Question

Sa se calculeze cos 23pi/12x sin pi/12

Answer 1

Bună! ⭐

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$cos(\frac{23\pi }{12}) = cos(\frac{\pi }{6} +\frac{7\pi }{4} )=$

$=cos(\frac{\pi }{6} )*cos(\frac{7\pi }{4}) -sin(\frac{\pi }{6} ) *sin(\frac{7\pi }{4} )=$

$=cos(30\°)*cos(315\°)-sin(30\°)*sin(315\°)=$

$=\frac{\sqrt{3} }{2} *\frac{\sqrt{2} }{2}-\frac{1}{2} } *(-\frac{\sqrt{2} }{2} )=$

$= \frac{\sqrt{6} }{4} +\frac{\sqrt{2} }{4} =\\\\={\boxed{\mathb{\frac{\sqrt{6} +\sqrt{2} }{4} }}$

$sin(\frac{\pi }{12} )= sin(\frac{\pi }{4} -\frac{\pi }{6} )=$

$= sin(\frac{\pi }{4})* cos(\frac{\pi }{6} )- cos(\frac{\pi }{4} ) * sin(\frac{\pi }{6} ) =$

$= sin(45\°)*cos(30\°)-cos(45\°)*sin(30\°) =$

$=\frac{\sqrt{2} }{2} *\frac{\sqrt{3} }{2} -\frac{\sqrt{2} }{2} *\frac{1}{2} =$

$=\frac{\sqrt{6} }{4} -\frac{\sqrt{2} }{4} =\\\\= {\boxed{\mathb{\frac{\sqrt{6}-\sqrt{2} }{4} }}$

$cos(\frac{23\pi }{12} )*sin(\frac{\pi }{12} ) =$

$=\frac{\sqrt{6} +\sqrt{2} }{4} *\frac{\sqrt{6} -\sqrt{2} }{4} =$

$=\frac{(\sqrt{6} +\sqrt{2})*(\sqrt{6}-\sqrt{2})}{16} =$

$=\frac{6-2}{16} =\frac{4}{16} ^{(4}= \\\\={\boxed{\boxed{\mathbf{\frac{1}{4} }}}$