三角関数公式

曲面面積

[例題1] 放物面S={(x,y,z)| z=x²+y²≦4} の曲面の面積S1を求めよ。
[解答]
z=x²+y² ...(1)より
　z_x=2x, z_y=2y ...(2)
S → D:{(x,y)|x²+y²≦4} ...(3)
S1=∬_S dS ...(4)
=∬_D √{1+z_x²+z_y²} dxdy
=∬_D √{1+4x²+4y²} dxdy (4-1)
x=r cosθ, y=r sinθ ...(5) とおくと
　z=r²≦4 ...(6-1)
　0≦r≦2, 0≦θ≦2π ...(6-2)

D → E:{(r,θ)|0≦r≦2, 0≦θ≦2π} ...(7)
√{1+4x²+4y²} dxdy=√(1+4r²) rdrdθ ...(8)
であるから
S1=∬_E r√(1+4r²) drdθ
=∫[θ:0→2π] dθ∫[r:0→2] r√(1+4r²) dr
=2π[(2/3)(1/8)(1+4r^2)^3/2][r:0→2]
={17(√17)-1}π/6

[別解]
x=r cosθ, y=r sinθ とおくと
0≦r, 0≦θ≦2π)
0≦z=r² ≦4
r=√z,dr/dz=(1/2)/√z
S1=2π∫[0→4] r√(1+(dr/dz)²)^1/2 dz
=2π∫[0→4] √(z+(1/4)) dz
=2π[(2/3)(z+(1/4))^3/2]∫[0→4]
=(4π/3)(17√17-1)/8
=(17√17-1)π/6

[例題2] 球面の一部S={(x,y,z)| z²=x²+y²≦4, z≧√3 }の曲面の面積S1を求めよ。 [解答]
r=(x,y,z)=(x,y,√(4-x²-y²))
r_x=(1,0,-x/√(4-x²-y²))
r_y=(0,1,-y/√(4-x²-y²))
r_x×r_y=(x/_{√(4-x²-y²),y/}_{√(4-x²-y²),1)}
|r_x×r_y|=2/√(4-x²-y²)
S={√3≦z=√(4-x2-y2)≦2}→D={(x,y):x²+y²≦1}
S1=∫_S dS
=∬_D 2/√(4-x²-y²) dxdy
x=2rcosθ, y=2rsinθとおくと
D→E={(r,θ)|0≦r≦1/2,0≦θ≦2π}
|J|=|∂(x,y)/∂(r,θ)|
=|∂x/∂r ∂x/∂θ|=|2cosθ, -2rsinθ|
|∂y/∂r ∂y/∂θ| |2sinθ , 2rcosθ|
=4r
S1=∬_E 1/√(1-r²) 4rdrdθ
　=4∫[0→2π] dθ∫[0→1/2] r/√(1-r²) dr
　=8π [-√(1-r²) ]∫[0→1/2]
　=4π(2-√3)

[例題3] S={(x,y,z)| x²+y²+z²=4,z≧0,x²+y²≦2x} の曲面の面積S1を求めよ。
[解答]
S={(x,y,z)| x²+y²+z²=4,z≧0,x²+y²≦2x}
曲面S上の任意点(x,y,z)の位置ベクトル
r=(x,y,z)=(x,y,√(4-x2-y2))

3次元体積

スカラ面積分

ベクトル面積分

オイラーの公式	e^ix = cos x + i sin x ,e^-ix = cos x - i sin x
cos x = { e^ix + e^-ix }/2	sin x = {. e^ix - e^-ix }/(2i)
三角関数の公式 △ABCの各頂点の角をA, B, C x, ｙは A, B, Cのいずれかの角 sin²x + cos²x = 1 1 + tan² x = sec²x =1/ cos²x 1 + cot² x = cosec²x ( csc²x ) = 1/sin²x tan x = sin x / cos x sec x = 1/ cos x cosec x = 1/sin x (csc x = 1/sin x) cot x = 1/tan x = cos x /sin x	cos (-x) = cos x sin (-x) = -sin x tan (-x) = -tan x cot (-x) = -cot x sin (x + π/2) = cos x sin (x - π/2) = -cos x sin (π/2 - x) = cos x cos (x + π/2) = -sin x cos (x - π/2) = sin x cos (π/2 - x) = sin x tan (x + π/2) = -cot x tan (x - π/2) = -cot x tan (π/2 - x) = cot x sin (π-x) = sin x sin (π+x) = -sin x cos (π±x) =-cos x tan (π-x) = -tan x tan (π+x) = tan x cos (x + 90°) = -sin x cos (x - 90°) = cos (90°- x) = sin x sin (180°-x) = sin x sin (180°+x) = -sin x cos(90°±x) = -cos x
三角関数の置換
t=tan(x/2) ∫1/cos(x) dx=∫2/(1-t²)dt =∫dt/(1+t)+∫dt/(1-t) =ln \|(1+t)/(1-t)\|+C =ln \|(1+tan(x/2))/(1-tan(x/2))\|+C =ln \|{cos(x/2)+sin(x/2)}/{cos(x/2)-sin(x/2)}\|+C =(1/2)ln \|{1+sin(x)}/{1-sin(x)\|+C =ln \|{1+sin(x)}/cos(x) \| +C	sin x = 2t/(1+t²), cos x =(1-t²)/(1+t²) dx=2dt/(1+t²)
t=tan(x/2) ∫[0,2π] 1/{5-3 cos(x)}² dx =2∫[0,π] 1/{5-3 cos(x)}² dx =4∫[0,∞] (1+t²)/{5(1+t²)-3(1-t²)}² dt =4∫[0,∞] (1+t²)/(2+8t²)² dt =∫[0,∞] (1+t²)/(1+4t²)² dt =(1/2)∫[-∞,∞] (1+t²)/(1+4t²)² dt =(1/2)∫_C (1+z²)/(1+4z²)² dz =(1/2)2πi Res{(1+z²)/(1+4z²)² ,z=i/2} =πi (-5i/32)=5π/32	sin x = 2t/(1+t²), cos x =(1-t²)/(1+t²) dx=2dt/(1+t²) z=t 複素積分経路C=C1+C2 C1:z=-∞→∞ C2:z=R*exp(iφ), φ=0～π, R→∞ 留数計算 Res{(1+z²)/(1+4z²)² ,z=i/2}={-5i/32} Laurent展開 (1+z²)/(1+4z²)²=-3/{64(z-i/2)²}-5i/{32(z-i/2)}+13/64+(i/4)(x-i/2)+O((x-i/2)²)
t=tan(x) ∫1/cos(x)² dx=∫dt=t+C=tan^-1(x) +C	dt=dx/cos(x)²
t=sin(x) ∫1/cos(x) dx=∫1/(1-t²)dt =(1/2){∫dt/(1+t)+∫dt/(1-t)dt} =(1/2)ln \|(1+t)/(1-t)\|+C =(1/2)ln \|(1+sin(x))/(1-sin(x)) \| +C =ln \|{1+sin(x)}/cos(x)\|+C	dt=cos(x)dx
t=cos(x)	dt=-sin(x)dx
主な三角関数の値
sin(-π/2) = -1 sin 0 = 0 sin(π/12)=cos(5π/12)=(√6-√2)/4 sin(π/10)= cos(2π/5)=(√5-1)/4 sin(π/8) = cos(3π/8) = {√(2-√2)}/2 sin(π/6) = cos(π/3) = 1/2 sin(π/4) = cos(π/4) = 1/√2 sin(π/3) = cos(π/6) = √3 / 2 sin(3π/8) = cos(π/8) = {√(2+√2)}/2 sin(5π/12)=cos(π/12)=(√6+√2)/4 sin(π/2) = 1 sin(±π) = 0 cos 0 = 1 cos(±π) = -1 cos(±π/2) = 0 tan(0)=0 tan(π/12)=2-√3 tan(π/8) = √2-1 tan(π/6) = 1/√3 tan(π/4) = 1 tan(π/3) = √3 tan(3π/8) = √2+1 tan(5π/12)=2+√3 tan(π/2)=±∞	sin(-90°)= -1 sin 0°= 0 sin15°=cos75°=sin165°=(√6-√2)/4 sin18°=cos72°=(√5-1)/4 sin22.5°=cos67.5°={√(2-√2)}/2 sin30°= cos60°=sin150°=1/2 sin45°= cos45°=sin135°= 1/√2 sin60°= cos30°=sin120°= √3 / 2 sin(67.5°)=cos(22.5°)={√(2+√2)}/2 sin75°=cos15°=sin105°=(√6+√2)/4 sin90°= 1 sin(±180°) = 0 cos 0°= 1 cos(±90°) = 0 cos(±180°)= -1 tan0°=0 tan15°=2-√3 tan(22.5°) = √2-1 tan30° = 1/√3 tan45°= 1 tan60°= √3 tan(67.5°) = √2+1 tan75°=2+√3 tan90°=±∞
4sin(π/18)+√3 tan(π/18)=1	4sin10°+√3 tan10°=1
[演習] 4sin(π/18)+√3 tan(π/18)=1を導け。 [解答] 4sin(π/18)cos(π/18)=2sin(π/9) =2sin(π/6-π/18) =2{sin(π/6)cos(π/18)-cos(π/6)sin(π/18)} =cos(π/18) -　√3 sin(π/18) 従って 4sin(π/18)cos(π/18) =cos(π/18) -√3 sin(π/18) 両辺をcos(π/18)で割れば 4sin(π/10)=1-√3 tan(π/18) 移項して ∴4sin(π/18)+√3 tan(π/18)=1	[演習] 4sin10°+√3 tan10°=1を導け。 [解答] 4sin10°cos10°=2sin20° =2sin(30°-10°)=2(sin30°cos10°-cos30°sin10°) =cos10°-(√3)sin10° 従って 4sin10°cos10°=cos10°-(√3)sin10° 両辺をcos10°で割れば 4sin10°=1-(√3)tan10° 移項して ∴4sin10°+(√3)tan10°=1
●2倍角の公式	sin 2x = 2 sin x cos x cos 2x = 2 cos²x -1 = 1 - 2 sin²x tan 2x = 2tan x / (1+tan²x)
●３倍角の公式	sin 3x = 3 sin x - 4 sin³x cos 3x = 4 cos³x - 3 cos x
●４倍角の公式	sin 4x = 4 (sin x cos³x - sin³x cos x) = 4sin x cos x(2cos²x -1) = 4sin x cos x(1-2sin^2 x) cos 4x = 8 sin⁴x -7
●5倍角の公式	cos(5x)=16 cos⁵(x)-20 cos³(x)+5 cos(x) sin(5x)=16 sin(x)cos⁴(x)-12 sin(x)cos²(x)+sin(x)
●6倍角の公式	cos(6x)=32 cos⁶(x)-48 cos⁴(x)+18 cos²(x)-1 sin(6x)=32 sin(x)cos⁵(x)-32 sin(x)cos³(x)+6 sin(x)cos(x)
[Maple10]での7倍角公式の導出＞ expand(sin(7x, trig); 　　expand(cos(7x), trig);	64 sin(x) cos⁶(x) - 80 sin(x) cos⁴(x) + 24 sin(x) cos²(x) - sin(x) 64 cos⁷(x) - 112 cos⁵(x) + 56 cos³(x) - 7 cos(x)
●半角の公式	sin²x = (1 - cos 2x) / 2 cos²x = (1 + cos 2x) / 2
●加法定理 sin (x+y) sin (x-y) sin(x)+sin(y) sin(x)-sin(y) cos (x+y) cos (x-y) cos(x)+cos(y) cos(x)-cos(y) tan(x+y)	sin(x+y) = sin x cos y + cos x sin y sin(x-y) = sin x cos y - cos x sin y sin(x)+sin(y)=2sin[(x+y)/2}cos{(x-y)/2} sin(x)-sin(y)=2cos[(x+y)/2}sin{(x-y)/2} cos (x+y) = cos x cos y - sin x sin y cos (x-y) = cos x cos y + sin x sin y cos(x)+cos(y)=2cos[(x+y)/2}cos{(x-y)/2} cos(x)-cos(y)=-2sin[(x+y)/2}sin{(x-y)/2} tan(x+y)={tan(x)+tan(y)}/{1-tan(x)tan(y)}
sin(A+B+C) cos(A+B+C) tan(A+B+C)	=cos(A)cos(B)sin(C)+cos(A)sin(B)cos(C)+sin(A)cos(B)cos(C)-sin(A)sin(B)sin(C) =cos(A)cos(B)cos(C)-cos(A)sin(B)sin(C)-sin(A)cos(B)sin(C)-sin(A)sin(B)cos(C) ={tan(C)+tan(B)+tan(A)-tan(A)tan(B)tan(C)}/{1-tan(B)tan(C)-tan(A)tan(C)-tan(A)*tan(B)}
●積和公式	2sin x cos y = sin(x+y) + sin(x-y) 2cos x sin y = sin(x+y) - sin(x-y) 2cos x cos y = cos(x+y) + cos(x-y) 2sin x sin y = cos(x-y) - cos(x+y)
cos(x)+cos(2x)+cos(3x)+…+cos(nx) sin(x)+sin(2x)+sin(3x)+…+sin(nx)	[sin{(n+(1/2))x}-sin(x/2)]/{2sin(x/2)} [cos(x/2)-cos{(n+(1/2))x}]/{2sin(x/2)}
sin (π/10)+sin(2π/10)+sin(3π/10)+…+sin(19π/10)=0 cos (π/10)+cos(2π/10)+cos(3π/10)+…+cos(19π/10)=-1	導出法: x = e^{i π/10} を次式に代入, {x²⁰ - 1}/(x-1) = (1+x+x²+…+x¹⁹ ) = 0
sin (π/10)-sin(2π/10)+sin(3π/10)-…-sin(9π/10)=0 cos (π/10)-cos(2π/10)+cos(3π/10)-…-cos(9π/10)=1	導出法: x = e^{i π/10}を次式に代入, { x¹⁰+1}/(x+1) = (1-x+x²-…+x⁹ ) = 0
sin (π/20)+sin(2π/20)+sin(3π/20)+…+sin(39π/20)=0 cos (π/20)+cos(2π/20)+cos(3π/20)+…+cos(39π/20)=-1	導出法: x = e^{i π/20} を次式に代入, { x⁴⁰-1}/(x+1) = (1+x+x²+…+x³⁹ ) = 0
sin (π/50)+sin(2π/50)+sin(3π/50)+…+sin(99π/50)=0 cos (π/50)+cos(2π/50)+cos(3π/50)+…+cos(99π/50)=-1	導出法: x = e^{i π/50} を次式に代入, { x¹⁰⁰-1}/(x+1) = (1+x+x²+…+x⁹⁹) = 0
sin (π/100)+sin(2π/100)+…+sin(199π/100)=0 cos (π/100)+cos(2π/100)+…+cos(199π/100)=-1	導出法: x = e^{i π/100} を次式に代入, {x²⁰⁰-1}/(x-1) = (1+x+x²+…+x¹⁹⁹ ) = 0
●三角関数の逆関数 sin^-1x : x=－π/2～π/2 cos^-1x : x= 0～π tan^-1x : x=－π/2～π/2 (a＞0)	sin^-1x + cos^-1x = π/2 sin^-1x = cos^-1√(1- x²) cos^-1x = sin^-1√(1-x²) sin^-1x/a = cos^-1(1/a)√(a² -x²) = tan^-1x/√(a² -x²)
sin^-1y/√(x²+y²) = cos^-1x/√(x²+y²) = tan^-1y/x	sin^-1x/√(x²+y²) = cos^-1y/√(x²+y²) = tan^-1x/y

座標系
一次元座標系（一変数積分・線積分・複素積分)	積分素 dx, dy, dt, dz, ds, dA
二次元座標系	面積素dS, dS =n dS
直交座標系 (x,y)	面積素dS=dxdy
極座標系円座表 (r,θ)：　x=r cosθ, y=r sinθ 　r=√(x²+y²), cosθ=x/r, sinθ=y/r 　r≧0, 0≦θ≦2π or -π≦θ≦π	面積素dS=r dr dθ \|J\|=\|∂(x,y)/∂(r,θ)\| J=\|∂x/∂r,∂x/∂θ\| =\|cosθ,-r sinθ \|= r \|∂y/∂r,∂y/∂θ\| \| sinθ , r cosθ\|
三次元座標系	体積素dV 面積積分：面積素dS=\|r_x×r_y\| dxdy, r=(x,y,z)は曲面上の点(x,y,z)の位置ベクトル
直交座標系 (x,y,z) 位置ベクトルr=(x,y,z)=xi +yj +zk	体積素dV=dxdydz 面積積分：面積素\|dS\|=dS=\|r_x×r_y\| dxdy=√(1+f_x²+f_y²) dxdy, z=f(x,y) 　z=f(x,y)のとき r=(x,y,z), r_x=(1,0,f_x), r_y=(0,1,f_y) 　r_x×r_y=(-f_x,-f_y,1),\|r_x×r_y\| =√(1+f_x²+f_y²) 　dS=ndS, n=(r_x×r_y)/\|r_x×r_y\|=(-f_x,-f_y,1)/√(1+f_x²+f_y²)
極座標：球座標系(r,θ,φ): x=r cosφsinθ, y=r sinφsinθ,z=r cosθ 　r=√(x²+y²+z²), cosθ=z/r, sinθ=√(x²+y²)/r 　cosφ=x/√(x²+y²), sinφ=y/√(x²+y²) 　r≧0,0≦θ≦π, 　0≦φ≦2π　or　-π≦φ≦π	体積素dV=\|J\|drdφdθ \|J\|:Jacobian 　ヤコビ行列Jの行列式の絶対値 J=∂(x,u,z)/∂(r,θ,φ)= \|cosφsinθ , sinφsinθ, cosθ\| \|rcosφcosθ,rsinφcosθ,-rsinθ\| \|-rsinφsinθ, rcosφsinθ, 0 \| \|cosφsinθ, sinφsinθ, cosθ \| \|J\|=r²sinθ \|cosφcosθ,sinφcosθ,-sinθ\| =r²sinθ \| -sinφ, cosφ, 0 \|
極座標：円柱座標(円筒座標)系(r,θ,z): 　x=r cosθ, y=r sinθ, z=z 　r=√(x²+y²), cosθ=x/r, sinθ=y/r 　r≧0, 0≦θ≦2π　or　-π≦θ≦π	体積素dV=\|J\|drdφdθ \|J\|:Jacobian 　ヤコビ行列[J]の行列式\|J\|の絶対値 [J]=∂(x,y,z)/∂(r,θ,z) \|cosθ , -r sinθ, 0\| \|J\|=\| sinθ, r cosθ, 0 \| = r \| 0, 0, 1 \|
回転体表面積S1=∬_SdS	x軸のまわりの回転体表面積　曲面半径y=f(x) or r=f(x) 　逆関数x=g(r) 　S1=2π∫[a→b] f(x)√(1+(f'(x))²)dx 　S1=∬_D √(1+(g'(r))²) rdrdθ z軸のまわりの回転体表面積　曲面半径x=f(z) or r=f(z) 　逆関数z=g(r) 　S1=2π∫[a→b] f(z)√(1+(f'(z))²)dz 　S1=∬_D √(1+(g'(r))²) rdrdθ
曲面面積S1=∬S dS 曲面を表す関数：z=f(x,y) 曲面上の任意点(x,y,z)の位置ベクトル：r=(x,y,z)	S1=∬_D √(1+z_x²+z_y²) dxdy=∬_D \|r_x×r_y\| dxdy
スカラ面積積曲面z=f(x,y) S2=∬S　φ(x,y,f(x,y)) dS	S2=∬_D φ(x,y,f(x,y))√(1+z_x²+z_y²) dxdy
ベクトル面積積分 A：ベクトル場, n：曲面S上の任意点の単位法線ベクトル S3=∬_S　A・dS=∬_S　(A・n)dS	r=(x,y,z)：曲面上の任意点の位置ベクトル r_x=(1,0,z_x), r_y=(0,1,z_y), S3=∬_D A・(r_x×r_y) dxdy 　=∬_D (A・n) √(1+z_x²+z_y²) dxdy

曲面面積
[例題1] 放物面S={(x,y,z)\| z=x²+y²≦4} の曲面の面積S1を求めよ。	[解答] z=x²+y² ...(1)より　z_x=2x, z_y=2y ...(2) S → D:{(x,y)\|x²+y²≦4} ...(3) S1=∬_S dS ...(4) =∬_D √{1+z_x²+z_y²} dxdy =∬_D √{1+4x²+4y²} dxdy (4-1) x=r cosθ, y=r sinθ ...(5) とおくと　z=r²≦4 ...(6-1) 　0≦r≦2, 0≦θ≦2π ...(6-2) D → E:{(r,θ)\|0≦r≦2, 0≦θ≦2π} ...(7) √{1+4x²+4y²} dxdy=√(1+4r²) rdrdθ ...(8) であるから S1=∬_E r√(1+4r²) drdθ =∫[θ:0→2π] dθ∫[r:0→2] r√(1+4r²) dr =2π[(2/3)(1/8)(1+4r^2)^3/2][r:0→2] ={17(√17)-1}π/6 [別解] x=r cosθ, y=r sinθ とおくと 0≦r, 0≦θ≦2π) 0≦z=r² ≦4 r=√z,dr/dz=(1/2)/√z S1=2π∫[0→4] r√(1+(dr/dz)²)^1/2 dz =2π∫[0→4] √(z+(1/4)) dz =2π[(2/3)(z+(1/4))^3/2]∫[0→4] =(4π/3)(17√17-1)/8 =(17√17-1)π/6
[例題2] 球面の一部S={(x,y,z)\| z²=x²+y²≦4, z≧√3 }の曲面の面積S1を求めよ。	[解答] r=(x,y,z)=(x,y,√(4-x²-y²)) r_x=(1,0,-x/√(4-x²-y²)) r_y=(0,1,-y/√(4-x²-y²)) r_x×r_y=(x/_{√(4-x²-y²),y/}_{√(4-x²-y²),1)} \|r_x×r_y\|=2/√(4-x²-y²) S={√3≦z=√(4-x2-y2)≦2}→D={(x,y):x²+y²≦1} S1=∫_S dS =∬_D 2/√(4-x²-y²) dxdy x=2rcosθ, y=2rsinθとおくと D→E={(r,θ)\|0≦r≦1/2,0≦θ≦2π} \|J\|=\|∂(x,y)/∂(r,θ)\| =\|∂x/∂r ∂x/∂θ\|=\|2cosθ, -2rsinθ\| \|∂y/∂r ∂y/∂θ\| \|2sinθ , 2rcosθ\| =4r S1=∬_E 1/√(1-r²) 4rdrdθ 　=4∫[0→2π] dθ∫[0→1/2] r/√(1-r²) dr 　=8π [-√(1-r²) ]∫[0→1/2] 　=4π(2-√3)
[例題3] S={(x,y,z)\| x²+y²+z²=4,z≧0,x²+y²≦2x} の曲面の面積S1を求めよ。	[解答] S={(x,y,z)\| x²+y²+z²=4,z≧0,x²+y²≦2x} 曲面S上の任意点(x,y,z)の位置ベクトル r=(x,y,z)=(x,y,√(4-x2-y2))

3次元体積