tan²θ-tanθ+根號3-根號3tanθ=0,則θ=? θ<90°θ=多少度?
tanθ(tanθ-1)-√3(tanθ-1)=0,
(tanθ-1)(tanθ-√3)=0,
tanθ=1,
θ=45°
tanθ=√3,
或,θ=60°.
根號3tan(pai/6-θ)tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=? 根3*tan(pai/6-θ)*tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=? 用三角的恒等變換
tan(π/6-θ)=[ tanπ/6-tanθ]/(1+tanπ/6tanθ)= [1/√3 - tanθ]/(1+tanθ/√3)=(1-√3tanθ)/(√3+tanθ)tan(π/6+θ)=(1+√3tanθ)/(√3-tanθ)√3tan(π/6-θ)tan(π/6+θ)+tan(π/6-θ)+tan(π/6+θ)=√…
當a=6倍根號2,b=根號12,求根號a+b乘根號a-b
a=6√2,b=√12=2√3
√(a+b)*√(a-b)=√((a+b)*(a-b))
=√(a^2-b^2)
=√(72-12)
=√60
=2√15
求根號6+根號8+根號12+根號24的值
3根號6+2根號2+2根號3
根號3(3根號3减根號12)求解
√3×(3√3-√12)
=√3×(3√3-2√3)
=√3×√3
=3
求解根號12-根號18+3根號2分之1-2分之3根號3分之4
根號12-根號18+3根號2分之1-2分之3根號3分之4
=2√3-3√2+3√2/2-√3
=√3-3√2/2
求根號下13^2-12^2*根號下3^2+4^2
√13²-12²=√(13+12)(13-12)=√25=5
√3²+4²=√9+16=√25 =5
半型公式cos^2(a/2)=(1+cos(a))/2如何推導
[1+cos(a)]/2
=1/2+1/2cos(a/2+a/2)
=1/2+1/2[cos^2(a/2)-sin^2(a/2)]
=1/2-1/2sin^2(a/2)+1/2cos^2(a/2)
=1/2cos^2(a/2)+1/2cos^2(a/2)
=cos^2(a/2)
累死我了.
半型的正弦公式推導過程 請以兩角差的正弦公式sin(x-y)=sinxcosy-cosxsiny為已知條件,1、推導兩角和的余弦公式,2、進而推導半型的正弦公式. 兩角和的余弦公式不用推導了,求第二問.
首先推導出兩角和公式:sin(x+y)=sinxcosy+cosxsiny令x=θ/2,y=θ/2sin(θ/2+θ/2)=sinθ/2cosθ/2+cosθ/2sinθ/2得到:cosθ/2=sinθ/2sinθ/2sin(x-y)=sinxcosy-cosxsiny令x=θ,y=θ/2sin(θ-θ/2)=sinθ…
半型公式的推導過程
正弦,余弦正切:首先推導出兩角和公式:sin(x+y)=sinxcosy+cosxsiny令x=θ/2,y=θ/2 sin(θ/2+θ/2)=sinθ/2cosθ/2+cosθ/2sinθ/2得到:cosθ/2=sinθ/2sinθ/2 sin(x-y)=sinxcosy-cosxsiny令x=θ,y=θ/2 sin(θ-θ/2)=sinθcosθ/2-cosθsinθ/2 sinθ/2=sinθ(sinθ/2sinθ/2)-cosθsinθ/2 sin²θ/2=sin²θ/2(1+cosθ)sin²θ/2=(1-cos²θ)/2(1+cosθ)sin²θ/2=(1+cosθ)/2 sinθ/2=±√(1+cosθ)/2 cos(a+b)=coacosb-sinasinb令a=b=d cos2d=(cosd)^2-(sind)^2=(cosd)^2-[1-(cosd)^2]=2(cosd)^2-1所以(cosd)^2=(cos2d+1)/2以d/2代d,開方有cosd/2=±√[(1+cosd)/2]而cos2d=(cosd)^2-(sind)^2=[1-(sind)^2]-(sind)^2=1-2(sind)^2所以(sind)^2=(1-cos2d)/2同樣的方法有sind/2=±√[(1-cosd)/2] tand/2=(sind/2)/(cosd/2)=±√[(1-cosd)/(1+ cosd/2)]還有一個是tand=sin2d/(1+cos2d)=(1-cos2d)/sin2d,推導如下:tand=sind/cosd=(2sindcosd)/(2cosdcosd)=sin2d/2(cosd)^2=sin2d/(1+cos2d)tand=sind/cosd=(2sindsind)/(2cosdsind)=2(sind)^2/sin2d=(1-cos2d)/sin2d [最後一步用了C(2d)的變形]