Question

最难得高一数学题不服来做

Answer 1

f（x）=(1+sin2x)*1+(sinx-cosx)*(sinx+cosx).
=1+sin2x-(cos^2x-sin^2x).
=1+sin2x-cos2x.
=1+√2sin(2x-π/4).
∴f(x)=√2sin(2x-π/4)+1.
【f（x）】max=2
x=3π/8+kπ，k∈Z.
cos2（π/4-2θ）=sin4θ
f(θ)=√2sin(2θ-π/4)+1=8/5.
√2sin(2θ-π/4)=3/5.
√2sin2θcos(π/4)-√2cos2θsin(π/4)=3/5.
sin2θ-cos2θ=3/5.
(sin2θ-cos2θ)^2=(3/5)^2.
sin^2(2θ)-2sin2θcos2θ+cos^2(2θ)=9/25.
1-sin4θ=9/25.
sin4θ=1-9/25.
∴sin4θ=16/25.

Answer 2

追答