 QUESTION

# To find the value of ${{\log }_{{{t}_{1}}}}(4\sin 9\cos 9)$, where ${{t}_{1}}$ = 4(sin 63)(cos 63).(a) $\dfrac{\sqrt{5}+1}{4}$(b) $\dfrac{\sqrt{5}-1}{4}$(c) 1(d) none of these

Hint: To solve this problem, we will first use the property that ${{\log }_{x}}y=\dfrac{{{\log }_{a}}y}{{{\log }_{a}}x}$. Here, a is the base of the logarithm (a is any positive number exclusive of 1) and x and y are positive numbers. In addition, we will also use the property ${{\log }_{a}}\left( \dfrac{x}{y} \right)={{\log }_{a}}x-{{\log }_{a}}y$. Further, we will use the trigonometric property that 2(sinx)(cosx) = sin 2x. After that, we will use the value of sin 18 = $\dfrac{\sqrt{5}-1}{4}$ and cos 36 = $\dfrac{\sqrt{5}+1}{4}$ in the obtained expressions to get the correct answer.

We first start by evaluating ${{\log }_{{{t}_{1}}}}(4\sin 9\cos 9)$ with ${{t}_{1}}$ = 4(sin 63)(cos 63). Now, using the property that ${{\log }_{x}}y=\dfrac{{{\log }_{a}}y}{{{\log }_{a}}x}$ (where, a, x and y are positive numbers and in addition a cannot be 1). Thus, we get,
= ${{\log }_{4\left( sin63 \right)\left( cos63 \right)}}(4\sin 9\cos 9)=\dfrac{{{\log }_{a}}(4\sin 9\cos 9)}{{{\log }_{a}}(4\sin 63\cos 63)}$
Now, we use the property that 2(sinx)(cosx) = sin 2x, thus, we get,
= $\dfrac{{{\log }_{a}}(2\sin 18)}{{{\log }_{a}}(2\sin 126)}$
Now, we already know the value of sin 18, but to find sin 126, we use the property, sin(90+x) = cosx. In this case, x = 36. Thus, we get,
= $\dfrac{{{\log }_{a}}(2\sin 18)}{{{\log }_{a}}(2\sin (90+36))}$
= $\dfrac{{{\log }_{a}}(2\sin 18)}{{{\log }_{a}}(2\cos 36)}$
Now, putting the values of sin 18 = $\dfrac{\sqrt{5}-1}{4}$ and cos 36 = $\dfrac{\sqrt{5}+1}{4}$ in this expression, we get,
= $\dfrac{{{\log }_{a}}\left( 2\times \left( \dfrac{\sqrt{5}-1}{4} \right) \right)}{{{\log }_{a}}\left( 2\times \left( \dfrac{\sqrt{5}+1}{4} \right) \right)}$
= $\dfrac{{{\log }_{a}}\left( \dfrac{\sqrt{5}-1}{2} \right)}{{{\log }_{a}}\left( \dfrac{\sqrt{5}+1}{2} \right)}$ -- (A)
Now, rationalizing the term $\left( \dfrac{\sqrt{5}-1}{2} \right)$, we get,
= $\left( \dfrac{\sqrt{5}-1}{2} \right)$ = $\left( \dfrac{\sqrt{5}-1}{2} \right)\left( \dfrac{\sqrt{5}+1}{\sqrt{5}+1} \right)$ = $\left( \dfrac{4}{2\left( \sqrt{5}+1 \right)} \right)=\left( \dfrac{2}{\left( \sqrt{5}+1 \right)} \right)$
Putting this in (A), we get,
= $\dfrac{{{\log }_{a}}\left( \dfrac{2}{\left( \sqrt{5}+1 \right)} \right)}{{{\log }_{a}}\left( \dfrac{\sqrt{5}+1}{2} \right)}$
Now, we can use the property that ${{\log }_{a}}\left( \dfrac{x}{y} \right)={{\log }_{a}}x-{{\log }_{a}}y$. Thus, we have,
= $\dfrac{{{\log }_{a}}2-{{\log }_{a}}\left( \sqrt{5}+1 \right)}{{{\log }_{a}}\left( \sqrt{5}+1 \right)-{{\log }_{a}}2}$
= -1
Hence, the correct answer is (d) none of these.

Note: In problems involving logarithms and trigonometry, one should try to bring the value inside the logarithm in terms of familiar terms. For example, in this problem, we brought the terms inside the logarithm in terms of sin 18 and cos 36, since, the value of these trigonometric angles are known and the values can be substituted. Once that can be done, it is much easier to figure out the rest of the steps as can be seen in the problem.