2cos(4pi/7)/sin(pi/14)

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вопрос

To simplify the expression 2cos(4π/7)/sin(π/14), we first need to recall the trigonometric identity: cos(2θ) = 2cos^2(θ) - 1.

We can rewrite our expression using this identity as follows:

2cos(4π/7) / sin(π/14) = 2cos(2 * 2π/7) / sin(π/14)

= 2(2cos^2(2π/7) - 1) / sin(π/14)

= 4cos^2(2π/7) / sin(π/14) - 2 / sin(π/14)

Next, we need to simplify the expression further using the double angle identity: cos(2θ) = 2cos^2(θ) - 1.

We know that cos(2π/7) = cos(π/7 + π/7) = cos(π/7)cos(π/7) - sin(π/7)sin(π/7) = cos^2(π/7) - sin^2(π/7).

Therefore, cos^2(2π/7) = (cos^2(π/7) - sin^2(π/7))^2 = cos^4(π/7) - 2cos^2(π/7)sin^2(π/7) + sin^4(π/7).

After substituting this back into our expression, we get:

4cos^4(π/7) - 8cos^2(π/7)sin^2(π/7) + 4sin^4(π/7) / sin(π/14) - 2 / sin(π/14)

Since sin(π/14) = sin^2(π/14) + cos^2(π/14) and sin^2(π/14) + cos^2(π/14) = 1, the expression becomes:

4cos^4(π/7) - 8cos^2(π/7)sin^2(π/7) + 4sin^4(π/7) / sin(π/14) - 2 / sin(π/14)
= 4cos^4(π/7) - 8sin^2(π/7)cos^2(π/7) + 4sin^4(π/7)

Therefore, the simplified expression is 4cos^4(π/7) - 8sin^2(π/7)cos^2(π/7) + 4sin^4(π/7).