積和公式と和積公式

本記事では,三角関数の積和公式と和積公式を証明します.
加法定理を使って証明しますので,加法定理がわからない人は以下の記事を参考にしてください.

積和公式

積和公式.
sinαcosβ=12{sin(α+β)+sin(αβ)}cosαsinβ=12{sin(α+β)sin(αβ)}cosαcosβ=12{cos(α+β)+cos(αβ)}sinαsinβ=12{cos(α+β)cos(αβ)} \begin{aligned} \sin{\alpha} \cos{\beta} &= \frac{1}{2} \{ \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)} \} \\ \cos{\alpha} \sin{\beta} &= \frac{1}{2} \{ \sin{(\alpha + \beta)} - \sin{(\alpha - \beta)} \} \\ \cos{\alpha} \cos{\beta} &= \frac{1}{2} \{ \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)} \} \\ \sin{\alpha} \sin{\beta} &= - \frac{1}{2} \{ \cos{(\alpha + \beta)} - \cos{(\alpha - \beta)} \} \\ \end{aligned}

sinαcosβ=12{sin(α+β)+sin(αβ)} \sin{\alpha} \cos{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)} \} の証明.
正弦の加法定理
sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta}

sin(αβ)=sinαcosβcosαsinβ\sin(\alpha - \beta) = \sin{\alpha} \cos{\beta} - \cos{\alpha} \sin{\beta}
を足して,
sin(α+β)+sin(αβ)=2sinαcosβ \begin{aligned} \sin(\alpha + \beta) + \sin(\alpha - \beta) = 2 \sin{\alpha} \cos{\beta} \end{aligned}
sinαcosβ=12{sin(α+β)+sin(αβ)} \begin{aligned} \sin{\alpha} \cos{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)} \} \end{aligned}

cosαsinβ=12{sin(α+β)sin(αβ)} \cos{\alpha} \sin{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} - \sin{(\alpha - \beta)} \} の証明.
正弦の加法定理
sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta}
から
sin(αβ)=sinαcosβcosαsinβ\sin(\alpha - \beta) = \sin{\alpha} \cos{\beta} - \cos{\alpha} \sin{\beta}
を引いて,
sin(α+β)sin(αβ)=2cosαsinβ \begin{aligned} \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2 \cos{\alpha} \sin{\beta} \end{aligned}
cosαsinβ=12{sin(α+β)sin(αβ)} \begin{aligned} \cos{\alpha} \sin{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} - \sin{(\alpha - \beta)} \} \end{aligned}

cosαcosβ=12{cos(α+β)+cos(αβ)} \cos{\alpha} \cos{\beta} = \frac{1}{2} \{ \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)} \} の証明.
余弦の加法定理
cos(α+β)=cosαcosβsinαsinβ\cos(\alpha + \beta) = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}

cos(αβ)=cosαcosβ+sinαsinβcos(\alpha - \beta) = \cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}
を足して,
cos(α+β)+cos(αβ)=2cosαcosβ \begin{aligned} \cos(\alpha + \beta) + \cos(\alpha - \beta) = 2 \cos{\alpha} \cos{\beta} \end{aligned}
cosαcosβ=12{cos(α+β)+cos(αβ)} \begin{aligned} \cos{\alpha} \cos{\beta} = \frac{1}{2} \{ \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)} \} \end{aligned}

sinαsinβ=12{cos(α+β)cos(αβ)} \sin{\alpha} \sin{\beta} = - \frac{1}{2} \{ \cos{(\alpha + \beta)} - \cos{(\alpha - \beta)} \} の証明.
余弦の加法定理
cos(α+β)=cosαcosβsinαsinβ\cos(\alpha + \beta) = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}
から
cos(αβ)=cosαcosβ+sinαsinβcos(\alpha - \beta) = \cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}
を引いて,
cos(α+β)cos(αβ)=2sinαsinβ \begin{aligned} \cos(\alpha + \beta) - \cos(\alpha - \beta) = - 2 \sin{\alpha} \sin{\beta} \end{aligned}
sinαsinβ=12{cos(α+β)cos(αβ)} \begin{aligned} \sin{\alpha} \sin{\beta} = - \frac{1}{2} \{ \cos{(\alpha + \beta)} - \cos{(\alpha - \beta)} \} \end{aligned}

和積公式

和積公式.
sinA+sinB=2sinA+B2cosAB2sinAsinB=2cosA+B2sinAB2cosA+cosB=2cosA+B2cosAB2cosAcosB=2sinA+B2sinAB2 \begin{aligned} \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} } \\ \end{aligned}

sinA+sinB=2sinA+B2cosAB2\sin{A} + \sin{B} = 2 \sin{ \frac{A+B}{2} } \cos{ \frac{A - B}{2} } の証明.
α+β=A \alpha + \beta = A αβ=B \alpha - \beta = B とおく.
すると,α=A+B2 \begin{aligned} \alpha = \frac{A + B}{2} \end{aligned} β=AB2 \begin{aligned} \beta = \frac{A - B}{2} \end{aligned} となる.
これらを積和公式
sinαcosβ=12{sin(α+β)+sin(αβ)} \begin{aligned} \sin{\alpha} \cos{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)} \} \end{aligned}
に代入して,
sinA+B2cosAB2=12(sinA+sinB) \begin{aligned} \sin{\frac{A + B}{2}} \cos{\frac{A - B}{2}} = \frac{1}{2} ( \sin{A} + \sin{B} ) \end{aligned}
sinA+sinB=2sinA+B2cosAB2 \begin{aligned} \sin{A} + \sin{B} = 2 \sin{\frac{A + B}{2}} \cos{\frac{A - B}{2}} \end{aligned}

sinAsinB=2cosA+B2sinAB2\sin{A} - \sin{B} = 2 \cos{ \frac{A+B}{2} } \sin{ \frac{A - B}{2} } の証明.
α+β=A \alpha + \beta = A αβ=B \alpha - \beta = B とおく.
すると,α=A+B2 \begin{aligned} \alpha = \frac{A + B}{2} \end{aligned} β=AB2 \begin{aligned} \beta = \frac{A - B}{2} \end{aligned} となる.
これらを積和公式
cosαsinβ=12{sin(α+β)sin(αβ)} \begin{aligned} \cos{\alpha} \sin{\beta} = \frac{1}{2} \{ \sin{(\alpha + \beta)} - \sin{(\alpha - \beta)} \} \end{aligned}
に代入して,
cosA+B2sinAB2=12(sinAsinB) \begin{aligned} \cos{\frac{A + B}{2}} \sin{\frac{A - B}{2}} = \frac{1}{2} ( \sin{A} - \sin{B} ) \end{aligned}
sinAsinB=2cosA+B2sinAB2 \begin{aligned} \sin{A} - \sin{B} = 2 \cos{\frac{A + B}{2}} \sin{\frac{A - B}{2}} \end{aligned}

cosA+cosB=2cosA+B2cosAB2\cos{A} + \cos{B} = 2 \cos{ \frac{A+B}{2} } \cos{ \frac{A - B}{2} } の証明.
α+β=A \alpha + \beta = A αβ=B \alpha - \beta = B とおく.
すると,α=A+B2 \begin{aligned} \alpha = \frac{A + B}{2} \end{aligned} β=AB2 \begin{aligned} \beta = \frac{A - B}{2} \end{aligned} となる.
これらを積和公式
cosαcosβ=12{cos(α+β)+cos(αβ)} \begin{aligned} \cos{\alpha} \cos{\beta} = \frac{1}{2} \{ \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)} \} \end{aligned}
に代入して,
cosA+B2cosAB2=12(cosA+cosB) \begin{aligned} \cos{\frac{A + B}{2}} \cos{\frac{A - B}{2}} = \frac{1}{2} ( \cos{A} + \cos{B} ) \end{aligned}
cosA+cosB=2cosA+B2cosAB2 \begin{aligned} \cos{A} + \cos{B} = 2 \cos{\frac{A + B}{2}} \cos{\frac{A - B}{2}} \end{aligned}

cosAcosB=2sinA+B2sinAB2\cos{A} - \cos{B} = - 2 \sin{ \frac{A+B}{2} } \sin{ \frac{A - B}{2} } の証明.
α+β=A \alpha + \beta = A αβ=B \alpha - \beta = B とおく.
すると,α=A+B2 \begin{aligned} \alpha = \frac{A + B}{2} \end{aligned} β=AB2 \begin{aligned} \beta = \frac{A - B}{2} \end{aligned} となる.
これらを積和公式
sinαsinβ=12{cos(α+β)cos(αβ)} \begin{aligned} \sin{\alpha} \sin{\beta} = - \frac{1}{2} \{ \cos{(\alpha + \beta)} - \cos{(\alpha - \beta)} \} \end{aligned}
に代入して,
sinA+B2sinAB2=12(cosAcosB) \begin{aligned} \sin{\frac{A + B}{2}} \sin{\frac{A - B}{2}} = - \frac{1}{2} ( \cos{A} - \cos{B} ) \end{aligned}
cosAcosB=2sinA+B2sinAB2 \begin{aligned} \cos{A} - \cos{B} = - 2 \sin{\frac{A + B}{2}} \sin{\frac{A - B}{2}} \end{aligned}

  

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