等式2cosπ/4=√2,2cosπ/8=√(2+√2),2cosπ/16=√(2+√(2+√2))…请从中归纳出一般结论.

等式2cosπ/4=√2,2cosπ/8=√(2+√2),2cosπ/16=√(2+√(2+√2))…请从中归纳出一般结论. 高二推理证明证明等式2cosπ/4=√2,2cosπ/8=√(2+√2),2cosπ/16=√(2+√(2+√2))…并从中归纳出一般结论 cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17) 计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) cos^2 π/8+cos ^2 3π/8+ cos^2 5π/8+cos^2 7π/8=? COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程 求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) 求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) cos 2π/7 +cos 4π/7 +cos 6π/7=? 请问cos(2π/7)+cos(4π/7)+cos(6π/7)=?, 是否存在α∈（-π/2,π/2）,β∈(0,π),使等式sin(3π-α)=√2cos(π/2-β),√3cos(-α)=-√2cos(π+β 证明下列等式,并从中归纳出一个一般性的结论 会的人帮下忙吧、2cos∏/4=√22cos∏/8=√2+√22cos∏/16=√2+√2+√2 cos(2x-π/2)=sin2x对吗?这个等式对吗?为什么 cos^6(π/8)-sin^6(π/8)=求值,[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-[sin^2(π/4)]/4]==7√2/16 数学 解析 有个步骤不明白S1:cos2α/sin(α-π/4)=cos²α-sin²α/((√2)/2*(sinα-cosα))=-(√2)/2 等式左边同时约去sinα-cosα,可得: S2:√2(sinα+cosα)=-(√2)/2 就是这步啦 S1怎么变到S2的~ 越详细越好 cosacosb=1/2[cos(a+b)+cos(a-b)]等式为什么成立 若cos(π/4-θ)cos(π/4+θ)=√2/6,且0