อ้างอิง:
ข้อความเดิมเขียนโดยคุณ FranceZii Siriseth
จงหาค่าของ
$$(\frac{1}{sin(20)})^2+(\frac{1}{sin(40)})^2+(\frac{1}{sin(80) })^2 $$
ปล.มุมเป็นองศา
|
$=\frac{(sin 40sin 80)^2+(sin 20sin80)^2+(sin20sin40)^2}{(sin 20sin 40sin80)^2} $
$=\frac{(sin 40cos 10)^2+(sin 20cos 10)^2+(sin20sin40)^2}{(sin 20sin 40cos 10)^2}$
$=\frac{({\frac{sin 50+sin30}{2}})^2 +({\frac{sin 30+sin10}{2}})^2+({\frac{cos 20+cos 60}{2}})^2}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}[(cos 40+\frac{1}{2})^2 +(\frac{1}{2}+sin 10)^2+(cos 20-\frac{1}{2})^2]}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}[(cos^2 40+cos 40+\frac{1}{4})+(sin^2 10+sin 10+\frac{1}{4})+(cos^2 20-cos 20+\frac{1}{4})]}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(cos^2 40+sin^2 10+cos^2 20+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(\frac{cos 80+1}{2} +\frac{1-cos 20}{2}+\frac{cos 40+1}{2}+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(\frac{sin 10+1}{2} +\frac{1-cos 20}{2}+\frac{cos 40+1}{2}+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{3}{8}(cos 40+sin 10-cos 20 +\frac{3}{2})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{3}{8}(0 +\frac{3}{2})}{\frac{3}{64}}$
$=12$