Giải bài 1.5 trang 7

Đề bài

Cho $\sin a+\cos a=m$. Hãy tính theo m.

a) $\sin a\cos a$                     

b) ${\sin }^{3}a+{\cos }^{3}a$                     

c) ${\sin }^{4}a+{\cos }^{4}a$.

Lời giải chi tiết

a) Ta có $\sin a+\cos a=m$ nên ${\left(\sin a+\cos a\right)}^2=m^2$

$\begin{array}{c}{\sin }^{2}a+2\sin a\cos a+{\cos }^{2}a=m^2\\ \Rightarrow \left({\sin }^{2}a+{\cos }^{2}a\right)+2\sin a\cos a=m^2\\ \Rightarrow 1+2\sin a\cos a=m^2\\ \Rightarrow 2\sin a\cos a=m^2-1\\ \Rightarrow \sin a\cos a=\frac{m^2-1}{2}.\end{array}$

b) Ta có

$\begin{array}{c}{\sin }^{3}a+{\cos }^{3}a\\ =(\sin a+\cos a)({\sin }^{2}a-\sin a\cos a+{\cos }^{2}a)\\ =m.\left[({\sin }^{2}a+{\cos }^{2}a)-\sin a\cos a\right]\\ =m.(1-\sin a\cos a)\\ =m\left(1-\frac{m^2-1}{2}\right)=m.\frac{2-m^2+1}{2}=m.\frac{3-m^2}{2}=\frac{3m-m^3}{2}.\end{array}$

c)

$\begin{array}{c}{\sin }^{4}a+{\cos }^{4}a\\ =\left({\sin }^{4}a+2{\sin }^{2}a{\cos }^{2}a+{\cos }^{4}a\right)-2{\sin }^{2}a{\cos }^{2}a\\ ={\left({\sin }^{2}a+{\cos }^{2}a\right)}^2-2(\sin a\mathrm{cosa}{)}^2\\ =1^2-2{\left(\frac{m^2-1}{2}\right)}^2=1-2.\frac{{\left(m^2-1\right)}^2}{4}=1-\frac{{\left(m^2-1\right)}^2}{2}.\end{array}$