sin(2兀/3一2∝)=3/5,sin²(∝十5兀/12〉
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您好，很高兴为您解答。
sin(2θ) = 2sin(θ)cos(θ)，sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)首先，让我们简化 sin(2π/3 - 2α)：sin(2π/3 - 2α) = sin(2π/3)cos(2α) - cos(2π/3)sin(2α) = (sqrt(3)/2)cos(2α) - (1/2)sin (2α) = (sqrt(3)/2)(2cos²(α) - 1) - (1/2)(2sin(α)cos(α)) = sqrt(3)cos²(α) - (3/4 )罪(2a)
sin(2θ) = 2sin(θ)cos(θ)，sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)首先，让我们简化 sin(2π/3 - 2α)：sin(2π/3 - 2α) = sin(2π/3)cos(2α) - cos(2π/3)sin(2α) = (sqrt(3)/2)cos(2α) - (1/2)sin (2α) = (sqrt(3)/2)(2cos²(α) - 1) - (1/2)(2sin(α)cos(α)) = sqrt(3)cos²(α) - (3/4 )罪(2a)
咨询记录 · 回答于2023-03-14
sin(2兀/3一2∝)=3/5,sin²(∝十5兀/12〉
您好，很高兴为您解答。sin(2θ) = 2sin(θ)cos(θ)，sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)首先，让我们简化 sin(2π/3 - 2α)：sin(2π/3 - 2α) = sin(2π/3)cos(2α) - cos(2π/3)sin(2α) = (sqrt(3)/2)cos(2α) - (1/2)sin (2α) = (sqrt(3)/2)(2cos²(α) - 1) - (1/2)(2sin(α)cos(α)) = sqrt(3)cos²(α) - (3/4 )罪(2a)
sin(2θ) = 2sin(θ)cos(θ)，sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)首先，让我们简化 sin(2π/3 - 2α)：sin(2π/3 - 2α) = sin(2π/3)cos(2α) - cos(2π/3)sin(2α) = (sqrt(3)/2)cos(2α) - (1/2)sin (2α) = (sqrt(3)/2)(2cos²(α) - 1) - (1/2)(2sin(α)cos(α)) = sqrt(3)cos²(α) - (3/4 )罪(2a)
我们还知道 sin(2π/3 - 2α) = 3/5，所以：sqrt(3)cos²(α) - (3/4)sin(2α) = 3/5，现在，让我们简化 sin²(α + 5π/12)：sin²(α + 5π/12) = (sin(α)cos(5π/12) + cos(α)sin(5π/12))² = (sin(α)(sqrt(2) + sqrt(6)) /4 + cos(α)(sqrt(2) - sqrt(6))/4)² = (sqrt(2)/4)(sin(α) + cos(α))² + (sqrt(6)/ 4)(sin(α) - cos(α))²，可以使用恒等式 sin²(θ) + cos²(θ) = 1 来根据彼此重写 sin(α) 和 cos(α)：sin²(α) = 1 - cos²(α)，cos²(α) = 1 - sin²(α)，将这些代入 sin²(α + 5π/12) 的表达式中，得到：sin²(α + 5π/12) = (sqrt(2)/4)(sqrt(2)cos(α) + sqrt(6)sin(α))² + (sqrt(6)/4)(sqrt(2 )sin(α) - sqrt(6)cos(α))² = 1/8(2cos²(α) + 6sin²(α) + 4sqrt(3)sin(α)cos(α) + 6cos²(α) - 4sqrt (3)sin(α)cos(α)) = 1/4(sin²(α) + cos²(α)) + 1/4(3sin²(α) - 3cos²(α)) = 1/4 - 1/4 (2cos²(α) - 2sin²(α)) = 1/4 - 1/2(cos(2α)。
亲亲选择bd哦
亲亲选择B哦
亲亲选择A哦
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得到：tanbb ≤atanb−7tanb≤ tanbb即：1 \leq a\tan^
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