Question

Sin2x-sin3x+sin8x=cos(7x+3Pi\2)

Answer 1

Решение на фото внизу

Answer 2

To verify the given trigonometric equation, we'll need to manipulate the left-hand side (LHS) and the right-hand side (RHS) to see if they are equal. Let's start by working with the LHS:

LHS = sin(2x) - sin(3x) + sin(8x)

Now, we'll use trigonometric identities to simplify this expression. The identity we need is:

sin(A) - sin(B) = 2 * cos((A + B) / 2) * sin((A - B) / 2)

Using this identity, we can rewrite the LHS as follows:

LHS = 2 * cos((2x + 3x) / 2) * sin((2x - 3x) / 2) + sin(8x)

LHS = 2 * cos(5x / 2) * sin(-x / 2) + sin(8x)

Now, let's use another identity:

sin(-x) = -sin(x)

LHS = 2 * cos(5x / 2) * (-sin(x / 2)) + sin(8x)

Now, we'll use the double angle formula for cosine:

cos(2θ) = 1 - 2 * sin^2(θ)

cos(5x / 2) = 1 - 2 * sin^2(5x / 4)

Now the LHS becomes:

LHS = 2 * (1 - 2 * sin^2(5x / 4)) * (-sin(x / 2)) + sin(8x)

LHS = -4 * sin^2(5x / 4) * sin(x / 2) + sin(8x)

Now, let's work on the RHS:

RHS = cos(7x + 3π / 2)

The cosine function evaluated at (7x + 3π / 2) is equivalent to the sine function evaluated at (7x). Hence:

RHS = sin(7x)

Now, we have:

LHS = -4 * sin^2(5x / 4) * sin(x / 2) + sin(8x) RHS = sin(7x)

To verify if the equation is true, we'll check if LHS equals RHS:

-4 * sin^2(5x / 4) * sin(x / 2) + sin(8x) = sin(7x)

It seems there is no direct trigonometric identity to simplify the left-hand side further to directly match the right-hand side. Therefore, we cannot conclude that the equation is true in its current form. It's possible that there may be an error in the equation or some additional information needed to find a solution. If you have any additional context or corrections, feel free to provide them, and I'll be happy to help further!