Câu 6.59 trang 206 Sách BT Đại số 10 Nâng cao:...

Cho $\sin \alpha  - \cos \alpha  = m$. Hãy tính theo $m$

a) $\sin \alpha \cos \alpha ;$

b) $\left| {\sin \alpha  + \cos \alpha } \right|;$

c) ${\sin ^3}\alpha  - {\cos ^3}\alpha ;$

d) ${\sin ^6}\alpha  + {\cos ^6}\alpha .$

Cho $\sin \alpha  - \cos \alpha  = m$ ta có

a)

$\begin{array}{l}\sin \alpha \cos \alpha  =  - \dfrac{1}{2}\left[ {{{\left( {\sin \alpha  - \cos \alpha } \right)}^2} - 1} \right]\\ = \dfrac{{1 - {m^2}}}{2}.\end{array}$

b)

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

Từ đó $\left| {\sin \alpha  + \cos \alpha } \right| = \sqrt {2 - {m^2}} .$

c)

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

d)

$\begin{array}{l}{\sin ^6}\alpha  + {\cos ^6}\alpha \\ = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha  + co{s^2}\alpha } \right)\\ = 1 - 3{\left( {\dfrac{{1 - {m^2}}}{2}} \right)^2}\\ = \dfrac{{ - 3{m^4} + 6{m^2} + 1}}{4}.\end{array}$