求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程

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求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程求(cosB)^2+[c

求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程
求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程

求(cosB)^2+[cos(2π/3-B)]^2=1-sin(2B-π/6)的化简过程
(cosB)^2+[cos(2π/3-B)]^2=(1+cos2B)/2+[1+cos(4π/3-2B)]/2=1+(cos2B+cos4π/3cos2B+sin4π/3sin2B)/2=1+(cos2B/2-根号3/2×sin2B)/2=1-sin(2B-π/6)/2