Question

lg(2cos15) / lg(2sin15)

Answer 1

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Answer 2

$\frac{lg(2cos15)}{lg(2sin15)} =\frac{lg(2*\sqrt{\frac{1+cos30}{2}})}{lg(2*\sqrt{\frac{1-cos30}{2}})} =\frac{lg(\sqrt{2+2cos30})}{lg(\sqrt{2-2cos30})} =\frac{lg(\sqrt{2+\sqrt{3}})}{lg(\sqrt{2-\sqrt{3}})} =\frac{lg(2+\sqrt{3})^{0.5}}{lg(2-\sqrt{3})^{0.5}} =\\ \\ = \frac{0.5lg(2+\sqrt{3})}{0.5lg(2-\sqrt{3})} =\frac{lg(2+\sqrt{3})}{lg(2-\sqrt{3})} =\frac{lg\frac{(2+\sqrt{3})(2-\sqrt{3})}{2-\sqrt{3}}}{lg(2-\sqrt{3})}=\frac{lg\frac{4-3}{2-\sqrt{3}}}{lg(2-\sqrt{3})}=$

$=\frac{lg\frac{1}{2-\sqrt{3}}}{lg(2-\sqrt{3})}=\frac{lg(2-\sqrt{3})^{-1}}{lg(2-\sqrt{3})} =\frac{-lg(2-\sqrt{3})}{lg(2-\sqrt{3})} =-1 \\ \\ OTBET: \ -1$