1.sina+cosa=(√2)⼀2(0<a<π)求cos2a 2.sina+cosa=1⼀5 (0<a<π)求tana

2024-11-16 06:46:37
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回答1:

sina+cosa=(√2)/2
cosa=(√2)/2-sina(平方)
cos²a=1/2-√2sina+sin²a
cos²a=1-sin²a
1/2-√2sina+sin²a=1-sin²a
2sin²a-√2sina-1/2=0
sin²a-√2sina/2-1/4=0
sin²a-√2sina/2+2/16-2/16-1/4=0
(sina-√2/4)^2-6/16=0
(sina-√2/4-√6/4)(sina-√2/4+√6/4)=0
sina=√2/4+√6/4或sina=√2/4-√6/4
由a的范围,sina>0
sina=√2/4+√6/4
cos2a=1-2sin²a
=1-2(√2/4+√6/4)
=1-2*(2/16+6/16+2√12/16)
=1-2*(1/2+√3/4)
=1-1-√3/2
=-√3/2

sina+cosa=1/5
(sina)^2+(cosa)^2=1
00
解得sina=4/5,cosa=-3/5
tga=sina/cosa=-4/3