MT300余弦的三倍角公式

Posted 2021-02-08 mathstudy

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$cos^ndfrac{pi}{9}+cos^ndfrac{3pi}{9}cos^ndfrac{5pi}{9}+cos^ndfrac{7pi}{9}=dfrac{1}{2}$