Question

1) 2) 3)

Answer 1

1) $\frac{cos2x}{cosx+sinx}=\frac{cos^{2}x-sin^{2}x}{cosx+sinx}= \\ =\frac{(cosx-sinx)(cosx+sinx)}{cosx+sinx}=cosx-sinx=cosx-cos(\frac{ \pi }{2}-x)=\\=-2sin(x-\frac{ \pi }{4})$
3)$sin10cos20cos40=\frac{(2cos10sin10)cos20cos40}{2cos10}=\\=\frac{sin20cos20cos40}{2cos10}=\frac{1/2sin40cos40}{2cos10}=\\=\frac{sin40cos40}{4cos10}=\frac{sin80}{8cos10}=\frac{sin(90-10)}{8cos10}=\\=\frac{cos10}{8cos10}=\frac{1}{8}$