化简(1)[1-根号2*sin(2a-pai/4)]/cosa (2) (2cos^4x-2cos^2x+1/2)/2t
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/18 21:56:09
化简(1)[1-根号2*sin(2a-pai/4)]/cosa (2) (2cos^4x-2cos^2x+1/2)/2tan(pai/4-x)sin^2(pai/4+x)
(1)sin(2a-π/4)=sin2acos(π/4)-cos2asin(π/4)=√2/2(sin2a-cos2a)
原式=[1-√2*√2/2(sin2a-cos2a)]/cosa
=(1-sin2a+cos2a)/cosa
=(2cos²a-2sinacosa)/cosa
=2cosa-2sina
=2√2(√2/2cosa-√2/2sina)
=2√2cos(a+π/4)
(2)分母=2tan(π/4-x)*sin^2(π/4+x)
=2cot(x+π/4)*sin^2(x+π/4)=2sin(x+π/4)cos(x+π/4)=sin(2x+π/2)=cos2x
分子=2cos²x(cos²x-1)+1/2
=1/2*4cos²x*(-sin²x)+1/2
=1/2(1-sin²2x)
=1/2cos²2x
原式=(1/2cos²2x)/cos2x=1/2cos2x
原式=[1-√2*√2/2(sin2a-cos2a)]/cosa
=(1-sin2a+cos2a)/cosa
=(2cos²a-2sinacosa)/cosa
=2cosa-2sina
=2√2(√2/2cosa-√2/2sina)
=2√2cos(a+π/4)
(2)分母=2tan(π/4-x)*sin^2(π/4+x)
=2cot(x+π/4)*sin^2(x+π/4)=2sin(x+π/4)cos(x+π/4)=sin(2x+π/2)=cos2x
分子=2cos²x(cos²x-1)+1/2
=1/2*4cos²x*(-sin²x)+1/2
=1/2(1-sin²2x)
=1/2cos²2x
原式=(1/2cos²2x)/cos2x=1/2cos2x
化简:1,sin(pai/4-x)sin(pai/4+x) 2,cosA+cos(120度-A)+cos(120度+A)
化简(1)[1-根号2*sin(2a-pai/4)]/cosa (2) (2cos^4x-2cos^2x+1/2)/2t
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