【数学I】例題4.1.6：三角比を含む方程式1（One More）★

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次の方程式を満たす$\theta$の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

(1) $2\sin\theta-\sqrt{2}=0$

(2) $2\cos\theta-1=0$

(3) $\sqrt{3}\tan\theta+1=0$

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(1) $2\sin\theta-\sqrt{2}=0$より，$\sin\theta=\frac{1}{\sqrt{2}}$

半径1の半円上で，$y$座標が$\frac{1}{\sqrt{2}}$となる点は，右の図の2点$\mathrm{P},\mathrm{P}^{\prime}$である．

よって，$\theta=45^{\circ},135^{\circ}$

(2) $2\cos\theta-1=0$より，$\cos\theta=\frac{1}{2}$

半径1の半円上で，$x$座標が$\frac{1}{2}$となる点は，右の図の点$\mathrm{P}$である．

よって，$\theta=60^{\circ}$

(3) $\sqrt{3}\tan\theta+1=0$より，$\tan\theta=-\frac{1}{\sqrt{3}}$

直線$x=1$上で，$y$座標が$-\frac{1}{\sqrt{3}}$となる点Tをとると，直線OTと半径1の半円上の交点は右の図の点Pである．

よって，$\theta=150^{\circ}$

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% 問題I4.1.6：三角比を含む方程式1（One More）★
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次の方程式を満たす$\theta$の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

(1) $2\sin\theta-1=0$

(2) $2\cos\theta+1=0$

(3) $\tan\theta+1=0$

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(1) $2\sin\theta-1=0$より，$\sin\theta=\frac{1}{2}$

半径1の半円上で，$y$座標が$\frac{1}{2}$となる点は，右の図の2点$\mathrm{P},\mathrm{P}^{\prime}$である．

よって，$\theta=30^{\circ},150^{\circ}$

(2) $2\cos\theta+1=0$より，$\cos\theta=-\frac{1}{2}$

半径1の半円上で，$x$座標が$-\frac{1}{2}$となる点は，右の図の点$\mathrm{P}$である．

よって，$\theta=120^{\circ}$

(3) $\tan\theta+1=0$より，$\tan\theta=-1$

直線$x=1$上で，$y$座標が$-1$となる点Tをとると，直線OTと半径1の半円上の交点は右の図の点Pである．

よって，$\theta=135^{\circ}$

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$\sin\theta=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

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原点を中心とする半径1の半円を利用して解く．半径1の半円を考えると，様々な問題が解きやすくなる．$\sin\theta=\frac{y}{r},\cos\theta=\frac{x}{r}$において，それぞれ$r=1$のとき，$\sin\theta=y$，$\cos\theta=x$となる．また，$\tan\theta=\frac{y}{x}$において，$x=1$のとき，$\tan\theta=y$，$x=-1$のとき，$\tan\theta=-y$となる．

よって，$\sin\theta$は半径1の半円上の点の$y$座標，$\cos\theta$は半径1の半円上の点の$x$座標，$\tan\theta$は直線$x=1$上における$y$座標，または直線$x=-1$上における$y$座標の符号を変えた値を考えればよい．

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$\sin\theta=k$は$y$座標，$\cos\theta=k$は$x$座標を考える．

$\tan\theta=k$は点$(1,k)$と原点を結ぶ直線を考える．