# Thread: solving this identity

1. ## solving this identity

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

i'm totally lost o.o

2. first, this is just an expression
not an identity we can prove
there is no = sign

otherwise ..

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

and

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

3. Originally Posted by alessandromangione
(sin (2x) +sin(4x)/sin(2x)-sin(4x)) + (tan(3x))/tanx)

i'm totally lost o.o
1. There's no such thing as "solving an identity". To solve means to find the value of $x$ (or whatever the unknown is) that satisfies an equation. An identity is an equation that is true no matter what the $x$ 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.

4. guys i'm really sorry...i forgot to put '=0'...i'm sorry

5. ## first step

given identity prove that:

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

use sum to product formulas

$\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?

6. thanks