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化简sin4x%sin2x+Cos2x=

sin2x+cos2x = √2·【cos45·sin2X + sin45·cos2X】 = √2·sin[2X + 45] 对吧 加油

cos4x-sin2x=1-2sin²2x-sin2x

D,你把sin2x先提出来,得到(sin2x+cos2x)sin2x+cos2x,然后接着做

sin2x+cos2x =√2(√2/2*sin2x+√2/2cos2x) =√2(sin2xcosπ/4+cos2xsinπ/4) =√2sin(2x+π/4)

因为:cos4x+sin4x+sin2xcos2xsin6x+cos6x+2sin2xcos2x=(sin 2x+cos 2x) 2?2sin 2xcos 2x +sin 2xcos 2x (sin 2x+cos 2x) [(sin 2x) 2?sin 2xcos 2x+(cos 2x ) 2]+ 2sin 2xcos 2x =1?sin 2xcos 2x (sin 2x) 2+(cos 2x) 2+sin 2xcos 2x =1?sin 2x...

方法1: 原式=∫sin⁴x cos²x =∫sin⁴x (1 - sin²x) dx =∫(sin⁴x - sin^6x) dx = ∫sin⁴x dx - ∫sin^6x dx 后面的看附图,自己整理吧 方法2: 原式=∫sin⁴x cos²x dx =∫sin²x (sinxcosx)² dx...

f(x)=sin(2x)+cos(2x) =√2[(√2/2)sin(2x)+(√2/2)cos(2x)] =√2[sin(2x)cos(π/4)+cos(2x)sin(π/4)] =√2sin(2x+π/4)

cos4x=cos²2x-sin²2x=cos2x+sin2x (cos2x+sin2x)(cos2x-sin2x)-(cos2x+sin2x)=0 (cos2x+sin2x)(cos2x-sin2x-1)=0 cos2x+sin2x=0,cos2x-sin2x=1 cos2x+sin2x=0 √2sin(2x+π/4)=0 2x+π/4=kπ x=kπ/2-π/8 cos2x-sin2x=1 sin2x-cos2x=-1 √2s...

(sin∧4xcos∧2x)的原函数是 ∫(sin∧4xcos∧2x)dx =∫[sin^2x(sinxcosx)^2dx =1/8∫(1-cos2x)(sin2x)^2dx =1/8∫(sin2x)^2dx-1/16∫(sin2x)^2d(sin2x) =1/16∫[1-cos4x]dx-1/48(sin2x)^3 =x/16-1/64*sin4x-1/48*(sin2x)^3+C 原函数的定义 primitive f...

思路,应用分部积分法 ∫(2xcos2x-sin2x)/4x^2dx =1/4∫1/xd(sin2x)-1/4∫sin2x/x^2dx =sin2x/(4x)+1/4∫sin2x/x^2dx-1/4∫sin2x/x^2dx+C =sin2x/(4x)+C

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