円周上の反時計回りの隣接弧 AB,BC,CA の中心角をそれぞれ ϕ A B \phi_{AB} ϕ A B ϕ B C \phi_{BC} ϕ B C ϕ C A \phi_{CA} ϕ C A ϕ A B = 2 ∠ A C B = 2 π 3 , ϕ C A = 2 ∠ A B C = π 2 . \displaystyle
\phi_{AB}=2\angle ACB=\frac{2\pi}{3},\qquad
\phi_{CA}=2\angle ABC=\frac{\pi}{2}. ϕ A B = 2∠ A C B = 3 2 π , ϕ C A = 2∠ A B C = 2 π . ϕ B C \phi_{BC} ϕ B C π \pi π 2 π 3 \displaystyle \frac{2\pi}{3} 3 2 π π 2 \displaystyle \frac{\pi}{2} 2 π 5 π 6 \displaystyle \frac{5\pi}{6} 6 5 π θ \theta θ θ \theta θ 2 π 3 \displaystyle \frac{2\pi}{3} 3 2 π θ \theta θ 3 π 2 \displaystyle \frac{3\pi}{2} 2 3 π
(i) 単位円の偏角 φ \varphi φ cos \cos cos φ \varphi φ sin \sin sin φ \varphi φ A = ( cos θ , sin θ ) , B = ( cos ( θ + 2 π 3 ) , sin ( θ + 2 π 3 ) ) , C = ( cos ( θ + 3 π 2 ) , sin ( θ + 3 π 2 ) ) = ( sin θ , − cos θ ) . \displaystyle
A=(\cos\theta,\sin\theta),\quad
B=\left(\cos\left(\theta+\frac{2\pi}{3}\right),\sin\left(\theta+\frac{2\pi}{3}\right)\right),\quad
C=\left(\cos\left(\theta+\frac{3\pi}{2}\right),\sin\left(\theta+\frac{3\pi}{2}\right)\right)
=(\sin\theta,-\cos\theta). A = ( cos θ , sin θ ) , B = ( cos ( θ + 3 2 π ) , sin ( θ + 3 2 π ) ) , C = ( cos ( θ + 2 3 π ) , sin ( θ + 2 3 π ) ) = ( sin θ , − cos θ ) .
(ii) 加法定理を用いると x A + x B + x C = cos θ + cos ( θ + 2 π 3 ) + cos ( θ + 3 π 2 ) = 1 2 ( cos θ + ( 2 − 3 ) sin θ ) . \displaystyle
x_A+x_B+x_C
=\cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{3\pi}{2}\right)
=\frac12\left(\cos\theta+(2-\sqrt3)\sin\theta\right). x A + x B + x C = cos θ + cos ( θ + 3 2 π ) + cos ( θ + 2 3 π ) = 2 1 ( cos θ + ( 2 − 3 ) sin θ ) . cos \cos cos θ \theta θ 3 \sqrt3 3 sin \sin sin θ \theta θ cos \cos cos θ \theta θ 3 \sqrt3 3 sin \sin sin θ \theta θ ≠ \ne = cos \cos cos θ \theta θ ≠ \ne = tan θ = − 1 2 − 3 = − ( 2 + 3 ) . \displaystyle
\tan\theta=-\frac1{2-\sqrt3}=-(2+\sqrt3). tan θ = − 2 − 3 1 = − ( 2 + 3 ) .
(iii) 位置ベクトルを a \mathbf a a O A → \overrightarrow{OA} O A b \mathbf b b O B → \overrightarrow{OB} O B c \mathbf c c O C → \overrightarrow{OC} O C a ⋅ b = cos 2 π 3 = − 1 2 , b ⋅ c = cos 5 π 6 = − 3 2 , c ⋅ a = cos π 2 = 0. \displaystyle
\mathbf a\cdot\mathbf b=\cos\frac{2\pi}{3}=-\frac12,\quad
\mathbf b\cdot\mathbf c=\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2},\quad
\mathbf c\cdot\mathbf a=\cos\frac{\pi}{2}=0. a ⋅ b = cos 3 2 π = − 2 1 , b ⋅ c = cos 6 5 π = − 2 3 , c ⋅ a = cos 2 π = 0. ( x A + x B + x C ) 2 + ( y A + y B + y C ) 2 = ∣ a + b + c ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 + ∣ c ∣ 2 + 2 ( a ⋅ b + b ⋅ c + c ⋅ a ) = 3 + 2 ( − 1 2 − 3 2 + 0 ) = 2 − 3 . \displaystyle
\begin{aligned}
(x_A+x_B+x_C)^2+(y_A+y_B+y_C)^2
&=|\mathbf a+\mathbf b+\mathbf c|^2\\
&=|\mathbf a|^2+|\mathbf b|^2+|\mathbf c|^2
+2(\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c+\mathbf c\cdot\mathbf a)\\
&=3+2\left(-\frac12-\frac{\sqrt3}{2}+0\right)=2-\sqrt3.
\end{aligned} ( x A + x B + x C ) 2 + ( y A + y B + y C ) 2 = ∣ a + b + c ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 + ∣ c ∣ 2 + 2 ( a ⋅ b + b ⋅ c + c ⋅ a ) = 3 + 2 ( − 2 1 − 2 3 + 0 ) = 2 − 3 .
