Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\frac{\sqrt{6} -\sqrt{2}}{4}\\ b.\ \frac{\sqrt{5} -1}{2}\\ \\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ cos795^{o} =cos\left( 75^{o} +2.360^{o}\right) =cos75^{o} =cos30^{o} cos45^{o} -sin30^{o} sin45^{o}\\ =\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} -\frac{1}{2}\frac{\sqrt{2}}{2} =\frac{\sqrt{6} -\sqrt{2}}{4}\\ b.\ Ta\ có:\ sin54^{o} =cos36^{o} =cos2.18^{o} =1-2sin^{2} 18^{o}\\ Ta\ có:\ sin54^{o} =sin\left( 18^{o} +36^{o}\right) =sin18^{o} cos36^{o} +cos18^{o} sin36^{o} =\ 3sin18^{o} -4sin^{3} 18^{o}\\ Do\ đó:\ 3sin18^{o} -4sin^{3} 18=1-2sin^{2} 18^{o}\\ \Leftrightarrow sin\ 18^{0} =1\ or\ sin\ 18^{0} =\frac{\sqrt{5} \pm 1}{2}\\ Do\ 0< sin\ 18^{0} < 1\Rightarrow sin\ 18^{0} =\frac{\sqrt{5} -1}{2}\\ \\ \end{array}$
