| オイラーの公式 |
eix = cos x + i sin x ,e-ix = cos x - i sin x |
| cos x = { eix + e-ix }/2 | sin x = {. eix - 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 |
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| do 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] 4*sin(π/18)*cos(π/18)=2sin(π/9) =2*sin(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 4*sin(π/18)*cos(π/18) =cos(π/18) -√3 sin(π/18) 両辺をcos(π/18)で割れば 4*sin(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 |
| (cos
x + i sin x)n=cos(nx)+i sin(nx) 公式の導出[Maxima] (%i1) f:(cos(x)+%i*sin(x))^6;realpart(f);imagpart(f); (%o1) (%i*sin(x)+cos(x))^6 (%o2) -sin(x)^6+15*cos(x)^2*sin(x)^4-15*cos(x)^4*sin(x)^2+cos(x)^6 (%o3) 6*cos(x)*sin(x)^5-20*cos(x)^3*sin(x)^3+6*cos(x)^5*sin(x) (%i4) f:(cos(x)+%i*sin(x))^7;realpart(f);imagpart(f); (%o4) (%i*sin(x)+cos(x))^7 (%o5) -7*cos(x)*sin(x)^6+35*cos(x)^3*sin(x)^4-21*cos(x)^5*sin(x)^2+cos(x)^7 (%o6) -sin(x)^7+21*cos(x)^2*sin(x)^5-35*cos(x)^4*sin(x)^3+7*cos(x)^6*sin(x) |
cos(nx)=cosnx -nC2cosn-2x
sin2x +...+(-1)knC2kcosn-2kx sin2kx+... +(-1)n/2sinnx (n=even) cos(nx)=cosnx -nC2cosn-2x sin2x +...+(-1)knC2kcosn-2kx sin2kx+... +(-1)(n-1)/2nCn-1cos x sinn-1x (n=odd) sin(nx)=nC1cosn-1x sin x -nC3cosn-3x sin3x +...+(-1)k-1nC2k-1cosn-2k-1x sin2k-1x+... +(-1)n/2nCn-1cos x sinn-1x (n=even) sin(nx)=nC1cosn-1x sin x +...+(-1)(k-1)nC2k-1cosn-2k+1x sin2k-1x+... +(-1)(n-1)/2sinnx (n=odd) |
| ●2倍角の公式 | sin 2x = 2 sin x cos x cos 2x = 2 cos2x -1 = 1 - 2 sin2x = cos2x - sin2x tan 2x = 2 tan x/(1- tan2x) |
| ●３倍角の公式 |
sin 3x = 3
sin x - 4 sin3x cos 3x = 4 cos3x - 3 cos x = cos3x -3 cos x sin2x tan 3x = (3 tan x - tan3x)/(1-3 tan2x) |
| ●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 - 2 sin2x) cos 4x = 8 cos4x - 8 cos2x + 1 = cos4x - 12 cos2x sin2x + sin4x |
| ●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) |
| ●7倍角の公式 [Maple10]での7倍角公式の導出 ＞ expand(sin(7*x, trig); expand(cos(7*x), 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) |
| ●8倍角の公式 [Maxima]での8倍角公式の導出 (%i1) trigexpand(sin(8*x)); (%i2) trigexpand(cos(8*x)); |
(%o1)
-8*cos(x)*sin(x)^7+56*cos(x)^3*sin(x)^5-56*cos(x)^5*sin(x)^3 +8*cos(x)^7*sin(x) (%o10) sin(x)^8-28*cos(x)^2*sin(x)^6+70*cos(x)^4*sin(x)^4 -28*cos(x)^6*sin(x)^2+cos(x)^8 |
| ●半角の公式 |
sin2
(x) = (1-cos(2x))/2 cos2 (x) = (1+cos(2x))/2 sin3 (x) = (3/4)sin(x)-(1/4)sin(3x)=sin(x)-sin(x)cos^2(x) cos3 (x) = (3/4)cos(x)+(1/4)cos(3x)=cos(x)-cos(x)sin^2(x) sin4 (x) = (1/8)cos(4x)-(1/2)cos(2x)+(3/8) cos4 (x) = (1/8)cos(4x)+(1/2)cos(2x)+(3/8) sin5 (x) = sin(x)-2sin(x)cos^2(x)+sin(x)cos^4(x) cos5 (x) = cos(x)-2cos(x)sin^2(x)+cos(x)sin^4(x) sin6 (x) = (5/16)-(1/2)cos(2x)+(1/8)cos(2x)sin^2(2x)+(3/16)cos(4x) cos6 (x) = (5/16)+(1/2)cos(2x)-(1/8)cos(2x)sin^2(2x)+(3/16)cos(4x) sin7 (x) = sin(x)-3sin(x)cos^3(x)+3sin(x)cos^4(x)-sin(x)cos^6(x) cos7 (x) = cos(x)-3cos(x)sin^2(x)+3cos(x)sin^4(x)-cos(x)sin^6(x) sin8 (x) =(35/128)-(1/2)cos(2x)+(1/4)cos(2x)sin^2(2x)+(7/32)cos(4x)+(1/128)cos(8x) cos8 (x) =(35/128)+(1/2)cos(2x)-(1/4)cos(2x)sin^2(2x)+(7/32)cos(4x)+(1/128)cos(8x) sin9 (x) =sin(x)-4sin(x)cos^2(x)+6sin(x)cos^4(x)-4cos(x)cos^6(x)+sin(x)cos^8(x) cos9 (x) =cos(x)-4cos(x)sin^2(x)+6cos(x)sin^4(x)-4cos(x)sin^6(x)+cos(x)sin^8(x) sin10 (x) =(63/256)-(1/2)cos(2x)+(3/8)cos(2x)sin^2(2x)-(1/32)cos(2x)sin^4(2x)+(15/64)cos(4x)+(5/256)cos(8x) cos10 (x) =(63/256)+(1/2)cos(2x)-(3/8)cos(2x)sin^2(2x)+(1/32)cos(2x)sin^4(2x)+(15/64)cos(4x)+(5/256)cos(8x) |
| ●積分のためのべき乗公式 cos2(x)=(1/2)+(1/2)cos(2x) cos3(x)=(3/4)cos(x)+(1/4)cos(3x) cos4(x)=(3/8)+(1/2)cos(2x)+(1/8)cos(4x) cos5(x)=(5/8)cos(x)+(5/16)cos(3x)+(1/16)cos(5x) cos6(x)=(5/16)+(15/32)cos(2x)+(3/16)cos(4x)+(1/32)cos(6x) |
sin2(x)=(1/2)-(1/2)cos(2x) sin3(x)=(3/4)sin(x)-(1/4)sin(3x) sin4(x)=(3/8)-(1/2)cos(2x)+(1/8)cos(4x) sin5(x)=(5/8)sin(x)-(5/16)sin(3x)+(1/16)sin(5x) sin6(x)=(5/16)-(15/32)cos(2x)+(3/16)cos(4x)-(1/32)cos(6x) |
| ●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) |
| 2sin(2x)cos(x)=sin(3x)+sin(x) 2cos(2x)sin(x)=sin(3x)-sin(x) 2cos(2x)cos(x)=cos(3x)+cos(x) 2sin(2x)sin(x)=cos(x)-cos(3x) |
2sin(3x)cos(x)=sin(4x)+sin(2x) 2cos(3x)sin(x)=sin(4x)-sin(2x) 2cos(3x)cos(x)=cos(4x)+cos(2x) 2sin(3x)sin(x)=cos(2x)-cos(4x) |
| 2sin(4x)cos(x)=sin(5x)+sin(3x) 2cos(4x)sin(x)=sin(5x)-sin(3x) 2cos(4x)cos(x)=cos(5x)+cos(3x) 2sin(4x)sin(x)=cos(3x)-cos(5x) |
2sin(5x)cos(x)=sin(6x)+sin(4x) 2cos(5x)sin(x)=sin(6x)-sin(4x) 2cos(5x)cos(x)=cos(6x)+cos(4x) 2sin(5x)sin(x)=cos(4x)-cos(6x) |
| 2sin(6x)cos(x)=sin(7x)+sin(5x) 2cos(6x)sin(x)=sin(7x)-sin(5x) 2cos(6x)cos(x)=cos(7x)+cos(5x) 2sin(6x)sin(x)=cos(5x)-cos(7x) |
2sin(7x)cos(x)=sin(8x)+sin(6x) 2cos(7x)sin(x)=sin(8x)-sin(6x) 2cos(7x)cos(x)=cos(8x)+cos(6x) 2sin(7x)sin(x)=cos(6x)-cos(8x) |
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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 = ei 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 = ei 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 = ei 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 = ei 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 = ei 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 |
| 【演習】 tan { sin-1(4/5)+cos-1(12/13)}
=tan { tan-1(4/3)+tan-1(1/12)} ={(4/3)+(1/12)}/{1-(4/3)(1/12)}=17*3/(36-4)=51/32 |
【演習】cos(sin-1x)sin(2sin-1x) = 2 sin(sin-1x) cos2(sin-1x) =2x(1-x2) |
| [演習] tan(tan-1(1/2)+tan-1(1/3)) =(1/2+1/3)/(1-(1/2)(1/3))=5/5 =1 |