all those things, I think I could solve it on much easier way:
If $$sinx+siny=a$$ then $$sinx+\sqrt{1-cos^2y}=a$$, from the other equation $$cosy=a-cosx$$:

$$sinx+\sqrt{1-(a-cosx)^2}=a$$

$$\sqrt{1-(a^2-2acosx+cos^2x)}=a-sinx$$

$$1-a^2+2acosx-cos^2x=a^2-2asinx+sin^2x$$

$$sin^2x-a^2+2acosx=a^2-2asinx+sin^2x$$

$$-a^2+2acosx=a^2-2asinx$$

$$2acosx+2asinx=2a^2$$

I think,now, out of here you know the answer. Regards.

14. Nov 18, 2008

### icystrike

Re: Trigonometry

thank you guys very much..
is there a way to rep you guys or something?

15. Nov 19, 2008

### tiny-tim

There's no repping on this forum …

just be extra kind to the next goldfish you meet!