加法定理を用いた等式・不等式の証明

数研出版｜数学Ⅱ[709] p.160 問題 8

[証明]（右辺）

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\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,\sin(\alpha-\beta)\,}{\,\sin(\alpha+\beta)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sin\alpha\cos\beta-\cos\alpha\sin\beta\,}{\,\sin\alpha\cos\beta+\cos\alpha\sin\beta\,}\end{eqnarray}\)

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分母分子を \(\cos\alpha\cos\beta\) で割ると、

\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,\displaystyle \frac{\,\sin\alpha\cos\beta\,}{\,\cos\alpha\cos\beta\,}-\displaystyle \frac{\,\cos\alpha\sin\beta\,}{\,\cos\alpha\cos\beta\,}\,}{\,\displaystyle \frac{\,\sin\alpha\cos\beta\,}{\,\cos\alpha\cos\beta\,}+\displaystyle \frac{\,\cos\alpha\sin\beta\,}{\,\cos\alpha\cos\beta\,}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\displaystyle \frac{\,\sin\alpha\,}{\,\cos\alpha\,}-\displaystyle \frac{\,\sin\beta\,}{\,\cos\beta\,}\,}{\,\displaystyle \frac{\,\sin\alpha\,}{\,\cos\alpha\,}+\displaystyle \frac{\,\sin\beta\,}{\,\cos\beta\,}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\tan\alpha-\tan\beta\,}{\,\tan\alpha+\tan\beta\,}\end{eqnarray}\)

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したがって、
　\(\displaystyle \frac{\,\tan\alpha-\tan\beta\,}{\,\tan\alpha+\tan\beta\,}=\displaystyle \frac{\,\sin(\alpha-\beta)\,}{\,\sin(\alpha+\beta)\,}\) [終]

数研出版｜高等学校数学Ⅱ[710] p.150 章末問題A 5
数研出版｜新編数学Ⅱ[711] p.146 章末問題A 6

\({\small (1)}~\)

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\(\sin(\pi-\theta)\) について、加法定理より、

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\(\begin{eqnarray}~~~\sin(\pi-\theta)&=&\sin\pi\cos\theta-\cos\pi\sin\theta
\\[5pt]~~~&=&0 \cdot \cos\theta-(-1) \cdot \sin\theta
\\[5pt]~~~&=&\sin\theta\end{eqnarray}\)
 
\(\cos(\pi-\theta)\) について、加法定理より、

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\(\begin{eqnarray}~~~\cos(\pi-\theta)&=&\cos\pi\cos\theta+\sin\pi\sin\theta
\\[5pt]~~~&=&(-1) \cdot \cos\theta+0 \cdot \sin\theta
\\[5pt]~~~&=&-\cos\theta\end{eqnarray}\)
 
\(\tan(\pi-\theta)\) について、

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\(\begin{eqnarray}~~~\tan(\pi-\theta)&=&\displaystyle \frac{\,\sin(\pi-\theta)\,}{\,\cos(\pi-\theta)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sin\theta\,}{\,-\cos\theta\,}
\\[5pt]~~~&=&-\tan\theta\end{eqnarray}\)

 
 

\({\small (2)}~\)

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\(\sin\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)\) について、加法定理より、

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\(\begin{eqnarray}~~~\sin\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)&=&\sin\displaystyle \frac{\,\pi\,}{\,2\,}\cos\theta-\cos\displaystyle \frac{\,\pi\,}{\,2\,}\sin\theta
\\[5pt]~~~&=&1 \cdot \cos\theta-0 \cdot \sin\theta
\\[5pt]~~~&=&\cos\theta\end{eqnarray}\)
 
\(\cos\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)\) について、加法定理より、

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\(\begin{eqnarray}~~~\cos\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)&=&\cos\displaystyle \frac{\,\pi\,}{\,2\,}\cos\theta+\sin\displaystyle \frac{\,\pi\,}{\,2\,}\sin\theta
\\[5pt]~~~&=&0 \cdot \cos\theta+1 \cdot \sin\theta
\\[5pt]~~~&=&\sin\theta\end{eqnarray}\)
 
\(\tan\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)\) について、

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\(\begin{eqnarray}~~~\tan\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)&=&\displaystyle \frac{\,\sin\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)\,}{\,\cos\left(\displaystyle \frac{\,\pi\,}{\,2\,}-\theta\right)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\cos\theta\,}{\,\sin\theta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,\tan\theta\,}\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.11 問

[証明]（左辺より）

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加法定理より、

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\(\begin{eqnarray}~~~\tan\left(\displaystyle \frac{\,\pi\,}{\,4\,}+\theta\right)&=&\displaystyle \frac{\,\tan\displaystyle \frac{\,\pi\,}{\,4\,}+\tan\theta\,}{\,1-\tan\displaystyle \frac{\,\pi\,}{\,4\,}\tan\theta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1+\tan\theta\,}{\,1-1 \cdot \tan\theta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1+\tan\theta\,}{\,1-\tan\theta\,}\end{eqnarray}\)

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したがって、
　\(\tan\left(\displaystyle \frac{\,\pi\,}{\,4\,}+\theta\right)=\displaystyle \frac{\,1+\tan\theta\,}{\,1-\tan\theta\,}\) [終]