Question

Lg(2cos15) / lg(2sin15)

Answer 1

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Answer 2

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

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

Answer 3

To simplify the expression $\frac{\log(2\cos(15^\circ))}{\log(2\sin(15^\circ))}$, you can use logarithm properties. One property of logarithms states that $\log(a) - \log(b) = \log\left(\frac{a}{b}\right)$. Using this property, you can rewrite the expression as follows:

$\frac{\log(2\cos(15^\circ))}{\log(2\sin(15^\circ))} = \log\left(\frac{2\cos(15^\circ)}{2\sin(15^\circ)}\right)$

Now, simplify the fraction inside the logarithm:

$\frac{2\cos(15^\circ)}{2\sin(15^\circ)} = \frac{2}{2} \cdot \frac{\cos(15^\circ)}{\sin(15^\circ)} = \frac{\cos(15^\circ)}{\sin(15^\circ)}$

Now, the expression becomes:

$\log\left(\frac{\cos(15^\circ)}{\sin(15^\circ)}\right)$

However, it's important to note that $\frac{\cos(15^\circ)}{\sin(15^\circ)}$ is equal to $\cot(15^\circ)$, which is the cotangent of $15^\circ$.

So, the simplified expression is:

$\log(\cot(15^\circ))$

You can leave it in this form, or if you want to calculate an approximate numerical value, you can evaluate $\cot(15^\circ)$ and then take the logarithm.