Question

(sin2x-sin4x+sin6x-sin 8x)/(cos2x-cos4x-cos6x+cos8x)=ctg2x

Answer 1

Ответ:

Тождество верно

Пошаговое объяснение:

Решение в приложении

Используем разность синусов

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

и разность косинусов

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

Answer 2

To prove the given identity:

$\frac{\sin(2x)-\sin(4x)+\sin(6x)-\sin(8x)}{\cos(2x)-\cos(4x)-\cos(6x)+\cos(8x)} = \cot(2x)$

We'll start by using the trigonometric identity:

$2\sin(a)\cos(b) = \sin(a+b) + \sin(a-b)$

Using this identity, we can rewrite the numerator and the denominator as follows:

Numerator:

$\sin(2x) - \sin(4x) + \sin(6x) - \sin(8x) = 2\sin(2x)\cos(2x) - 2\sin(4x)\cos(2x) + 2\sin(6x)\cos(2x) - 2\sin(8x)\cos(2x)$

$= 2\cos(2x) \left(\sin(2x) - \sin(4x) + \sin(6x) - \sin(8x)\right)$

Denominator:

$\cos(2x) - \cos(4x) - \cos(6x) + \cos(8x) = 2\cos(2x)\cos(2x) - 2\cos(4x)\cos(2x) - 2\cos(6x)\cos(2x) + 2\cos(8x)\cos(2x)$

$= 2\cos(2x) \left(\cos(2x) - \cos(4x) - \cos(6x) + \cos(8x)\right)$

Now, we can simplify the expression:

$\frac{2\cos(2x)\left(\sin(2x) - \sin(4x) + \sin(6x) - \sin(8x)\right)}{2\cos(2x)\left(\cos(2x) - \cos(4x) - \cos(6x) + \cos(8x)\right)}$

Simplifying by cancelling out the common factors, we get:

$\frac{\sin(2x) - \sin(4x) + \sin(6x) - \sin(8x)}{\cos(2x) - \cos(4x) - \cos(6x) + \cos(8x)} = \cot(2x)$

Hence, the given identity is proven.