利用三角公式sin(a+b)+sin(a-b)=2*sin(a)cos(b),令a=2x,b=3x,则sin(5x)+sin(-x) = 2sin(2x)cos(3x),即sin(2x)cos(3x)=1/2*[sin(5x)+sin(-x)].
证 sin2x * cos3x =(1/2) * [sin5x + sin(-x)]
1个回答
相关问题
-
sinx+sin3x+sin5x+.sin(2n-1)/cosx+cos3x+cos5x+.cos(2n-1)= sin
-
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
-
（sin3xsinx）sin2x+（cos3xcosx）cos2x =1/ 2 [（cos2x-cos4x）sin2x+
-
(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
-
若sin(x-π)=2cos(2π-x),求(sin(π-x)+5cos(2π-x))/(3cos(π-x)-sin(-
-
sin(5x)-sin(3x)= 根号2cos(4x)
-
化简cos3/2x*sin3/2x-cos1/2x*sin1/2x
-
(1+sin2x-cos2x)/1+sin2x+cos2x
-
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin