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Câu hỏi: Biết cosα +cosβ =a; sinα+sinβ =b (a, b là hằng số và a2 + b2 ≠ 0)
Hãy tính sin(α + β) theo a và b.
Hãy tính sin(α + β) theo a và b.
Lời giải chi tiết
Ta có:
\(\left. \matrix{
a = 2\cos {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} \hfill \cr
b = 2\sin {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} \hfill \cr} \right\} \)
\(\begin{array}{l}
\Rightarrow ab = 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}. 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\
= \left({2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha + \beta }}{2}} \right). 2{\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 2\sin \left({\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\
{a^2} + {b^2}\\
= 4{\cos ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\
+ 4{\sin ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 4{\cos ^2}\frac{{\alpha - \beta }}{2}\left({{{\cos }^2}\frac{{\alpha + \beta }}{2} + {{\sin }^2}\frac{{\alpha + \beta }}{2}} \right)\\
= 4{\cos ^2}\frac{{\alpha - \beta }}{2}\\
\Rightarrow {\cos ^2}\frac{{\alpha - \beta }}{2} = \frac{{{a^2} + {b^2}}}{4}\\
\Rightarrow ab = 2\sin \left({\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 2\sin \left({\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{4}\\
= \sin \left({\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{2}\\
\Rightarrow \sin \left({\alpha + \beta } \right) = \frac{{2ab}}{{{a^2} + {b^2}}}
\end{array}\)
Cách khác:
Từ giả thiết ta có:
(I): $\left\{\begin{array}{l}\cos \alpha+\cos \beta=a \\ \sin \alpha+\sin \beta=b\end{array}\right.$
$=>\left\{\begin{array}{l}\cos ^{2} \alpha+2 \cos \alpha \cdot \cos +\cos ^{2} \beta=a^{2} \\ \sin ^{2} \alpha+2 \sin \alpha \cdot \sin +\sin ^{2} \beta=b^{2}\end{array}\right.$
$=>\cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+\cos ^{2} \beta$
$+\sin ^{2} \alpha+2 \sin \alpha \cdot \sin \beta+\sin ^{2} \beta=b^{2}+a^{2}$
$=>2+2 \cos (\alpha-\beta)=a^{2}+b^{2}=>\cos (\alpha-\beta)=\frac{a^{2}+b^{2}-2}{2}$ (1)
Mặt khác từ $(\mathrm{I})=>(\cos \alpha+\cos \beta)(\sin \alpha+\sin \beta)=a \cdot b$
$=>\sin \alpha \cdot \cos \alpha+\sin (\alpha+\beta)+\sin \cdot \cos \beta=a \cdot b$
$\Rightarrow>(\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta)+\sin (\alpha+\beta)=\boldsymbol{a} \cdot \boldsymbol{b}$
$=>\frac{1}{2}(\sin 2 \alpha+\sin 2 \beta)+\sin (\alpha+\beta)=a \cdot b$
$=>\sin (\alpha+\beta)[1+\cos (\alpha-\beta)]=a \cdot b$ (2)
Từ (1) và (2) ta có:
$\sin \alpha+\beta=\frac{a \cdot b}{1+\frac{a^{2}+b^{2}-2}{2}}=\frac{2 a \cdot b}{a^{2}+b^{2}}\left(a^{2}+b^{2} \neq 0\right)$
Ta có:
\(\left. \matrix{
a = 2\cos {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} \hfill \cr
b = 2\sin {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} \hfill \cr} \right\} \)
\(\begin{array}{l}
\Rightarrow ab = 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}. 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\
= \left({2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha + \beta }}{2}} \right). 2{\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 2\sin \left({\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\
{a^2} + {b^2}\\
= 4{\cos ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\
+ 4{\sin ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 4{\cos ^2}\frac{{\alpha - \beta }}{2}\left({{{\cos }^2}\frac{{\alpha + \beta }}{2} + {{\sin }^2}\frac{{\alpha + \beta }}{2}} \right)\\
= 4{\cos ^2}\frac{{\alpha - \beta }}{2}\\
\Rightarrow {\cos ^2}\frac{{\alpha - \beta }}{2} = \frac{{{a^2} + {b^2}}}{4}\\
\Rightarrow ab = 2\sin \left({\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\
= 2\sin \left({\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{4}\\
= \sin \left({\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{2}\\
\Rightarrow \sin \left({\alpha + \beta } \right) = \frac{{2ab}}{{{a^2} + {b^2}}}
\end{array}\)
Cách khác:
Từ giả thiết ta có:
(I): $\left\{\begin{array}{l}\cos \alpha+\cos \beta=a \\ \sin \alpha+\sin \beta=b\end{array}\right.$
$=>\left\{\begin{array}{l}\cos ^{2} \alpha+2 \cos \alpha \cdot \cos +\cos ^{2} \beta=a^{2} \\ \sin ^{2} \alpha+2 \sin \alpha \cdot \sin +\sin ^{2} \beta=b^{2}\end{array}\right.$
$=>\cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+\cos ^{2} \beta$
$+\sin ^{2} \alpha+2 \sin \alpha \cdot \sin \beta+\sin ^{2} \beta=b^{2}+a^{2}$
$=>2+2 \cos (\alpha-\beta)=a^{2}+b^{2}=>\cos (\alpha-\beta)=\frac{a^{2}+b^{2}-2}{2}$ (1)
Mặt khác từ $(\mathrm{I})=>(\cos \alpha+\cos \beta)(\sin \alpha+\sin \beta)=a \cdot b$
$=>\sin \alpha \cdot \cos \alpha+\sin (\alpha+\beta)+\sin \cdot \cos \beta=a \cdot b$
$\Rightarrow>(\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta)+\sin (\alpha+\beta)=\boldsymbol{a} \cdot \boldsymbol{b}$
$=>\frac{1}{2}(\sin 2 \alpha+\sin 2 \beta)+\sin (\alpha+\beta)=a \cdot b$
$=>\sin (\alpha+\beta)[1+\cos (\alpha-\beta)]=a \cdot b$ (2)
Từ (1) và (2) ta có:
$\sin \alpha+\beta=\frac{a \cdot b}{1+\frac{a^{2}+b^{2}-2}{2}}=\frac{2 a \cdot b}{a^{2}+b^{2}}\left(a^{2}+b^{2} \neq 0\right)$