一道困扰N久的数学题...1/sin2x + 1/sin4x + 1/sin8x + 1/sin16x + 1/sin3
4个回答
可用这公式来算吧?
sin2x=2sinxcosx
相关问题
数学归纳法的证明题用数学归纳法证明:1 sin x+2 sin 2x+…+n sin nx=sin[(n+1)x]/4s
sinx+sin3x+sin5x+.sin(2n-1)/cosx+cos3x+cos5x+.cos(2n-1)= sin
-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
为什么f(x)=sin2x-2sin²x=sin2x+(1-2sin²x)-1=sin2x+cos2
(1+cos x-sin x)/(1-sin x-cos x)+(1-cos x-sin x)/(1-sin x+cos
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
化简√1+sin x /1-sin x -√1-sin x /1+sin x ,其中x为第2象限角
求证:(sin x+cos x−1)(sin x−cos x+1)sin
证 sin2x * cos3x =(1/2) * [sin5x + sin(-x)]