Question

Сократите дробь:
$1) \frac{ \cos80 {}^{0} }{ \cos40 {}^{0} - \sin40 {}^{0} } \\ 2) \frac{ \cos36 {}^{0} + \sin {}^{2}18 {}^{ {0}^{} } }{ \cos18 {}^{0} }$
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Answer 1

Ответ:

Объяснение:

Формула :

$\boldsymbol{ \boxed{\boxed{\sf \cos2a= \cos^2a -\sin^2a}}}$


Решение :

$1) \ \ \displaystyle \sf \frac{\cos 80 }{\cos 40 - \sin 40} =\frac{\cos^240-\sin ^240}{\cos 40 - \sin 40} = \\\\\\\frac{(\cos 40 - \sin 40)(\cos 40 + \sin 40)}{\cos 40 - \sin 40} = \cos 40 +\sin 40$



$\sf 2) ~~~\displaystyle \frac{\cos 36 +\sin^2 18}{\cos 18 }=\frac{\cos^218 -\sin ^218 + \sin ^218}{\cos18 } =\frac{\cos^2 18 }{\cos 18 } = \cos 18$

Answer 2

$\displaystyle\bf\\1)\\\\\frac{Cos80^\circ}{Cos40^\circ-Sin40^\circ } =\frac{Cos(2\cdot 40^\circ)}{Cos40^\circ-Sin40^\circ } =\frac{Cos^{2} 40^\circ-Sin^{2} 40^\circ }{Cos40^\circ-Sin^40^\circ } =\\\\\\=\frac{(Cos40^\circ-Sin40^\circ)(Cos40^\circ+Sin40^\circ) }{Cos40^\circ-Sin^40^\circ } =Cos40^\circ+Sin40^\circ$

$\displaystyle\bf\\2)\\\\\frac{Cos36^\circ+Sin^{2}18^\circ }{Cos18^\circ} =\frac{Cos^{2} 18^\circ-Sin^{2} 18^\circ+Sin^{2}18^\circ }{Cos18^\circ} =\\\\\\=\frac{Cos^{2} 18^\circ }{Cos18^\circ} =Cos18^\circ$