$$\frac{1}{cos^{2}10^{\circ}}+\frac{1}{sin^{2}20^{\circ}}+\frac{1}{sin^{2}40^{\circ}}=\frac{1}{\frac{1+cos20^{\circ}}{2}}+\frac{ 1}{1-cos^{2}20^{\circ}}+\frac{1}{(2sin20^{\circ}cos20^{\circ})^{2}}$$
$$=\frac{2}{1+cos20^{\circ}}+\frac{1}{1-cos^{2}20^{\circ}}+\frac{1}{4(1-cos^{2}20^{\circ})cos^{2}20^{\circ}}$$
ให้ $a=cos20^\circ$ จะได้ว่า
$$=\frac{2}{1+a}+\frac{1}{1-a^{2}}+\frac{1}{4(1-a^{2})a^{2}}$$
$$=\frac{2(4a^{2})(1-a)+(4a^2)+1}{4(1-a^{2})a^{2}}$$
$$=\frac{-8a^3+12a^2+1}{4(1-a^{2})a^{2}}$$
$$=\frac{-8cos^{3}20^{\circ}+12cos^{2}20^{\circ}+1}{sin^{2}40^{\circ}}$$
เพราะ $cos60^{\circ}=4cos^{3}20^{\circ}-3cos20^{\circ}$
จึงได้ว่า $-8cos^{3}20^{\circ}=-2cos60^{\circ}-6cos20^{\circ}=-1-6cos20^{\circ}$
นั่นคือ $$\frac{-8cos^{3}20^{\circ}+12cos^{2}20^{\circ}+1}{sin^{2}40^{\circ}}=\frac{(-1-6cos20^{\circ})+12cos^{2}20^{\circ}+1}{(sin40^{\circ})(2sin20^{\circ}cos20^{\circ})}$$
$$=\frac{6(2cos20^{\circ}-1)}{2sin40^{\circ}sin20^{\circ}}$$
$$=\frac{6(2cos20^{\circ}-1)}{cos20^{\circ}-cos60^{\circ}}$$
$$=\frac{6(2cos20^{\circ}-1)}{cos20^{\circ}-\frac{1}{2}}$$
$$=12$$
:great:

ให้ $z=\cos\left(\dfrac{\pi}{9}\right)+i\sin\left(\dfrac{\pi}{9}\right)$

จะได้ว่า

$\sin 20^{\circ}=\dfrac{z-z^{-1}}{2i}$

$\sin 40^{\circ}=\dfrac{z^2-z^{-2}}{2i}$

$\sin 80^{\circ}=\dfrac{z^4-z^{-4}}{2i}$

และ $z^9=-1$

$(z^3+1)(z^6-z^3+1)=0$

$z^6=z^3-1$

$z^7=z^4-z$

$z^8=z^5-z^2$

$z^9=-1$

จึงได้

$\dfrac{1}{\sin^2 20^{\circ}}+\dfrac{1}{\sin^2 40^{\circ}}+\dfrac{1}{\sin^2 80^{\circ}}=\dfrac{-4}{(z-z^{-1})^2}+\dfrac{-4}{(z^2-z^{-2})^2}+\dfrac{-4}{(z^4-z^{-4})^2}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{14}+3z^{12}+3z^{10}+7z^8+3z^6+3z^4+z^2}{(z^8-1)^2}\right]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{14}+3z^{12}+3z^{10}+7z^8+3z^6+3z^4+z^2}{(-z^{-1}-1)^2}\right]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{16}+3z^{14}+3z^{12}+7z^{10}+3z^8+3z^6+z^4}{(z+1)^2}\right]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{-(z^4-z)-3z^{5}-3z^{3}-7z+3(z^5-z^2)+3(z^3-1)+z^4}{(z+1)^2}\right]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{-3z^{2}-6z-3}{(z+1)^2}\right]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=12$

ข้อความเดิมเขียนโดยคุณ bell18 (ข้อความที่ 128146) #50 เยี่ยมครับ ผมชอบวิธีของคุณ doraemon_j ดูแล้วน่าตื่นเต้นดีจัง ^_^