三角関数の公式(加法定理から積和・和積公式、三角関数の合成まで)

三角関数の公式は量が多いため,それ自体をそのまま覚えるのでなく,加法定理からその他の公式を導くように理解しよう.自分で公式を証明する訓練を続けていくと,自然に公式をアウトプットできるようになる.

また倍角と三倍角の公式は数学Ⅲの微分積分で頻繁に使うことになるので,(一部の学部を除く)理系志望の方は加法定理から三倍角の公式までを必ず理解しよう.

参考ページ
三角関数はどうすれば理解できるか?三角関数の勉強法と参考書

三角関数の定理

加法定理

[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta]

[\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta]

[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta]

[\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta]

[\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}]

[\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}]

倍角公式

[\sin2\alpha=2\sin\alpha\cos\beta]

\begin{eqnarray}
\cos2\alpha&=&\cos^{2}\alpha-\sin^{2}\alpha\
&=&2\cos^{2}\alpha-1\
&=&1-2\sin^{2}\alpha
\end{eqnarray
}

[\tan2\alpha=\frac{2\tan\alpha}{1-\tan^{2}\alpha}]

三倍角公式

[\sin3\alpha=-4\sin^{3}\alpha+3\sin\alpha]

[\cos3\alpha=4\cos^{3}\alpha-3\cos\alpha]

半角公式

[\sin^{2}\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}]

[\cos^{2}\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}]

[\tan^{2}\frac{\alpha}{2}=\frac{1-\cos\alpha}{1+\cos\alpha}]

積和公式

[\sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}]

[\cos\alpha\cos\beta=\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}]

[\sin\alpha\sin\beta=\frac{\cos(\alpha+\beta)-\cos(\alpha-\beta)}{2}]

和積公式

[\sin{A}+\sin{B}=2 \sin\frac{A+B}{2}\cos\frac{A-B}{2}]

[\sin{A}-\sin{B}=2 \cos\frac{A+B}{2}\sin\frac{A-B}{2}]

[\cos{A}+\cos{B}=2 \cos\frac{A+B}{2}\cos\frac{A-B}{2}]

[\cos{A}-\cos{B}=-2 \sin\frac{A+B}{2}\sin\frac{A-B}{2}]

三角関数の合成

[
a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+\phi)
]
ここで
[
\left{\begin{array}{l}
\sin\phi=\dfrac{b}{\sqrt{a^{2}+b^{2}}}\
\cos\phi=\dfrac{a}{\sqrt{a^{2}+b^{2}}}
\end{array}\right.
]
である.

三角関数の計算

加法定理の計算

1次の値を求めなさい.

((1)) (\sin{105^{\circ}})

((2)) (\cos{105^{\circ}})

((3)) (\tan{105^{\circ}})

((4)) (\sin{15^{\circ}})

((5)) (\cos{15^{\circ}})

((6)) (\tan{15^{\circ}})

[su_accordion]
[su_spoiler title="解答" style="fancy"]

1

((1))
\begin{eqnarray}
&\enspace&\sin{105^{\circ}}\
&=&\sin{(45^{\circ}+60^{\circ})}\
&=&\sin{45^{\circ}}\cos{60^{\circ}}+\cos{45^{\circ}}\sin{60^{\circ}}\
&=&\dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2}\
&=&\dfrac{1+\sqrt{3}}{2\sqrt{2}}\
&=&\dfrac{\sqrt{2}+\sqrt{6}}{4}
\end{eqnarray
}

((2))
\begin{eqnarray}
&\enspace&\cos{105^{\circ}}\
&=&\cos{(45^{\circ}+60^{\circ})}\
&=&\cos{45^{\circ}}\cos{60^{\circ}}-\sin{45^{\circ}}\sin{60^{\circ}}\
&=&\dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2}\
&=&\dfrac{1-\sqrt{3}}{2\sqrt{2}}\
&=&\dfrac{\sqrt{2}-\sqrt{6}}{4}
\end{eqnarray
}

((3))
\begin{eqnarray}
&\enspace&\tan{105^{\circ}}\
&=&\dfrac{\sin{105^{\circ}}}{\cos{105^{\circ}}}\
&=&\dfrac{1+\sqrt{3}}{1-\sqrt{3}}\
&=&\dfrac{\left(1+\sqrt{3}\right)^{2}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\
&=&\dfrac{4+2\sqrt{3}}{-2}\
&=&-2-\sqrt{3}
\end{eqnarray
}

((4))
\begin{eqnarray}
&\enspace&\sin{15^{\circ}}\
&=&-\cos{105^{\circ}}\
&=&\dfrac{\sqrt{6}-\sqrt{2}}{4}
\end{eqnarray
}

((5))
\begin{eqnarray}
&\enspace&\cos{15^{\circ}}\
&=&\sin{105^{\circ}}\
&=&\dfrac{\sqrt{6}+\sqrt{2}}{4}
\end{eqnarray
}

((6))
\begin{eqnarray}
&\enspace&\tan{15^{\circ}}\
&=&\dfrac{\sin{15^{\circ}}}{\cos{15^{\circ}}}\
&=&\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}\
&=&\dfrac{\left(\sqrt{6}-\sqrt{2}\right)^{2}}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\
&=&\dfrac{6+2-2\sqrt{12}}{4}\
&=&\dfrac{8-4\sqrt{3}}{4}\
&=&2-\sqrt{3}
\end{eqnarray
}

[/su_spoiler]
[/su_accordion]

三角関数の合成

1次の三角関数の式を合成しなさい.

((1)) (\sin\theta+\sqrt{3}\cos\theta)

((2)) (3\sin\theta+4\cos\theta)

((3)) (\sin\theta-\cos\theta)

((4)) (-\sqrt{2}\sin\theta+\sqrt{6}\cos\theta)

[su_accordion]
[su_spoiler title="解答" style="fancy"]

1

((1))
[
\sqrt{1^{2}+{\left(\sqrt{3}\right)}^{2}}=\sqrt{4}=2
]
\begin{eqnarray}
&\enspace&\sin\theta+\sqrt{3}\cos\theta\
&=&2\left(\sin\cdot\dfrac{1}{2}+\cos\theta\cdot\dfrac{\sqrt{3}}{2}\right)\
&=&2\left(\sin\theta\cos\dfrac{\pi}{3}+\cos\theta\sin\dfrac{\pi}{3}\right)\
&=&2\sin{\left(\theta+\dfrac{\pi}{3}\right)}
\end{eqnarray
}

((2))
[
\sqrt{3^{2}+4^{2}}=\sqrt{25}=5
]
\begin{eqnarray}
&\enspace&3\sin\theta+4\cos\theta\
&=&5\left(\sin\theta\cdot\dfrac{3}{5}+\cos\theta\cdot\dfrac{4}{5}\right)\
&=&5(\sin\theta\cos\alpha+\cos\theta\sin\alpha)\
&=&5\sin{(\theta+\alpha)}\ \ \left(\cos\alpha=\dfrac{3}{5},\ \sin\alpha=\dfrac{4}{5}\right)
\end{eqnarray
}

((3))
[
\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}
]
\begin{eqnarray}
&\enspace&\sin\theta-\cos\theta\
&=&\sqrt{2}\left(\sin\theta\cdot\dfrac{1}{\sqrt{2}}+\cos\theta\cdot\left(-\dfrac{1}{\sqrt{2}}\right)\right)\
&=&\sqrt{2}\left(\sin\theta\cos\left(-\dfrac{\pi}{4}\right)+\cos\theta\sin\left(-\dfrac{\pi}{4}\right)\right)\
&=&\sqrt{2}\sin{\left(\theta+\left(-\dfrac{\pi}{4}\right)\right)}\
&=&\sqrt{2}\sin{\left(\theta-\dfrac{\pi}{4}\right)}
\end{eqnarray
}

((4))
[
\sqrt{\left(-\sqrt{2}\right)^{2}+\left(\sqrt{6}\right)^{2}}=\sqrt{2+6}=2\sqrt{2}
]
\begin{eqnarray}
&\enspace&-\sqrt{2}\sin\theta+\sqrt{6}\cos\theta\
&=&2\sqrt{2}\left(\sin\theta\cdot\left(-\dfrac{1}{2}\right)+\cos\theta\cdot\dfrac{\sqrt{6}}{2\sqrt{2}}\right)\
&=&2\sqrt{2}\left(\sin\theta\cdot\left(-\dfrac{1}{2}\right)+\cos\theta\cdot\dfrac{\sqrt{3}}{2}\right)\
&=&2\sqrt{2}\left(\sin\theta\cos\dfrac{2\pi}{3}+\cos\theta\sin\dfrac{2\pi}{3}\right)\
&=&2\sqrt{2}\sin\left(\theta+\dfrac{2\pi}{3}\right)
\end{eqnarray
}

[/su_spoiler]
[/su_accordion]