Sin(П/3-2x)cos(П/3-2x)≥-√3/4

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вопрос

To prove the inequality sin(π/3 - 2x)cos(π/3 - 2x) ≥ -√3/4, we can use the double angle formula for sine and cosine:

sin(2θ) = 2sin(θ)cos(θ
cos(2θ) = cos^2(θ) - sin^2(θ)

Let's first substitute θ = π/6 - x into the double angle formulas:

sin(π/3 - 2x) = 2sin(π/6 - x)cos(π/6 - x
cos(π/3 - 2x) = cos^2(π/6 - x) - sin^2(π/6 - x)

Now, substitute these expressions back into the original inequality:

2sin(π/6 - x)cos(π/6 - x)(cos^2(π/6 - x) - sin^2(π/6 - x)) ≥ -√3/4

Expand the expression and simplify:

2sin(π/6 - x)cos(π/6 - x)cos^2(π/6 - x) - 2sin^3(π/6 - x) ≥ -√3/4

Since sin(π/6) = 1/2 and cos(π/6) = √3/2, the inequality simplifies to:

2(1/2 - sin^3(π/6 - x))√3/2 - 2sin^3(π/6 - x) ≥ -√3/4

√3 - 3sin^3(π/6 - x)√3 - 2sin^3(π/6 - x) ≥ -√3/4

√3 - 5sin^3(π/6 - x)√3 ≥ -√3/4

1 - 5sin^3(π/6 - x) ≥ -1/4

5sin^3(π/6 - x) ≤ 5/4

sin^3(π/6 - x) ≤ 1/4

Since -1 ≤ sin(x) ≤ 1, and sin(π/6 - x) is in the range of sin(x), we have established that sin^3(π/6 - x) ≤ 1/4. Thus, the inequality sin(π/3 - 2x)cos(π/3 - 2x) ≥ -√3/4 holds true.