sin^2(x-π/4)=[1-cos2（x-π/

1个回答

cos2a=1-2sin²a
所以sin²a=(1-cos2a)/2
第一个等号就是这样来的
cos[2(x-π/4)]=cos(2x-π/2)=sin2x
所以最后应该等于(1-sin2x)/2

(1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin
fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x
√2sin(x/2+π/4)cos(x/2)=√2/2sin(x+π/4)+ 1/2 怎么得到的
已知cos(π/4+x)=4/5,x∈(-π/2,-π/4).求（sin2x-2sin²x)/(1+tanx)
y=sin(π/4+x/2)sin(π/4-x/2) =sin(π/4+x/2)sin[π/2-(π/4+x/2)]
(1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1 这个是怎么化简得的?