Question

Найдите синус косинус тангенс котангенс 22³30

Answer 1

$22^030'=22,5^0=\frac{45^0}{2};\\ a=22,5^0;\\ \sin22^030',\ \cos22^030',\ tg22^030'-?\\ \left \{ {{\sin2a=2\sin a\cos a} \atop {\cos2a=\cos^2a-\sin^2a=2\cos^2a-1=1-2\sin^2a}} \right.\\ \sin^2a =\frac{1-\cos2a}{2};\\ \cos^2a=\frac{1+\cos2a}{2};\\ \sin a=\pm\sqrt{\frac{1-\cos2a}{2}};\\ \cos a=\pm\sqrt{\frac{1+\cos2a}{2}};\\ a=22,5^0==>\sin a,\ \cos a,\ tga>0;\\ \sin a=\sqrt{\frac{1-\cos2a}{2}};\\ \cos a=\sqrt{\frac{1+\cos2a}{2}};\\ tg a=\frac{\sin a}{\cos a}=\sqrt{\frac{1-\cos2a}{1+\cos2a}};$
$\cos45^0=\frac{1}{\sqrt2}=\frac{\sqrt{2}}{2}\\ \sin22^030'=\sqrt{\frac{1-\cos45^0}{2}}=\sqrt{\frac{1-\frac{\sqrt2}{2}}{2}}=\sqrt{\frac{2-\sqrt2}{4}}=\frac{\sqrt{2-\sqrt{2}}}{2};\\ \cos22^030'=\sqrt{\frac{1+\cos45^0}{2}}=\sqrt{\frac{1+\frac{\sqrt2}{2}}{2}}=\sqrt{\frac{2+\sqrt2}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2};\\ tg22^030'=\frac{\sin22^030'}{\cos22^030'}=\sqrt{\frac{1-\cos45^0}{1+\cos45^0}}=\sqrt{\frac{1-\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}}=$
$=\\ =\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}=\sqrt{\frac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}}=\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}$


Ответ:
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