sry for the trouble! another trigo problem!!

**cyy** · February 18th 2008, 07:26 AM

Hi sry for the previous double post, below is another porblem i faced:
given: 0 <= x < pi ; 0 <= y; < pi
...

sin^2x = sin^2y
(sinx + siny )( sinx - siny ) = 0
thus, how to get from above equation to form below 2 statements?
x = y+2pi( k ) //where is the pi & k come from?
x = -y+2pi( k )

Thank You!

**topsquark** · February 18th 2008, 10:20 AM

Originally Posted by cyy

Hi sry for the previous double post, below is another porblem i faced:
given: 0 <= x < pi ; 0 <= y; < pi
...

sin^2x = sin^2y
(sinx + siny )( sinx - siny ) = 0
thus, how to get from above equation to form below 2 statements?
x = y+2pi( k ) //where is the pi & k come from?
x = -y+2pi( k )

Thank You!

There is a problem with one of the solutions:

Given two real numbers a and b, if ab = 0 then a = 0 and/or b = 0.

So starting with
$(sin(x) + sin(y) )( sin(x) - sin(y) ) = 0$

We see that
$sin(x) + sin(y) = 0$
or
$sin(x) - sin(y) = 0$

The second is (almost) trivial, so I'll do the first.
$sin(x) + sin(y) = 0$

$sin(x) = -sin(y)$

This can only happen when $x = y + (2k + 1) \pi$ . If you don't see why then expand it:
$sin(y + (2k + 1) \pi) = sin(x)~cos((2k + 1) \pi) + sin((2k + 1) \pi)~cos(y)$

Now, 2k + 1 is odd, so $cos((2k +1) \pi) = -1$ and $sin((2k + 1)\pi) = 0$ .

Thus
$x = y + (2k + 1) \pi$

The other solution
$x = y + 2k \pi$
is correct.

-Dan

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