sin ^2 2a+sin 2acos a-cos 2a=1,a∈(0,2/π)求tan a
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sin ^2 2a+sin 2acos a-cos 2a=1,a∈(0,2/π)求tan a
sin ^2 2a+sin 2acos 2a-cos 2a=1;
sin 2acos 2a-cos 2a=1-sin ^2 2a=cos^2 2a ;
sin 2a-1=cos 2a,或cos 2a=0;
当cos 2a=0时,a=π/4;则tan a=1.;
当sin 2a-1=cos 2a时,
sin2a+cos2a=1;
√2sin(2a+π/4)=1;
sin(2a+π/4)=√2/2;
2a+π/4=3π/4;
仍然是a=π/4;则tan a=1
sin 2acos 2a-cos 2a=1-sin ^2 2a=cos^2 2a ;
sin 2a-1=cos 2a,或cos 2a=0;
当cos 2a=0时,a=π/4;则tan a=1.;
当sin 2a-1=cos 2a时,
sin2a+cos2a=1;
√2sin(2a+π/4)=1;
sin(2a+π/4)=√2/2;
2a+π/4=3π/4;
仍然是a=π/4;则tan a=1
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