6cos(2kπ+π/3)-2sin(2kπ+π/6)+3tan(2kπ) k€Z

cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα 化简sin(kπ+π/3)cos(3kπ+2π/3) 已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)sin平方θ+(2/5)cos平方θ已知sin(θ+kπ)=-2cos (θ+kπ)求⑴ 4sinθ-2cosθ/5cosθ+3sinθ⑵(1/4)sin平方θ+(2/5)cos平方θ 已知sin(θ+kπ)=-2cos(θ+kπ),k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ①4sinθ-2cosθ/5cosθ+3sinθ ②sin²θ+2/5cos²θ 化简：sin(kπ-2)cos[(k-1)π-2]/sin[(k+1)π+2]cos(kπ+2),k属于Z 6cos(2kπ+π/3)-2sin(2kπ+π/6)+3tan(2kπ) k€Z [sinπ(/2+α)-cos(3π/2-α)]/[tan(2kπ -α)+cot(-kπ+α)]=[sin(4kπ-α)sin(π/2 -α)]/[cos(5π+α)-cos(π/2+α) 已知sin(θ+kπ)=2cos[θ+(k+1)π],k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ的值 已知sin(θ＋kπ)=2cos[θ＋（k+1）π],k∈Ζ,求4sinθ-2cosθ/5cosθ+3sinθ的值 [sin(a+2kπ)+cos(π/2+a)+tan(3π-a)]/[sin(a-π)+cos(a-π/2)+cos(π/2-a). 求证sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)-cos(π/2 + α)谁帮我做做 2(sin a)^2+(2sin a*cos a)/(1+tan a)=k试用k表示sin a-cos aa∈(π/4，π/2） 已知sin(θ+kπ)=-2cos(θ+kπ)(k∈Z) 求1/4sin^2θ+2/5cos^2θ 利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ sin(a+kπ)=2cos[a+(k+1)π].求5cosa+3sina/4sina-2cosa sin(a+kπ)=2cos[a+(k+1)π].求5cosa+3sina/4sina-2cosa 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) 已知sin(α-3π)+cos(π/2+α)=m,求证cos(α-7π/2)+2sin(2kπ-α)=3m/2