이 문서는 삼각함수의 덧셈 정리에 대해 설명한다.

${\displaystyle \sin \left(x+y\right)=\sin x\cos y+\cos x\sin y}$

${\displaystyle \sin \left(x-y\right)=\sin x\cos y-\cos x\sin y}$

${\displaystyle \cos \left(x+y\right)=\cos x\cos y-\sin x\sin y}$

${\displaystyle \cos \left(x-y\right)=\cos x\cos y+\sin x\sin y}$

${\displaystyle \tan \left(x+y\right)={{\tan x+\tan y} \over {1-\tan x\tan y}}}$

${\displaystyle \tan \left(x-y\right)={{\tan x-\tan y} \over {1+\tan x\tan y}}}$

예각 삼각형 ${\displaystyle ABC}$의 넓이 ${\displaystyle \triangle ABC}$에 대해서,^[1][2]

${\displaystyle \triangle ABC=\triangle AHB+\triangle AHC}$

${\displaystyle \triangle ABC={1 \over {2}}{b}c\sin(\alpha +\beta )}$

${\displaystyle \triangle AHB+\triangle AHC}$

${\displaystyle ={1 \over {2}}\;{\overline {BH}}\;{\overline {AH}}+{1 \over {2}}\;{\overline {CH}}\;{\overline {AH}}}$

${\displaystyle ={1 \over {2}}c\sin \alpha \;b\cos \beta +{1 \over {2}}b\sin \beta \;c\cos \alpha }$

${\displaystyle ={1 \over {2}}cb(\sin \alpha \;\cos \beta +\sin \beta \;\cos \alpha )}$

따라서,

${\displaystyle {1 \over {2}}{b}c\sin(\alpha +\beta )={1 \over {2}}cb(\sin \alpha \;\cos \beta +\sin \beta \;\cos \alpha )}$

${\displaystyle \sin(\alpha +\beta )=(\sin \alpha \;\cos \beta +\sin \beta \;\cos \alpha )}$

둔각삼각형 ${\displaystyle \triangle {ABC}}$에서${\displaystyle ,\;{\overline {BD}}}$의 임의의 한점 ${\displaystyle C}$에 대해서,^{[3][4] [5]}

${\displaystyle {\overline {BD}}^{2}={\overline {BC}}^{2}+{\overline {CD}}^{2}+2\left({\overline {BC}}\cdot {\overline {CD}}\right)}$

${\displaystyle {\overline {AB}}^{2}={\overline {BD}}^{2}+{\overline {AD}}^{2}\;\;}$

그리고,

${\displaystyle {\overline {AC}}^{2}={\overline {CD}}^{2}+{\overline {AD}}^{2}\;\;}$

${\displaystyle {\overline {AC}}^{2}-{\overline {CD}}^{2}={\overline {AD}}^{2}\;\;}$

따라서,

${\displaystyle {\overline {AB}}^{2}={\overline {BD}}^{2}+{\overline {AD}}^{2}}$

${\displaystyle {\overline {AB}}^{2}=\left({\overline {BC}}^{2}+{\overline {CD}}^{2}+2\left({\overline {BC}}\cdot {\overline {CD}}\right)\right)+\left({\overline {AC}}^{2}-{\overline {CD}}^{2}\right)}$

${\displaystyle {\overline {AB}}^{2}={\overline {BC}}^{2}+{\overline {AC}}^{2}+2\left({\overline {BC}}\cdot {\overline {CD}}\right)}$

그리고

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

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

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

따라서,

${\displaystyle {\overline {AB}}^{2}={\overline {BC}}^{2}+{\overline {AC}}^{2}+2\left({\overline {BC}}\cdot {\overline {CD}}\right)}$

${\displaystyle {\overline {AB}}^{2}={\overline {BC}}^{2}+{\overline {AC}}^{2}-2\left({\overline {BC}}\cdot {{\overline {AC}}\cos \alpha }\right)}$

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

이것은 제2코사인법칙이고,

유클리드 원론 3권 법칙3 에서,[6] 두 점 사이의 거리를 가정하면, ${\displaystyle l^{2}={({x_{2}}-{x_{1}})^{2}+({y_{2}}-{y_{1}})^{2}}}$이므로, ${\displaystyle P=(\cos \;\alpha ,\sin \;\alpha )\;\;,\;\;Q=(\cos \beta ,\sin \beta )}$ 일때, ${\displaystyle {\overline {PQ}}^{2}=(\cos \beta -\cos \;\alpha )^{2}+(\sin \beta -\sin \;\alpha )^{2}}$ ${\displaystyle =\left((\cos \beta -\cos \;\alpha )\cdot (\cos \beta -\cos \;\alpha )\right)+\left((\sin \beta -\sin \;\alpha )\cdot (\sin \beta -\sin \;\alpha )\right)}$ ${\displaystyle =\left((\cos \beta )^{2}-2\cos \alpha \cos \beta +(\cos \;\alpha )^{2}\right)+\left((\sin \beta )^{2}-2\sin \alpha \sin \beta +(\sin \alpha )^{2}\right)}$ ${\displaystyle =(\cos \beta )^{2}+(\cos \;\alpha )^{2}+(\sin \beta )^{2}+(\sin \alpha )^{2}-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta }$ ${\displaystyle =(\cos ^{2}\beta +\sin ^{2}\beta )+(\cos ^{2}\;\alpha +\sin ^{2}\alpha )-2\left(\cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}$ 그리고 삼각 함수 항등식의 피타고라스 정리에서, ${\displaystyle \sin ^{2}{x}+\cos ^{2}{x}=1}$ 따라서, ${\displaystyle =1+1-2\left(\cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}$ ${\displaystyle {\overline {PQ}}^{2}=2-2\left(\cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}$

한편,

이것은, 제2코사인법칙에서는,

${\displaystyle {\overline {PQ}}^{2}={\overline {OP}}^{2}+{\overline {OQ}}^{2}-2\left({\overline {OP}}\cdot {{\overline {OQ}}\cos(\alpha -\beta )}\right)}$

${\displaystyle {\overline {PQ}}^{2}=1^{2}+1^{2}-2\left(1\cdot {1\cos(\alpha -\beta )}\right)}$

${\displaystyle {\overline {PQ}}^{2}=2-2\left({\cos(\alpha -\beta )}\right)}$

그리고 두 점 사이의 거리에서,

${\displaystyle {\overline {PQ}}^{2}=2-2\left({\cos(\alpha -\beta )}\right)=2-2\left(\cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}$

따라서,

${\displaystyle {\cos(\alpha -\beta )}=\left(\cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}$

${\displaystyle \tan x={{\sin x} \over {\cos x}}}$

${\displaystyle \tan \left(x+y\right)={{\sin \left(x+y\right)} \over {\cos \left(x+y\right)}}}$

${\displaystyle \tan \left(x+y\right)={{\sin x\cos y+\cos x\sin y} \over {\cos x\cos y-\sin x\sin y}}}$

${\displaystyle \tan \left(x+y\right)={{{{\sin x\cos y} \over {\cos x\cos y}}+{{\cos x\sin y} \over {\cos x\cos y}}} \over {{{\cos x\cos y} \over {\cos x\cos y}}-{{\sin x\sin y} \over {\cos x\cos y}}}}}$

${\displaystyle \tan \left(x+y\right)={{{\tan x}+{\tan y}} \over {1-{\tan x\tan y}}}}$

${\displaystyle \sin(-\theta )=-\sin \theta \;\;,\;\;\cos(-\theta )=\cos \theta }$

따라서,

${\displaystyle \sin \left(x+(-y)\right)=\sin x\cos(-y)+\cos x\sin(-y)}$

${\displaystyle \sin \left(x-y\right)=\sin x\cos y-\cos x\sin y}$

그리고,

${\displaystyle \cos \left(x-(-y)\right)=\cos x\cos(-y)+\sin x\sin(-y)}$

${\displaystyle \cos \left(x+y\right)=\cos x\cos y-\sin x\sin y}$

그리고,

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

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

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

정삼각형에서 ${\displaystyle {\overline {AE}}=1}$을 예약하고,[7] ${\displaystyle {{\overline {AF}} \over {1}}={\cos \alpha }}$ ${\displaystyle {{\overline {EF}} \over {1}}={\sin \alpha }}$ ${\displaystyle {{\overline {FD}} \over {AF}}={\sin \beta }}$ ${\displaystyle {\overline {FD}}={\overline {GC}}={\sin \beta {\overline {AF}}}={\sin \beta \cos \alpha }}$ ${\displaystyle {{\overline {EG}} \over {\overline {EF}}}={\cos \beta }}$ ${\displaystyle {\overline {EG}}={\cos \beta }{\overline {EF}}={\cos \beta }{\sin \alpha }}$ ${\displaystyle {\sin(\alpha +\beta )}={{{\overline {EG}}+{\overline {GC}}} \over {1}}}$ ${\displaystyle {\sin(\alpha +\beta )}={{\cos \beta }{\sin \alpha }}+{\sin \beta \cos \alpha }}$ 이것은 사인함수이다. 한편,예약된 정삼각형에서,[8] ${\displaystyle {\overline {AE}}=1,\angle A=\angle E=\angle B,\angle A=\angle \alpha +\angle \beta ,\angle \alpha =\angle \beta }$ ${\displaystyle {{\overline {EF}} \over {1}}={\sin \alpha }}$ ${\displaystyle {{\overline {AF}} \over {1}}={\cos \alpha }}$ ${\displaystyle {{\overline {AD}} \over {AF}}={\cos \beta }}$ ${\displaystyle {\overline {AD}}={\cos \beta }{\overline {AF}}}$ ${\displaystyle {\overline {AD}}={\cos \beta }{\cos \alpha }}$ ${\displaystyle {{\overline {GF}} \over {\overline {EF}}}={\sin \beta }}$ ${\displaystyle {\overline {GF}}={\sin \beta }{\overline {EF}}}$ ${\displaystyle {\overline {GF}}={\sin \beta }{\sin \alpha }}$ ${\displaystyle {\overline {GF}}={\overline {CD}}={\sin \beta }{\sin \alpha }}$ ${\displaystyle {\cos(\alpha +\beta )}={{\overline {AC}} \over {1}}}$ ${\displaystyle {\cos(\alpha +\beta )}={{\overline {AD}}-{\overline {CD}}}}$ ${\displaystyle {\cos(\alpha +\beta )}={{\cos \beta }{\cos \alpha }}-{\sin \beta \sin \alpha }}$ 이것은 코사인함수이다.

: ${\displaystyle \sin(-\theta )=-\sin \theta \;\;,\;\;\cos(-\theta )=\cos \theta }$이고, 삼각함수의 점의 좌표 ${\displaystyle ({\text{cosine}},{\text{sine}})}$에서, ${\displaystyle P=(x,y)\;\;,\;\;P'=(x',y')}$ ${\displaystyle P=(\cos ,\sin )\;\;,\;\;P'=(\cos ',\sin ')}$이고, ${\displaystyle \theta =90^{\circ }}$로, ${\displaystyle P'=(\alpha +\theta ,\alpha +\theta )}$을 예약해보면,[9][10] ${\displaystyle P'=(\alpha +{{\pi } \over {2}},\alpha +{{\pi } \over {2}})}$ ${\displaystyle P'=(x\cos \alpha +{x}\cos \theta ,y\cos \alpha +{y}\sin \theta )}$ ${\displaystyle P'=\left(x\cos \alpha +{x}\cos {{\pi } \over {2}},y\cos \alpha +{y}\sin {{\pi } \over {2}}\right)}$ 삼각함수의 값에서, ${\displaystyle P'=(x\cos \alpha +{y}\sin(-\theta ),y\cos \alpha +{x}\cos(-\theta ))}$ ${\displaystyle P'=(x\cos \alpha -{y}\sin \theta ,y\cos \alpha +{x}\cos \theta )}$ 그리고, ${\displaystyle x'={x}\cos \alpha -{y}\sin \theta }$ ${\displaystyle y'={x}\sin \alpha +{y}\cos \theta }$ 따라서, ${\displaystyle \alpha =90^{\circ }}$로 변형해서, ${\displaystyle \theta =\alpha =90^{\circ }}$ ${\displaystyle {x}\cos \theta -{y}\sin \theta =x'}$ ${\displaystyle {x}\sin \theta +{y}\cos \theta =y'}$ ${\displaystyle {\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}x'\\y'\end{pmatrix}}}$ ${\displaystyle {\begin{pmatrix}\cos y&-\sin y\\\sin y&\cos y\end{pmatrix}}{\begin{pmatrix}\cos x\\\sin x\end{pmatrix}}={\begin{pmatrix}\cos(x+y)\\\sin(x+y)\end{pmatrix}}}$ ${\displaystyle cosx\cos y-\sin x\sin y=\cos \left(x+y\right)}$ ${\displaystyle \cos x\sin y+\sin x\cos y=\sin \left(x+y\right)}$

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