ここで勉強すれば数学検定１級の壁は超えられるか。

$\cos \alpha =\sqrt {\dfrac {1}{2}+\dfrac {1}{2\sqrt {5}}},\cos \beta =\sqrt {\dfrac {1}{2}+\dfrac {1}{\sqrt {5}}}$

$0<\alpha ,\beta <\dfrac {\pi }{2}$　のとき、

$\cos \left( \alpha +\beta \right)$の値を求める。

$cos \alpha =\sqrt {\dfrac {\sqrt {5}+1}{2\sqrt {5}}}\Rightarrow \sin \alpha =\sqrt {\dfrac {\sqrt {5}-1}{2\sqrt {5}}}$

$\cos \beta =\sqrt {\dfrac {\sqrt {5}+2}{2\sqrt {5}}}\Rightarrow \sin \beta =\sqrt {\dfrac {\sqrt {5}-2}{2\sqrt {5}}}$

$cos \alpha \cos \beta =\sqrt {\dfrac {14+2\sqrt {45}}{40}}=\dfrac {3+\sqrt {5}}{2\sqrt {10}}$

$sin \alpha \sin \beta = \sqrt {\dfrac {14-2\sqrt {45}}{40}} = \dfrac {3-\sqrt {5}}{2\sqrt {10}}$

$cos \left( \alpha +\beta \right) =\cos \alpha \cos \beta -\sin \alpha \sin \beta$

$=\dfrac {3+\sqrt {5}}{2\sqrt {10}}-\dfrac {3-\sqrt {5}}{2\sqrt {10}}=\dfrac {2\sqrt {5}}{2\sqrt {10}}=\dfrac {1}{\sqrt {2}}$・・・答え