∫[-π 3,π 3]4cosθ的5次方dcosθ 用分部积分法求此定积分

π/2≤θ≤3π/4sinθ>0>cosθ所以cosθ-sinθ

tanα+cotα=-10/33(tanα)^2+10tanα+3=0(3tanα+1)(tanα+3)=0tanα=-1/3,或tanα=-3而：3π/4

tan(a-π)=tan(a+π)=tanatan(2π-a)=-tan(a-2π)=-tana所以有2+tana/1+tana=-4也即是2+tana=-4tana-4移项得5tana=-6tana

π/2

∵cos（x-π/6）=-√3/3∴cosx+cos（x-π/3）=cosx+cosxcosπ/3+sinxsinπ/3=cosx+（1/2）cosx+（√3/2）sinx=（3/2）cosx+（√3

sin(θ+kπ)=2cos[θ+(k+1)π]=-2cos(θ+kπ)tan(θ+kπ)=sin(θ+kπ)/(-2cos(θ+kπ))=-1/2tanθ=-1/24sinθ-2cosθ/5cosθ

已知tanθ=3(3cosθ-5sin^2θcosθ)/sin(π-θ)=-1/2楼上的：(3cosθ-5sin^2θcosθ)/sin(π-θ)=(3cosθ-5sin^2θcosθ)/sinθ=(

tanθ=√3/3,θ=kπ+π/6,k为偶数时cos^2θ-sin(θ+π/6)cosθ=3/4-√3/2*√3/2=0k为奇数时cos^2θ-sin(θ+π/6)cosθ=3/4-3/4=0

sin(θ+kπ)=-2cos(θ+kπ),可得tanQ=-24sinθ-2cosθ/5cosθ+3sinθ（分子分母同时除以cosQ）=10⑵(1/4)sin平方θ+(2/5)cos平方θ（分子分母

cos（π/6+θ）=cos（-（π/6+θ））=-cos（-（π/6+θ）+π）=-cos（5π/6-θ）所以：cos（5π/6-θ）=-cos（π/6+θ）=-根号3/3.

tan(θ+π/4)=3(1+tanθ)/(1-tanθ)=3tanθ=1/25sin^2θ-3sinθcosθ+2cos^2θ=(5sin^2θ-3sinθcosθ+2cos^2θ)/(sin^2θ

(cosa)^2=(cos2a+1)/2(cosa)^2=(cos2a+1)^2/4cos(π/8))^4=(cosπ/4+1)^2/4=[3+2√2]/8cos(3π/8))^4=(cos3π/4+

对没错~cosx的图像中,π/2是平衡点就是(0,0)因为0到π是减函数π/40,3π/5>π/2所以

=cosα+cos2π/3*cosα-sin2π/3*sinα+cos4π/3*cosα-sin4π/3*sinα=cosα-1/2cosα-（根号3）/2sinα-1/2cosα+（根号3）/2si

θ∈(0,π/2)sinθcosθθ∈(0,π/2)cosθ∈(0,1)又sinθ

解tan2θ=(2tanθ)/(1-tan^2θ)=3/4即3(1-tan^2θ)=8tanθ即3tan^2θ+8tanθ-3=0(3tanθ-1)(tanθ+3)=0∵π/2

cos^2θ+cos^2(θ+2π/3)+cos^2(θ-2π/3)=(1+cos2θ)/2+[1+cos(2θ+4π/3)]/2+[1+cos(2θ-4π/3)]/2=3/2+[cos2θ+cos(

[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2

原式=（-cotθcosθsin²θ）/（tanθcos²θ）=（-cotθsin²θ）/（tanθcosθ）=（-cotθsinθtanθ）/tanθ=-cotθsin