若sinα-cosα=2√2,(0扫码下载作业帮拍照答疑一拍即得

答

解
∵0∴-π/2∴0∴sin(b-a)>0
∵cos(b-a)=√2/10
∴sin(b-a)=7√2/10
∵sina
=2sina/2cosa/2
=(2sina/2cosa/2)/(sin^2a/2+cos^2a/2)
=(2tana/2)/(tan^2a/2+1)
=(2×1/2)/(1/4+1)
=4/5
∴cosa=√1-sin^2a=3/5
∴cosb
=cos[(b-a)+a]
=cos(b-a)cosa- sin(b-a)sina
= -√2/2
∵0∴b=135°
是否可以解决您的问题？

答

解
∵0∴-π/2∴0∴sin(b-a)>0
∵cos(b-a)=√2/10
∴sin(b-a)=7√2/10
∵sina
=2sina/2cosa/2
=(2sina/2cosa/2)/(sin^2a/2+cos^2a/2)
=(2tana/2)/(tan^2a/2+1)
=(2×1/2)/(1/4+1)
=4/5
∴cosa=√1-sin^2a=3/5
∴cosb
=cos[(b-a)+a]
=cos(b-a)cosa- sin(b-a)sina
= -√2/2
∵0∴b=135°

答

解
∵0

若sinα-cosα=2√2,(0扫码下载作业帮拍照答疑一拍即得

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