『急求』1/2sin2x+1/2cos2x怎么等于

『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)

1个回答

(1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin
y=sin2x-cos2x+1为什么等于√2sin（2x-π/4）+1
sin^2(x-π/4)=[1-cos2（x-π/4）]/2=(1+sin2x)/2
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
√2sin(x/2+π/4)cos(x/2)=√2/2sin(x+π/4)+ 1/2 怎么得到的
(1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1 这个是怎么化简得的?
1.y=cos^4x+sin^4x 求周期 2.y=（sin2x+sin（2x+π/3)）/（ cos2x+cos（2x
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(
已知tan(x-π/4）=2 求（sin2x+cos2x）/（2cos²x-3sin2x-1)