삼각 함수의 덧셈 정리

틀:위키데이터 속성 추적 이 문서는 삼각함수의 덧셈 정리에 대해 설명한다.

$\sin (x+y)=\sin x\cos y+\cos x\sin y$

$\sin (x-y)=\sin x\cos y-\cos x\sin y$

$\cos (x+y)=\cos x\cos y-\sin x\sin y$

$\cos (x-y)=\cos x\cos y+\sin x\sin y$

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

사인함수의 덧셈정리

예각 삼각형 $ABC$의 넓이 $\mathrm{△}ABC$에 대해서,

$\mathrm{△}ABC=\mathrm{△}AHB+\mathrm{△}AHC$

$\mathrm{△}ABC=\frac{1}{2}bc\sin (\alpha +\beta )$

$\mathrm{△}AHB+\mathrm{△}AHC$

$=\frac{1}{2}\overline{BH}\overline{AH}+\frac{1}{2}\overline{CH}\overline{AH}$

$=\frac{1}{2}c\sin \alpha b\cos \beta +\frac{1}{2}b\sin \beta c\cos \alpha$

$=\frac{1}{2}cb(\sin \alpha \cos \beta +\sin \beta \cos \alpha )$

따라서,

$\frac{1}{2}bc\sin (\alpha +\beta )=\frac{1}{2}cb(\sin \alpha \cos \beta +\sin \beta \cos \alpha )$

$\sin (\alpha +\beta )=(\sin \alpha \cos \beta +\sin \beta \cos \alpha )$

코사인의 덧셈정리

둔각삼각형 $\mathrm{△}ABC$에서$,\overline{BD}$의 임의의 한점 $C$에 대해서,

$\stackrel{2}{\stackrel{‾}{BD}}=\stackrel{2}{\stackrel{‾}{BC}}+\stackrel{2}{\stackrel{‾}{CD}}+2(\overline{BC}\cdot \overline{CD})$

$\stackrel{2}{\stackrel{‾}{AB}}=\stackrel{2}{\stackrel{‾}{BD}}+\stackrel{2}{\stackrel{‾}{AD}}$

그리고,

$\stackrel{2}{\stackrel{‾}{AC}}=\stackrel{2}{\stackrel{‾}{CD}}+\stackrel{2}{\stackrel{‾}{AD}}$

$\stackrel{2}{\stackrel{‾}{AC}}-\stackrel{2}{\stackrel{‾}{CD}}=\stackrel{2}{\stackrel{‾}{AD}}$

따라서,

$\stackrel{2}{\stackrel{‾}{AB}}=\stackrel{2}{\stackrel{‾}{BD}}+\stackrel{2}{\stackrel{‾}{AD}}$

$\stackrel{2}{\stackrel{‾}{AB}}=(\stackrel{2}{\stackrel{‾}{BC}}+\stackrel{2}{\stackrel{‾}{CD}}+2(\overline{BC}\cdot \overline{CD}))+(\stackrel{2}{\stackrel{‾}{AC}}-\stackrel{2}{\stackrel{‾}{CD}})$

$\stackrel{2}{\stackrel{‾}{AB}}=\stackrel{2}{\stackrel{‾}{BC}}+\stackrel{2}{\stackrel{‾}{AC}}+2(\overline{BC}\cdot \overline{CD})$

그리고

$\cos (\pi -\alpha )=\frac{\overline{CD}}{\overline{AC}}$

$\overline{CD}=\overline{AC}\cos (\pi -\alpha )$

$\overline{CD}=-\overline{AC}\cos \alpha$

따라서,

$\stackrel{2}{\stackrel{‾}{AB}}=\stackrel{2}{\stackrel{‾}{BC}}+\stackrel{2}{\stackrel{‾}{AC}}+2(\overline{BC}\cdot \overline{CD})$

$\stackrel{2}{\stackrel{‾}{AB}}=\stackrel{2}{\stackrel{‾}{BC}}+\stackrel{2}{\stackrel{‾}{AC}}-2(\overline{BC}\cdot \overline{AC}\cos \alpha )$

$c^2=a^2+b^2-2ab\cos \alpha$

이것은 제2코사인법칙이고,

유클리드 원론 3권 법칙3 에서, 두 점 사이의 거리를 가정하면, $l^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$이므로, $P=(\cos \alpha ,\sin \alpha ),Q=(\cos \beta ,\sin \beta )$ 일때, $\stackrel{2}{\stackrel{‾}{PQ}}=(\cos \beta -\cos \alpha {)}^2+(\sin \beta -\sin \alpha {)}^2$ $=((\cos \beta -\cos \alpha )\cdot (\cos \beta -\cos \alpha ))+((\sin \beta -\sin \alpha )\cdot (\sin \beta -\sin \alpha ))$ $=((\cos \beta {)}^2-2\cos \alpha \cos \beta +(\cos \alpha {)}^2)+((\sin \beta {)}^2-2\sin \alpha \sin \beta +(\sin \alpha {)}^2)$ $=(\cos \beta {)}^2+(\cos \alpha {)}^2+(\sin \beta {)}^2+(\sin \alpha {)}^2-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta$ $=({\cos }^2\beta +{\sin }^2\beta )+({\cos }^2\alpha +{\sin }^2\alpha )-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$ 그리고 삼각 함수 항등식의 피타고라스 정리에서, ${\sin }^{2}x+{\cos }^{2}x=1$ 따라서, $=1+1-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$ $\stackrel{2}{\stackrel{‾}{PQ}}=2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$

한편,

이것은, 제2코사인법칙에서는,

$\stackrel{2}{\stackrel{‾}{PQ}}=\stackrel{2}{\stackrel{‾}{OP}}+\stackrel{2}{\stackrel{‾}{OQ}}-2(\overline{OP}\cdot \overline{OQ}\cos (\alpha -\beta ))$

$\stackrel{2}{\stackrel{‾}{PQ}}=1^2+1^2-2(1\cdot 1\cos (\alpha -\beta ))$

$\stackrel{2}{\stackrel{‾}{PQ}}=2-2\left(\cos (\alpha -\beta )\right)$

그리고 두 점 사이의 거리에서,

$\stackrel{2}{\stackrel{‾}{PQ}}=2-2\left(\cos (\alpha -\beta )\right)=2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$

따라서,

$\cos (\alpha -\beta )=(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$

탄젠트의 덧셈정리

$\tan x=\frac{\sin x}{\cos x}$

$\tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}$

$\tan (x+y)=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

$\tan (x+y)=\frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}$

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

덧셈정리의 변형

$\sin (-\theta )=-\sin \theta ,\cos (-\theta )=\cos \theta$

따라서,

$\sin (x+(-y))=\sin x\cos (-y)+\cos x\sin (-y)$

$\sin (x-y)=\sin x\cos y-\cos x\sin y$

그리고,

$\cos (x-(-y))=\cos x\cos (-y)+\sin x\sin (-y)$

$\cos (x+y)=\cos x\cos y-\sin x\sin y$

그리고,

$\tan (-\theta )=\frac{\sin (-\theta )}{\cos (-\theta )}=\frac{-\sin \theta }{\cos \theta }=-\tan \theta$

$\tan (x+(-y))=\frac{\tan x+\tan (-y)}{1-\tan x\tan (-y)}$

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

유클리드 기하학 원론 2권 법칙9를 이용한 정리

정삼각형에서 $\overline{AE}=1$을 예약하고, $\frac{\overline{AF}}{1}=\cos \alpha$ $\frac{\overline{EF}}{1}=\sin \alpha$ $\frac{\overline{FD}}{AF}=\sin \beta$ $\overline{FD}=\overline{GC}=\sin \beta \overline{AF}=\sin \beta \cos \alpha$ $\frac{\overline{EG}}{\overline{EF}}=\cos \beta$ $\overline{EG}=\cos \beta \overline{EF}=\cos \beta \sin \alpha$ $\sin (\alpha +\beta )=\frac{\overline{EG}+\overline{GC}}{1}$ $\sin (\alpha +\beta )=\cos \beta \sin \alpha +\sin \beta \cos \alpha$ 이것은 사인함수이다. 한편,예약된 정삼각형에서, $\overline{AE}=1,\mathrm{\angle }A=\mathrm{\angle }E=\mathrm{\angle }B,\mathrm{\angle }A=\mathrm{\angle }\alpha +\mathrm{\angle }\beta ,\mathrm{\angle }\alpha =\mathrm{\angle }\beta$ $\frac{\overline{EF}}{1}=\sin \alpha$ $\frac{\overline{AF}}{1}=\cos \alpha$ $\frac{\overline{AD}}{AF}=\cos \beta$ $\overline{AD}=\cos \beta \overline{AF}$ $\overline{AD}=\cos \beta \cos \alpha$ $\frac{\overline{GF}}{\overline{EF}}=\sin \beta$ $\overline{GF}=\sin \beta \overline{EF}$ $\overline{GF}=\sin \beta \sin \alpha$ $\overline{GF}=\overline{CD}=\sin \beta \sin \alpha$ $\cos (\alpha +\beta )=\frac{\overline{AC}}{1}$ $\cos (\alpha +\beta )=\overline{AD}-\overline{CD}$ $\cos (\alpha +\beta )=\cos \beta \cos \alpha -\sin \beta \sin \alpha$ 이것은 코사인함수이다.

행렬을 이용한 정리

: $\sin (-\theta )=-\sin \theta ,\cos (-\theta )=\cos \theta$이고, 삼각함수의 점의 좌표 $(\text{cosine},\text{sine})$에서, $P=(x,y),P^{\prime }=(x^{\prime },y^{\prime })$ $P=(\cos ,\sin ),P^{\prime }=({\cos }^{\prime },{\sin }^{\prime })$이고, $\theta =90^{\circ }$로, $P^{\prime }=(\alpha +\theta ,\alpha +\theta )$을 예약해보면, $P^{\prime }=(\alpha +\frac{\pi }{2},\alpha +\frac{\pi }{2})$ $P^{\prime }=(x\cos \alpha +x\cos \theta ,y\cos \alpha +y\sin \theta )$ $P^{\prime }=(x\cos \alpha +x\cos \frac{\pi }{2},y\cos \alpha +y\sin \frac{\pi }{2})$ 삼각함수의 값에서, $P^{\prime }=(x\cos \alpha +y\sin (-\theta ),y\cos \alpha +x\cos (-\theta ))$ $P^{\prime }=(x\cos \alpha -y\sin \theta ,y\cos \alpha +x\cos \theta )$ 그리고, $x^{\prime }=x\cos \alpha -y\sin \theta$ $y^{\prime }=x\sin \alpha +y\cos \theta$ 따라서, $\alpha =90^{\circ }$로 변형해서, $\theta =\alpha =90^{\circ }$ $x\cos \theta -y\sin \theta =x^{\prime }$ $x\sin \theta +y\cos \theta =y^{\prime }$ $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$ $\left(\begin{array}{cc}\cos y& -\sin y\\ \sin y& \cos y\end{array}\right)\left(\begin{array}{c}\cos x\\ \sin x\end{array}\right)=\left(\begin{array}{c}\cos (x+y)\\ \sin (x+y)\end{array}\right)$ $cosx\cos y-\sin x\sin y=\cos (x+y)$ $\cos x\sin y+\sin x\cos y=\sin (x+y)$

같이 보기

• 삼각 함수 항등식
• 예각삼각형
• 둔각삼각형
• 삼각함수
• 회전행렬
• 유클리드 원론