求证(1)sin4平方a+sin²acos²a+cos²a=1(2)tan²a-sin²a=tan²asin²a
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![求证(1)sin4平方a+sin²acos²a+cos²a=1(2)tan²a-sin²a=tan²asin²a](/uploads/image/z/1963159-7-9.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%881%EF%BC%89sin4%E5%B9%B3%E6%96%B9a%2Bsin%26%23178%3Bacos%26%23178%3Ba%2Bcos%26%23178%3Ba%3D1%EF%BC%882%EF%BC%89tan%26%23178%3Ba-sin%26%23178%3Ba%3Dtan%26%23178%3Basin%26%23178%3Ba)
求证(1)sin4平方a+sin²acos²a+cos²a=1(2)tan²a-sin²a=tan²asin²a
求证(1)sin4平方a+sin²acos²a+cos²a=1(2)tan²a-sin²a=tan²asin²a
求证(1)sin4平方a+sin²acos²a+cos²a=1(2)tan²a-sin²a=tan²asin²a
证明∶
(1)
sin⁴a+sin²acos²a+cos²a
=sin²a(sin²a+cos²a)+cos²a
=sin²a×1+cos²a
=sin²a+cos²a
=1
得证;
(2)
tan²a-sin²a
=sin²a/cos²a-sin²a
=sin²a(1/cos²a-1)
=sin²a[(1-cos²a)/cos²a]
=sin²a[sin²a)/cos²a]
=sin²atan²a
=tan²asin²a
得证;