求解3角函数题需详细过程

1)
三角形ABC种，已知cos(π/4+A）=3/5>0
0<A<π/4
0<2A<π/2

cos(π/4+A）
=cosπ/4cosA-sinπ/4sinA
=(cosA-sinA)*√2/2
=3/5
(cosA-sinA)=3√2/5
(cosA-sinA)^2=18/25
1-2sinAcosA=18/25
sin2A=1-18/25=7/25
cos2A=√(1-sin^2 2A)=√(1-(7/25)^2)=24/25

2.
cos2xcosπ/3+sin2xsinπ/3-cos2x
=1/2*cos2x+sin2xsinπ/3-cos2x
=sin2xsinπ/3-1/2*cos2x
=sin2xsinπ/3-cosπ/3cos2x
=-cos(2x+π/3)

1. cos(π/4+A)=cosπ/4*cosA-sinπ/4*sinA
=√2/2*cosA-√2/2*sinA=3/5
所以 cosA-sinA=3√2/5
又因为 (cosA)^2+(sinA)^2=1
两式联立可以解得sinA
然后cos2A=1-2(sinA)^2 就知道答案了你自己算下sinA
2.π/3=60°
所以原式=cos2x*1/2+sin2x*√3/2-cos2x
=sin2x*√3/2-cos2x*1/2
=-cos(2x+π/3)