የድምር ወይም የልዩነት ቀመሮች (Product-to-Sum and Sum-to-Product Identities)
የትሪጎኖሜትሪ ፋንክሽኖችን ምርት ወደ ድምር ወይም ልዩነት የሚቀይሩ ቀመሮችን እንመለከታለን፣ እነሱንም ከመደመር ቀመሮች እንዴት እንደምናወጣ እናሳያለን። ከዚያም ድምር ወይም ልዩነትን ወደ ምርት የሚቀይሩ ቀመሮችንም እናወጣለን።
አጭር ማጠቃለያ
ምርትን ወደ ድምር ወይም ልዩነት የሚቀይሩ ቀመሮች (Product-to-Sum Identities)
- \[\sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}\]
- \[\cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}\]
- \[\cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) + \cos ( \alpha - \beta )\}\]
- \[\sin \alpha \sin \beta = - \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) - \cos ( \alpha - \beta ) \}\]
ድምርን ወይም ልዩነትን ወደ ምርት የሚቀይሩ ቀመሮች (Sum-to-Product Identities)
- \[\sin A + \sin B = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}\]
- \[\sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}\]
- \[\cos A + \cos B = 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\]
- \[\cos A - \cos B = -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}\]
ቀመሮቹን ብቻ ሳይሆን የማውጣት ሂደታቸውንም አብሮ መማር ጥሩ ነው።
ቅድመ መስፈርቶች
ምርትን ወደ ድምር ወይም ልዩነት የሚቀይሩ ቀመሮች (Product-to-Sum Identities)
- \[\sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}\]
- \[\cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}\]
- \[\cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) + \cos ( \alpha - \beta )\}\]
- \[\sin \alpha \sin \beta = - \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) - \cos ( \alpha - \beta ) \}\]
ማውጣት
\[\begin{align} \sin(\alpha+\beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \tag{1}\label{eqn:sin_add}\\ \sin(\alpha-\beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{2}\label{eqn:sin_dif} \end{align}\]እንጠቀማለን።
($\ref{eqn:sin_add}$)+($\ref{eqn:sin_dif}$) ካደረግን
\[\sin(\alpha+\beta) + \sin(\alpha-\beta) = 2 \sin \alpha \cos \beta \tag{3}\label{sin_product_to_sum}\] \[\therefore \sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}.\]($\ref{eqn:sin_add}$)-($\ref{eqn:sin_dif}$) ካደረግን
\[\sin(\alpha+\beta) - \sin(\alpha-\beta) = 2 \cos \alpha \sin \beta \tag{4}\label{cos_product_to_dif}\] \[\therefore \cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}.\]በተመሳሳይ መንገድ
\[\begin{align} \cos(\alpha+\beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{5}\label{eqn:cos_add} \\ \cos(\alpha-\beta ) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \tag{6}\label{eqn:cos_dif} \end{align}\]ከዚህ
($\ref{eqn:cos_add}$)+($\ref{eqn:cos_dif}$) ካደረግን
\[\cos(\alpha+\beta) + \cos(\alpha-\beta) = 2 \cos \alpha \cos \beta \tag{7}\label{cos_product_to_sum}\] \[\therefore \cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos(\alpha+\beta) + \cos(\alpha-\beta) \}.\]($\ref{eqn:cos_add}$)-($\ref{eqn:cos_dif}$) ካደረግን
\[\cos(\alpha+\beta) - \cos(\alpha-\beta) = -2 \sin \alpha \sin \beta \tag{8}\label{sin_product_to_dif}\] \[\therefore \sin \alpha \sin \beta = -\frac { 1 } { 2 } \{ \cos(\alpha+\beta) - \cos(\alpha-\beta) \}.\]ድምርን ወይም ልዩነትን ወደ ምርት የሚቀይሩ ቀመሮች (Sum-to-Product Identities)
- \[\sin A + \sin B = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}\]
- \[\sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}\]
- \[\cos A + \cos B = 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\]
- \[\cos A - \cos B = -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}\]
ማውጣት
ምርትን ወደ ድምር ወይም ልዩነት የሚቀይሩ ቀመሮች (Product-to-Sum Identities) ከሚባሉት መነሳት፣ ድምርን ወይም ልዩነትን ወደ ምርት የሚቀይሩ ቀመሮች (Sum-to-Product Identities) እንዲሁም ማውጣት ይቻላል።
\[\alpha + \beta = A, \quad \alpha - \beta = B\]ብለን ካስቀመጥን እና ሁለቱን ስሌቶች በ $\alpha$, $\beta$ ላይ በአንድነት ካፈታን
\[\alpha = \frac{A+B}{2}, \quad \beta = \frac{A-B}{2}.\]ይህንንም ቀደም ሲል ባሉት ($\ref{sin_product_to_sum}$), ($\ref{cos_product_to_dif}$), ($\ref{cos_product_to_sum}$), ($\ref{sin_product_to_dif}$) ውስጥ በየተራ በመተካት የሚከተሉትን ቀመሮች እናገኛለን።
\[\begin{align*} \sin A + \sin B &= 2\sin \frac{A+B}{2}\cos \frac{A-B}{2} \\ \sin A - \sin B &= 2\cos \frac{A+B}{2}\sin \frac{A-B}{2} \\ \cos A + \cos B &= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2} \\ \cos A - \cos B &= -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}. \end{align*}\]