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  • MT【300】余弦的三倍角公式

    2017清华大学THUSSAT附加学科测试数学(二测)
    $cos^5dfrac{pi}{9}+cos^5dfrac{5pi}{9}+cos^5dfrac{7pi}{9}$ 的值为_____
    A.$frac{15}{32}$

    B.$frac{15}{16}$

    C.$frac{8}{15}$

    D.$frac{16}{15}$


    解答:注意到$cos3 heta=4cos^3 heta-3cos heta,cos3 heta=dfrac{1}{2}$的三个根为$dfrac{pi}{9},dfrac{5pi}{9},dfrac{7pi}{9}$故$cosdfrac{pi}{9},cosdfrac{5pi}{9},cosdfrac{7pi}{9}$ 为$4cos^3 heta-3cos heta-dfrac{1}{2}=0$的三个根,即$cos^3 heta=dfrac{3}{4}cos heta+dfrac{1}{8}$;
    故$cos^5 heta=cos^2 hetaleft(dfrac{3}{4}cos heta+dfrac{1}{8} ight)=dfrac{3}{4}cos^3 heta+dfrac{1}{8}cos^2 heta=dfrac{9}{16}cos heta+dfrac{3}{32}+dfrac{1}{8}cos^2 heta$
    故$cos^5dfrac{pi}{9}+cos^5dfrac{5pi}{9}+cos^5dfrac{7pi}{9}$

    $=dfrac{1}{8}(cosdfrac{pi}{9}+cosdfrac{5pi}{9}+cosdfrac{7pi}{9})^2-dfrac{1}{4}(cosdfrac{pi}{9}cdotcosdfrac{5pi}{9}+cosdfrac{5pi}{9}cdotcosdfrac{7pi}{9}+cosdfrac{7pi}{9}cdotcosdfrac{pi}{9})+dfrac{9}{32}$
    $=dfrac{1}{8}cdot0^2-dfrac{1}{4}cdot(-dfrac{3}{4})+dfrac{9}{32}=dfrac{15}{32}$

    注:也可以用正余弦的快速降幂公式去做

    注:一般的$cos^ndfrac{pi}{9}+cos^ndfrac{3pi}{9}+cos^ndfrac{5pi}{9}+cos^ndfrac{7pi}{9}=dfrac{1}{2}$这里 n 为奇数.

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  • 原文地址:https://www.cnblogs.com/mathstudy/p/10355544.html
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