Đáp án: $B=2$
Giải thích các bước giải:
Ta có:
$\cos 36^o = \sin 54^o$
$\Leftrightarrow 1-2\sin^2 18^o= 3\sin 18^o-4\sin^3 18^o$
$\Leftrightarrow 4t^3 -2t^2 -3t +1=0$ với $t = \sin 18^o$
$\to t\in\{1,\dfrac14(-\sqrt{5}-1),\dfrac14(\sqrt{5}-1)\}$
Mà $-1\le t\le 1, t=\sin18^o\to t=\dfrac14(\sqrt{5}-1)$
Ta có:
$B=\dfrac{1}{\sin(\dfrac{\pi}{10})}-\dfrac{1}{\sin(\dfrac{3\pi}{10})}$
$\to B=\dfrac{1}{\sin(18^o)}-\dfrac{1}{\sin(54^o)}$
$\to B=\dfrac{1}{\sin(18^o)}-\dfrac{1}{\cos(36^o)}$
$\to B=\dfrac{1}{\sin(18^o)}-\dfrac{1}{1-2\sin^2(18^o)}$
$\to B=\dfrac{1}{\dfrac14(\sqrt{5}-1)}-\dfrac{1}{1-2\cdot (\dfrac14(\sqrt{5}-1))^2}$
$\to B=2$
