Question

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24,6 и 24,8​

Answer 1

$24.6\ \ 1)\ \ \dfrac{sin(a+\beta)-sin\beta \cdot cosa}{sin(a-\beta )+sin\beta \cdot cosa}=\dfrac{sina\cdot cos\beta +cosa\cdot sin\beta -sin\beta \cdot cosa}{sina\cdot cos\beta -cosa\cdot sin\beta +sin\beta \cdot cosa}=\\\\\\=\dfrac{sina\cdot cos\beta }{sina\cdot cos\beta }=1$

$2)\ \ \dfrac{\sqrt2\cdot cosa-2sin(45^\circ -a)}{2sin(60^\circ +a)-\sqrt3\cdot cosa}=\dfrac{\sqrt2\cdot cosa -2(sin45^\circ cosa-cos45^\circ sina)}{2(sin60^\circ cosa+cos60^\circ sina)-\sqrt3\cdot cosa}=\\\\\\=\dfrac{\sqrt2\cdot cosa-\sqrt2cosa+\sqrt2sina}{\sqrt3cosa+sina-\sqrt3\cdot cosa}=\dfrac{\sqrt2\, sina}{sina}=\sqrt2$

$24.8.\ \ \ cosa=-0,6\\\\180^\circ <a<270^\circ \ \ \Rightarrow \ \ \ sina<0\ \ \to \ \ sina=-\sqrt{1-cos^2a}\\\\sina=-\sqrt{1-0,36}=-\sqrt{0,64}=-0,8\\\\cos(60^\circ -a)=cos60^\circ \, cosa+sin60^\circ \, sina=\dfrac{1}{2}\cdot (-0,6)+\dfrac{\sqrt3}{2}\cdot (-0,8)=\\\\=-0,3-\sqrt3\cdot 0,4=-\dfrac{3+4\sqrt3}{10}=-\dfrac{\sqrt3\cdot (\sqrt3+4)}{10}$