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# Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

## Problem - Trigonometry (SMO Test)

Find the value of $(log_{\sqrt 2}(cos 20^\circ) + log_{\sqrt 2} (cos 40^\circ) + log_{\sqrt 2}(cos 80^\circ))^2$

• 32
• 15
• 36
• 20

### Key Concepts

Trigonometry

Log Function

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

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)

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