1+sin^2x)^1/2-1如何化简为1/2sin^2x
在mathematica里输入Plot[Sin[x] Sin[x + 2] - Sin[x + 1]Sin[x + 1]
如何将sin(2x)*cos(x)+sin^2(x) 展开为sin(x)的表达式?或者如何化简?
x*(1+sin^2 x )/sin^2x 不定积分
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1)
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
泰勒公式的为什么㏑( 1 + sin X ) = sin X - ( sin X )²/2 +(sin X )
1/sin^2x的不定积分谢谢!(是1/sin x * sin x)
s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)
求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2
三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x
5sin^2(X)+sin(2X)-cos^2(X)=1, 求解X