Trigonometry Problem from SMO, 2008 | Problem No.17

This one is a very simple. We can start from here :

As all are in the function of log with \(\sqrt 2\) as base so we can take it as common such that

\(log_{\sqrt 2}(cos 20 ^\circ . cos 40 ^\circ . cos 80^\circ)\)

Now as you can see we dont know the exact value of \(cos 20^\circ\) or \(cos 40^\circ\) or \(cos 80^\circ\) values.

But theres a formula that we can use which is

cosA.cos B = \(\frac {1}{2} (cos (A+B) + cos (A-B))\)

Now try apply this formula in the above expression and try to solve.........

Now those who did not get the answer yet try this:

If we apply the formula in the expression mentioned in the last hint :

\(log_{\sqrt 2}(cos 20 ^\circ . \frac {1}{2}(cos 120 ^\circ . cos 40^\circ)\)

\(log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{2} cos 40 ^\circ . cos 20^\circ)\)

\(log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{4} (cos 60 ^\circ + cos 20^\circ)\)

We need to do the rest of the calculation.Try to do that .......................

Continue from the last hint:

\(log_{\sqrt 2} \frac {1}{8} = - \frac {log_{2} 8}{log_{2}(2^{\frac {1}{2}})} = -6 \)

So squaring this answer = \((-6)^2 = 36\) ..........................(Answer)