sin^2Xsin^2B+cos^2Xcos^2B-1/2cos2Xcos2B

zdt8002092022-10-04 11:39:541条回答

sin^2Xsin^2B+cos^2Xcos^2B-1/2cos2Xcos2B
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19117171 共回答了16个问题 | 采纳率93.8%: sin^2Xsin^2B+cos^2Xcos^2B-1/2cos2Xcos2B
=sin^2Xsin^2B+cos^2Xcos^2B-1/2(cos^2x-sin^2x)(cos^2B-sin^2B)
= sin^2Xsin^2B+cos^2Xcos^2B-1/2(cos^2xcos^2B-cos^2xsin^2B-sin^2xcos^2B+sin^2xsin^2B)
=1/2 sin^2Xsin^2B+1/2cos^2Xcos^2B+1/2cos^2xsin^2B+1/2sin^2xcos^2B
=1/2sin^2x(sin^2B+cos^2B)+1/2cos^2x(cos^2B+sin^2B)
=1/2sin^2x+1/2cos^2x
=1/2; 1年前