作业帮已知sina-cosa=1/2,则sin²a-cos²a=?
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已知sina-cosa=1/2,则sin²a-cos²a=?
已知sina-cosa=1/2,则sin²a-cos²a=?
已知sina-cosa=1/2,则sin²a-cos²a=?
sina-cosa=1/2
平方
sin²a+cos²a-2sinacosa=1/4
1-2sinacosa=1/4
2sinacosa=3/4
(sina+cosa)²
=sin²a+cos²a+2sinacosa
=1+3/4
=7/4
sina+cosa=±√7/2
所以原式=(sina+cosa)(sina-cosa)=±√7/4
(cosa-cosb)^2=1/4 (sina-sinb)^2=1/9
两式相加得
cos^2a-2cosacosb+cos^2b+sin^2a-2sinasinb+sin^2b=13/36
2-2cos(a-b)=13/36
cos(a-b)=1-13/72=59/72
由已知 cosa=c...
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(cosa-cosb)^2=1/4 (sina-sinb)^2=1/9
两式相加得
cos^2a-2cosacosb+cos^2b+sin^2a-2sinasinb+sin^2b=13/36
2-2cos(a-b)=13/36
cos(a-b)=1-13/72=59/72
由已知 cosa=cosb+1/2 cos^2a=cos^2b+cosb+1/4 (1)
sina=sinb-1/3 sin^2a=sin^2b-2sinb/3+1/9 (2)
(1)+(2)得 1=1+cosb-2sinb/3+13/36
sinb=3cosb/2+13/24
sina=3cosb/2+13/24-1/3=3cosb/2+5/24
所以 1-cos^2b=9cos^2b/4+39cosb/24+169/576
13cos^2b/4+13cosb/8=407/576
2cosb^2+cosb=407/936
sin(a+b)=sinacosb+sinbcosa=(3cosb/2+5/24)cosb
+(3cosb/2+13/24)*(cosb+1/2)
=3cos^2b+3cosb/2+13/48
=3/2*(2cos^2b+cosb)+13/48
=407/624+13/48=(407+13*13)/624=576/624=12/13
cos(a+b)=cosacosb-sinasinb=9(cosb+1/2)cosb-(3cosb/2
+5/24)(3cosb/2+13/24)
=-(5/8)(2cos^2b+cosb)-65/576
=-5/13
收起
(sina-cosa)^2=1/4 2sinacosa=3/4
(sina+cosa)^2=7/4 sina+cosa=±√7/2
sin²a-cos²a=(sina+cosa)(sina-cosa)=±√7/4