Riêng câu nè xin 1 ít sò nhá :))
Đặt $\cos \dfrac{\pi}{32}=x \Rightarrow \cos^2 \dfrac{\pi}{32} =x^2$ hạ bậc ta có
$1+\cos \dfrac{\pi}{16}=2x^2 \Rightarrow \cos^2 \dfrac{\pi}{16} =(2x^2 -1)^2$ lại hạ bậc tiếp
$1+\cos \dfrac{\pi}{8}= 2(2x^2-1)^2 \Rightarrow \cos^2 \dfrac{\pi}{8}=\bigg [ 2(2x^2-1)^2 -1 \bigg ] ^2$ lại hạ bấc phát chót
$1+\cos \dfrac{\pi}{4}=2 \bigg [ 2(2x^2-1)^2 -1 \bigg ] ^2$
$\Leftrightarrow 1+\dfrac{\sqrt 2}{2}=2 \bigg [ 2(2x^2-1)^2 -1 \bigg ] ^2$
$2+\sqrt 2= 4 \bigg [ 2(2x^2-1)^2 -1 \bigg ] ^2$
$2 \bigg [ 2(2x^2-1)^2 -1 \bigg ] =\sqrt{2+\sqrt 2}$
$\Leftrightarrow 4(2x^2-1)^2 = 2+\sqrt{2+\sqrt 2}$
$\Leftrightarrow 2 (2x^2-1) =\sqrt {2+\sqrt{2+\sqrt 2}}$
$\Leftrightarrow 4 x^2 =2+\sqrt {2+\sqrt{2+\sqrt 2}}$
$\Leftrightarrow 2x=2\cos \dfrac{\pi}{32} = \sqrt{2+\sqrt {2+\sqrt{2+\sqrt 2}}}$
Vậy $\cos \dfrac{\pi}{32} =\dfrac{1}{2} \sqrt{2+\sqrt {2+\sqrt{2+\sqrt 2}}}$