Question

Cosx+cos2x+cos6x+cos7x=4cosx/2cos5x/2cos4

Answer 1

Ответ:0=0

Пошаговое объяснение:

1) cosx+cos7x=2cos4x*cos3x

2)cos2x+cos6x=2cos4x*cos2x

3)2cos4x(cos3x+cos2x)=2cos4x(2cos5x/2*cosx/2)=4cos4x*cos(5x/2)*cos(x/2)

4)4cos4x*cos(5x/2)*cos(x/2)=4cos4x*cos(5x/2)*cos(x/2)

Answer 2

To prove the given trigonometric equation:

cos(x) + cos(2x) + cos(6x) + cos(7x) = 4cos(x)/2cos(5x)/2cos(4x)

Let's work on the left-hand side (LHS) and the right-hand side (RHS) separately and simplify both sides to see if they are equal.

Step 1: Simplify the left-hand side (LHS)

cos(x) + cos(2x) + cos(6x) + cos(7x)

Now, let's use the trigonometric identity: cos(A) + cos(B) = 2 * cos((A + B) / 2) * cos((A - B) / 2)

Applying this identity to the given equation:

cos(x) + cos(2x) = 2 * cos((x + 2x) / 2) * cos((x - 2x) / 2) = 2 * cos(3x/2) * cos(-x/2) = 2 * cos(3x/2) * cos(x/2)

cos(6x) + cos(7x) = 2 * cos((6x + 7x) / 2) * cos((6x - 7x) / 2) = 2 * cos(13x/2) * cos(-x/2) = 2 * cos(13x/2) * cos(x/2)

Substituting back into the original expression:

cos(x) + cos(2x) + cos(6x) + cos(7x) = 2 * cos(3x/2) * cos(x/2) + 2 * cos(13x/2) * cos(x/2)

Now, we can use the following identity: cos(A) * cos(B) = (1/2) * [cos(A + B) + cos(A - B)]

Apply the above identity to the expression:

2 * cos(3x/2) * cos(x/2) = (1/2) * [cos(3x/2 + x/2) + cos(3x/2 - x/2)] = (1/2) * [cos(2x) + cos(x)] = (1/2) * [2 * cos(x) + cos(x)] = 3/2 * cos(x)

2 * cos(13x/2) * cos(x/2) = (1/2) * [cos(13x/2 + x/2) + cos(13x/2 - x/2)] = (1/2) * [cos(7x) + cos(6x)] = (1/2) * [2 * cos(7x) + cos(6x)] = 3/2 * cos(7x)

Substitute back into the expression:

cos(x) + cos(2x) + cos(6x) + cos(7x) = 3/2 * cos(x) + 3/2 * cos(7x)

Step 2: Simplify the right-hand side (RHS)

4 * cos(x) / [2 * cos(5x) * 2 * cos(4x)]

= 4 * cos(x) / (4 * cos(5x) * cos(4x))

= cos(x) / (cos(5x) * cos(4x))

Step 3: Compare LHS and RHS

Now, let's compare the simplified LHS and RHS:

LHS: cos(x) + cos(2x) + cos(6x) + cos(7x) = 3/2 * cos(x) + 3/2 * cos(7x)

RHS: cos(x) / (cos(5x) * cos(4x))

It is evident that the LHS and RHS are not the same. So, the given equation is not true:

cos(x) + cos(2x) + cos(6x) + cos(7x) ≠ 4cos(x)/2cos(5x)/2cos(4x)

The statement in question is incorrect. There might be a typographical error or a misunderstanding of the trigonometric equation involved. Please double-check the original equation and ensure there are no mistakes or omissions.