Chứng minh rằng: ( sin³ x . sin³ x ) + ( cos³ x . cos³ x ) = 1 - 3sin² x . cos² x
2 câu trả lời
Ta xét
VT=(sin3x.sin3x)+(cos3x.cos3x)
=sin6x+cos6x
=(sin2x)3+(cos2x)3
=(sin2x+cos2x)(sin4x+cos4x−sin2xcos2x)
=sin4x+cos4x−sin2xcos2x
=(sin2x)2+(cos2x)2+2sin2xcos2x−2sin2xcos2x−sin2xcos2x
=(sin2x+cos2x)2−3sin2xcos2x
=1−3sin2xcos2x=VP.
VT=(sin3a.sin3a)+(cos3a.cos3a)
=cos6a+sin6a
=(cos2a)3+(sin2a)3
=(cos2a+sin2a)(cos4a−cos2a.sin2a+sin4a)
=cos4a+sin4a−sin2acos2a
=(cos2a+sin2a)2−2sin2acos2a−sin2acos2a
=1−3sin2acos2a
=VP
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