Question

证明(sinA)^2+[sin(A+2π/3)]^2+[sin(A-2π/3)]^2=3/2

Answer 1

证明：因为(sinA)^2+[sin(A+2π/3)]^2+[sin(A-2π/3)]^2

=(sinA)^2+[sinAcos120度+sin120度cosA]^2+[sinAcos120度-sin120度cosA]^2

=(sinA)^2+2(sinAcos120度)^2+2(sin120度cosA)^2

=(sinA)^2+2(-1/2sinA)^2+2(√3/2cosA)^2

=(sinA)^2+1/2(sinA)^2+3/2(cosA)^2

=3/2(sinA)^2+3/2(cosA)^2

=3/2[(sinA)^2+(cosA)^2]

因为(sinA)^2+(cosA)^2=1

所以原式=3/2[(sinA)^2+(cosA)^2]=3/2

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