Question

(cos6x+sin6x)²+(cos6x-sin6x)²=2​

Answer 1

Ответ:

${( \cos(6x) + \sin(6x) ) }^{2} + {( \cos(6x) - \sin(6x)) }^{2} = { \cos(6x) }^{2} + 2 \sin(6x) \cos(6x) + { \sin(6x) }^{2} + { \cos(6x) }^{2} - 2 \sin(6x) \cos(6x) + { \sin(6x) }^{2} = 2 { \cos(6x) }^{2} + 2 { \sin(6x) }^{2} = 2( \cos(6x) + \sin(6x)) = 2$

Answer 2

Let's simplify the given expression step by step:

Starting with the left-hand side (LHS): (cos(6x) + sin(6x))^2 + (cos(6x) - sin(6x))^2

Expanding both terms individually: (cos(6x))^2 + 2*cos(6x)sin(6x) + (sin(6x))^2 + (cos(6x))^2 - 2cos(6x)*sin(6x) + (sin(6x))^2

Using the identity cos^2(x) + sin^2(x) = 1: 1 + 2*cos(6x)sin(6x) + 1 - 2cos(6x)*sin(6x)

Combining like terms: 2 + 0*cos(6x)*sin(6x)

Since the middle term is zero, the simplified expression becomes: 2

Therefore, the given equation (cos(6x) + sin(6x))^2 + (cos(6x) - sin(6x))^2 = 2 holds true.