Lời giải
Ta có cos 2α = 2 cos2 α – 1 = \(2.{\left( {\frac{3}{4}} \right)^2} - 1 = \frac{1}{8}\).
Ta có sin2 α = 1 – cos2 a = \(1 - {\left( {\frac{3}{4}} \right)^2}\)= \(\frac{7}{{16}}\).
Lại do sin α > 0 nên sin α = \(\frac{{\sqrt 7 }}{4}\).
Suy ra sin 2α = 2 sin α cos α = \(2.\frac{{\sqrt 7 }}{4}.\frac{3}{4} = \frac{{3\sqrt 7 }}{8}\).
Ta có cos 2β = 1 – 2 sin2 β = \(1 - 2.{\left( {\frac{3}{5}} \right)^2}\) = \(\frac{7}{{25}}\).
Ta có cos2 β = 1 – sin2 β = \(1 - {\left( {\frac{3}{5}} \right)^2}\)= \(\frac{{16}}{{25}}\).
Lại do \(\beta \in \left( {\frac{{9\pi }}{2};\,\,5\pi } \right)\) nên cos β < 0, do đó \(\cos \beta = - \frac{4}{5}\).
Suy ra sin 2β = 2 sin β cos β = \(2.\frac{3}{5}.\left( { - \frac{4}{5}} \right) = - \frac{{24}}{{25}}\).
Ta có
cos(α + β) = cos α cos β – sin α sin β = \(\frac{3}{4}.\left( { - \frac{4}{5}} \right) - \frac{{\sqrt 7 }}{4}.\frac{3}{5} = \frac{{ - 12 - 3\sqrt 7 }}{{20}}\).
sin(α – β) = sin α cos β – cos α sin β = \(\frac{{\sqrt 7 }}{4}.\left( { - \frac{4}{5}} \right) - \frac{3}{4}.\frac{3}{5} = \frac{{ - 9 - 4\sqrt 7 }}{{20}}\).