Solve Trig Factor Law: sin x + cos x from a Given Equation

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The discussion revolves around solving the equation involving trigonometric functions: sin x + sin y = a and cos x + cos y = a, with the goal of expressing sin x + cos x in terms of a. Participants suggest using transformations involving (x + y)/2 and (x - y)/2 to simplify the equations. There is a debate about whether to consider angles beyond the standard 0 to 360 degrees, as it could yield different solutions. Ultimately, one participant derives a more straightforward approach to reach the answer, emphasizing the importance of careful algebraic manipulation. The conversation highlights the collaborative nature of problem-solving in mathematics.
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Homework Statement


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\neq0 , find sin x + cos x in terms of a.

Homework Equations





The Attempt at a Solution

 
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icystrike said:
Given that sin x + sin y = a and cos x + cos y = a. where a\neq0 , find sin x + cos x in terms of a.

Hi icystrike! Welcome to PF! :smile:

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

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


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

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

No, you shouldn't have got that :confused:

show us your full calculation. :smile:
 


http://img143.imageshack.us/img143/2867/lalalodw5.jpg
 
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ah … you went a bit doo-lally :rolleyes: 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 = … ? :smile:
 


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

Go on then … so x+y = … ?

and so sinx + cosx = … ? :smile:
 


eh. x+y=90,450 deg

sinx+cosx
i can't find..
 
  • #10
icystrike said:
eh. x+y=90,450 deg

sinx+cosx
i can't find..

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

ii] so x + y = 90º … so siny = … ? :smile:
 
  • #11


i think can't 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
icystrike said:
theefore sinx+cosx=sinx+siny=a

Yup! :biggrin:
i think can't 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..

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

multiples of 360 make no difference. :wink:
siny=cosx or 2(siny/2)(cosy/2)

uhh? :confused:

2(siny/2)(cosy/2) = siny whatever x is … how does that help?
 
  • #13


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


thank you guys very much..:smile:
is there a way to rep you guys or something?
 
  • #15
icystrike said:
thank you guys very much..:smile:
is there a way to rep you guys or something?

There's no repping on this forum …

just be extra kind to the next goldfish you meet! :biggrin:
 
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