Question

高一数学问题，需要详细过程，谢谢

cos^2 θ+cos^2 (θ+120°)+cos^2 (θ+240°)的值是？答案是3/2

Answer 1

你好
原式=cos^2 θ+cos^2 (θ+120°)+cos^2 (θ-120°)
=cos^2 θ+(-cosθ/2-√3sinθ/2)^2+(-cosθ/2+√3sinθ/2)^2（和分角公式）
=cos^2 θ+cos^2 θ/4+3sin^2 θ/4+cos^2 θ/4+3sin^2 θ/4(中间项消去)
=3cos^2 θ/2+3sin^2 θ/2
=3/2
希望能帮到你

Answer 2

cos^2a+cos^2(a+120°)+cos^2(a+240°)
=(cosa)^2+{-cos[180-(a+120)]}^2+{-cos[180-(a+240)]}^2
=(cosa)^2+[cos(60-a)]^2+[cos(60+a)]^2
=(cosa)^2+(cos60cosa+sin60sina)^2+(cos60cosa-sin60sina)^2
=(cosa)^2+(1/2*cosa+√3/2*sina)^2+(1/2*cosa-√3/2*sina)^2
=(cosa)^2+1/4(cosa)^2+√3/2sinacosa+3/4(sina)^2+1/4(cosa)^2-√3/2sinacosa+3/4(sina)^2
=3/2*(cosa)^2+3/2*(sina)^2
=3/2*[(sina)^2+(cosa)^2]
=3/2