(sin (2x) +sin(4x)/sin(2x)-sin(4x)) + (tan(3x))/tanx)

i'm totally lost o.o

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- Dec 28th 2009, 08:21 PMalessandromangionesolving this identity
(sin (2x) +sin(4x)/sin(2x)-sin(4x)) + (tan(3x))/tanx)

i'm totally lost o.o - Dec 28th 2009, 08:35 PMbigwave
first, this is just an expression

not an identity we can prove

there is no = sign

otherwise ..

$\displaystyle

\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\c os\frac{\alpha-\beta}{2}

$

and

$\displaystyle

\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\al pha-\beta}{2}

$ - Dec 28th 2009, 08:44 PMProve It
1. There's no such thing as "solving an identity". To solve means to find the value of $\displaystyle x$ (or whatever the unknown is) that satisfies an equation. An identity is an equation that is true no matter what the $\displaystyle x$ value.

2. You can PROVE an identity - i.e. show that two expressions are identically equal. Is that what you are trying to do? What are you trying to show that this is the same as? Or are you trying to ESTABLISH an identity? What was said above - there is not an equals sign so this is not an identity. - Dec 28th 2009, 09:21 PMalessandromangione
guys i'm really sorry...i forgot to put '=0'...i'm sorry

- Dec 28th 2009, 09:35 PMbigwavefirst step
given identity prove that:

$\displaystyle

\frac{\sin{2x} +\sin{4x}}{\sin{2x}-\sin{4x}} + \frac{\tan{3x}}{\tan{x}}=0$

use sum to product formulas

$\displaystyle \frac

{2\sin\frac{2x+4x}{2}\cos\frac{2x-4x}{2}}

{2\cos\frac{2x+4x}{2}\sin\frac{2x-4x}{2}}

+\frac{\tan{3x}}{tan{x}}=0

$

can you finish from here? - Dec 29th 2009, 02:24 PMalessandromangione
thanks