Question

高一数学三角恒等变换练习题

已知0<a<π/4,B为f(x)=cos(2x+π/8)的最小正周期，a=(tan(a+1/4B),1),b=(cosa,2),且a*b=m,求2cos2a+sin2(a+B)/cosa-sina的值

Answer 1

b=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)

Answer 2

b=π
cosatan(a+b/4)=m+2
a∈(0,π/4)
原式=(2cosa^2+sin(2a+2π))/(cosa-sina)
=(2cosa^2+sin2a)/(cosa-sina)
=(2cosa(cosa+sina))/(cosa-sina)
=2cosa*(1+tana)/(1-tana)
=2cosatan(a+π/4)
=2*(2+m)
（07,安徽高考题）