Bài tập 35 trang 207 SGK Toán 10 NC

Ta có:

\(\begin{array}{*{20}{l}}
{{{\sin }^3}\alpha  - {{\cos }^3}\alpha }\\
\begin{array}{l}
 = \left( {\sin \alpha  - \cos \alpha } \right)({\sin ^2}\alpha \\
 + \sin \alpha \cos \alpha  + {\cos ^2}\alpha )
\end{array}\\
{ = m\left( {1 + \sin \alpha \cos \alpha } \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( 1 \right)}
\end{array}\)

Từ \(\sin \alpha  - \cos \alpha  = m \)

\(\Rightarrow 1 - 2\sin \alpha \cos \alpha  = {m^2}\)

\( \Rightarrow \sin \alpha \cos \alpha  = \frac{{1 - {m^2}}}{2}\,\,\,\,\,\,\left( 2 \right)\)

Thay (2) vào (1) ta được:

\(\begin{array}{l}
{\sin ^3}\alpha  - {\cos ^3}\alpha \\
 = m\left( {1 + \frac{{1 - {m^2}}}{2}} \right) = \frac{m}{2}\left( {3 - {m^2}} \right)
\end{array}\)

-- Mod Toán 10 HỌC247