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Question Number 123385 by help last updated on 25/Nov/20

Answered by Dwaipayan Shikari last updated on 25/Nov/20

$${sin}^{6}1°+{cos}^{6}1°+{sin}^{6}2°+{cos}^{6}2°+...+{sin}^{6}n°+{cos}^{6}n°$$$$(1-3{sin}^{2}1°{cos}^{2}1°+1-3{sin}^{2}2°{cos}^{2}2°+...)+1$$$$=45-\frac{3}{4}({sin}^{2}2°+{sin}^{2}4°+...+{sin}^{2}90°)$$$$=45-\frac{3}{8}(1-cos4°+1-cos8°+....)$$$$=\frac{225}{8}+\frac{3}{8}\left(\underset{n=1}{\sum ^{45}}cos4n°\right)=\frac{222}{8}=\frac{111}{4}=27.75$$$$\underset{n=1}{\sum ^n}cosan°=cosa°+cos2a°+..$$$$=\frac{1}{2sin2°}(sin(a+2)°-sin(a-2)°+...+sin(na+2)-sin(na-2))$$$$=\frac{1}{2sin2°}(sin(na+2)-sin(a-2))$$$$=\frac{1}{sin2°}(cos\left((n+1)\frac{a}{2}\right)sin((n-1)\frac{a}{2}+2)$$$$a=4°n=45$$$$=-1$$$$$$$$$$

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