证明这题左边等于右边,谢谢
2个回答
展开全部
证:
cos(45°-x)/[1+sin(2x)]
=sin[90°-(45°-x)]/(sin²x+cos²x+2sinxcosx)
=sin(x+45°)/(sinx+cosx)²
=sin(x+45°)/ {2[(√2/2)sinx+(√2/2)cosx]²}
=sin(x+45°)/{2(sinxcos45°+cosxsin45°)²}
=sin(x+45°)/[2sin(x+45°)²]
=1/[2sin(x+45°)]
=½csc(x+45°)
cos(45°-x)/[1+sin(2x)]
=sin[90°-(45°-x)]/(sin²x+cos²x+2sinxcosx)
=sin(x+45°)/(sinx+cosx)²
=sin(x+45°)/ {2[(√2/2)sinx+(√2/2)cosx]²}
=sin(x+45°)/{2(sinxcos45°+cosxsin45°)²}
=sin(x+45°)/[2sin(x+45°)²]
=1/[2sin(x+45°)]
=½csc(x+45°)
追问
第三个等于号那里就不明白了
追答
就是构造sin(x+45°)
sinx+cosx
=√2[(√2/2)sinx+(√2/2)cosx]
=√2(sinxcos45°+cosxsin45°)
=√2sin(x+45°) (和差角公式)
(sinx+cosx)²=[√2sin(x+45°)]²=2sin²(x+45°)
cos(45°-x)/[1+sin(2x)]
=sin(x+45°)/[2sin²(x+45°)]
=1/[2sin(x+45°)]
=½csc(x+45°)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询