证明以下等式(tanπ/9)^2+(tan2π/9)^2+(tan4π/9)^2=33[tan(π/9)]^2*[tan
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证明以下等式
(tanπ/9)^2+(tan2π/9)^2+(tan4π/9)^2=33
[tan(π/9)]^2*[tan(2π/9)]^2*[tan(4π/9)]^2=3.
(tanπ/9)^2+(tan2π/9)^2+(tan4π/9)^2=33
[tan(π/9)]^2*[tan(2π/9)]^2*[tan(4π/9)]^2=3.
[tan(π/9)]² * [tan(2π/9)]² * [tan(4π/9)]²
= [tan(π/9) * tan(2π/9) * tan(4π/9)]²
= [tan(π/9) * tan(π/3 - π/9) * tan(π/3 + π/9)]²
= [tan(π/9) * (√3 - tanπ/9)/(1 + √3tanπ/9) * (√3 + tanπ/9)/(1 - √3tanπ/9)]²
= {tan(π/9) * [3 - (tanπ/9)²]/[1 - 3(tanπ/9)²]}²
= [3tan(π/9) - (tanπ/9)^3]/[1 - 3(tanπ/9)²]²
= [tan(π/3)]² (三倍角公式)
= 3
再问: 谢谢!第一个等式能帮忙吗?
= [tan(π/9) * tan(2π/9) * tan(4π/9)]²
= [tan(π/9) * tan(π/3 - π/9) * tan(π/3 + π/9)]²
= [tan(π/9) * (√3 - tanπ/9)/(1 + √3tanπ/9) * (√3 + tanπ/9)/(1 - √3tanπ/9)]²
= {tan(π/9) * [3 - (tanπ/9)²]/[1 - 3(tanπ/9)²]}²
= [3tan(π/9) - (tanπ/9)^3]/[1 - 3(tanπ/9)²]²
= [tan(π/3)]² (三倍角公式)
= 3
再问: 谢谢!第一个等式能帮忙吗?
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