- #1

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__cos x - cos y__

__sin x - sin y__

sin x + sin y + cos x + cos y = 0

or to see better i guess....

(cos x - cos y)/(sin x + sin y) + (sin x - siny)/(cos x + cos y) = 0

can you guys help me? i'm really stuck on this!

- Thread starter InFiNiTeX
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- #1

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- 0

sin x + sin y + cos x + cos y = 0

or to see better i guess....

(cos x - cos y)/(sin x + sin y) + (sin x - siny)/(cos x + cos y) = 0

can you guys help me? i'm really stuck on this!

- #2

- 506

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What have you tried and where are you stuck ?InFiNiTeX said:cos x - cos ysin x - sin y

sin x + sin y + cos x + cos y = 0

or to see better i guess....

(cos x - cos y)/(sin x + sin y) + (sin x - siny)/(cos x + cos y) = 0

can you guys help me? i'm really stuck on this!

- #3

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- #4

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How about just combining the fractions?

- #5

- 506

- 1

There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know.robphy said:

- #6

- 5,757

- 1,052

While this may be true, this exponential method never fails, of course.hypermorphism said:There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know.

- #7

Integral

Staff Emeritus

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In fact it is nearly trivial, obvious to casual inspection.

OP , What have you tried?

OP , What have you tried?

- #8

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well this is what i tried to do....

first i tried to combine them

(cos x - cos y) ( sin x - sin y)/ (sin x + sin y) (cos x + cos y)

then i got

(cos^2 x ) - (cos^2 y)/(sin x + sin y) (cos x + cos y) + (sin^2 x)(sin^2 y)/ (sin x + sin y) (cos x + cos y)

and after that part i kinda just confused myself but i just thought of this....

that part changes to

(1 - sin^2 x) - (1 - sin^2 y) / (sin x + sin y) (cos x + cos y) + (1 - cos^2 x) (1 - cos y) / (sin x + sin y) (cos x + cos y)

which i can turn into

(1 - sin x) - (1 - sin y) / cos x + cos y + (1 - cos x) (1 - cos y) / (sin x + sin y)

now those fractions cancel out right? cause (1 - sin x) and (1 - sin y) is the same as cos x and cos y as (1 - cos x) and (1 - cos y) are the same as (sin x) and (sin y) right? so if all that cancels out i get 0!

while in the shower it struck me that 1 - sin x = cos x , so i went from there..

i hope that is the right answer, if not could you please help me go in the right direction? thanks for your guys much appreciated help!

first i tried to combine them

(cos x - cos y) ( sin x - sin y)/ (sin x + sin y) (cos x + cos y)

then i got

(cos^2 x ) - (cos^2 y)/(sin x + sin y) (cos x + cos y) + (sin^2 x)(sin^2 y)/ (sin x + sin y) (cos x + cos y)

and after that part i kinda just confused myself but i just thought of this....

that part changes to

(1 - sin^2 x) - (1 - sin^2 y) / (sin x + sin y) (cos x + cos y) + (1 - cos^2 x) (1 - cos y) / (sin x + sin y) (cos x + cos y)

which i can turn into

(1 - sin x) - (1 - sin y) / cos x + cos y + (1 - cos x) (1 - cos y) / (sin x + sin y)

now those fractions cancel out right? cause (1 - sin x) and (1 - sin y) is the same as cos x and cos y as (1 - cos x) and (1 - cos y) are the same as (sin x) and (sin y) right? so if all that cancels out i get 0!

while in the shower it struck me that 1 - sin x = cos x , so i went from there..

i hope that is the right answer, if not could you please help me go in the right direction? thanks for your guys much appreciated help!

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- #9

- 506

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This isn't correct. The equation you may be attempting to reference is sinInFiNiTeX said:while in the shower it struck me that 1 - sin x = cos x , so i went from there..

Regarding your original equation, try putting the fractions on opposite sides of the equation:

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} = 0[/tex]

is the same as

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = -\frac{\sin x - \sin y}{\cos x + \cos y}[/tex]

or

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}[/tex]

Do you see a way to get rid of the fractions from here ?

- #10

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- #11

Integral

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(cos^2 x ) - (cos^2 y)/(sin x + sin y) (cos x + cos y) +

go back and double check where this came from, it is not correct. Everything else is ok

- #12

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ok thanks for your help Integral! and hypermorphism, sorry i just can't see how to get rid of the fractions... i'm sure its very easy and i will feel really dumb though....

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- #13

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You have this equation, from hypermorphism's last post:

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}[/tex]

Do you remember cross multiplication?

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}[/tex]

Do you remember cross multiplication?

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- #14

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OHH!!!!! see i told you it would probably be very easy.... thanks for your help!

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