=2π÷2=π
a.b=cosa*tan(a+π/4)+2
=cosa*(1+tana)/(1-tana)+2
=cosa*(cosa+sina)/(cosa-sina)+2
=m
2cos2a+sin2(a+π)
=2cos2a+sin2a
=2cos^2a-2sin^2a+2sinacosa
=2cosa(cosa+sina)-2sin^2a
2cos2a+sin2(a+B)/cosa-sina
=m-2-2sin^2a/(cosa-sina)
高一数学三角恒等变换练习题已知0
=2π÷2=π
a.b=cosa*tan(a+π/4)+2
=cosa*(1+tana)/(1-tana)+2
=cosa*(cosa+sina)/(cosa-sina)+2
=m
2cos2a+sin2(a+π)
=2cos2a+sin2a
=2cos^2a-2sin^2a+2sinacosa
=2cosa(cosa+sina)-2sin^2a
2cos2a+sin2(a+B)/cosa-sina
=m-2-2sin^2a/(cosa-sina)