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Singapore Math Olympiad

Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry. You may use sequential hints to solve the problem.

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

Check the Answer


But try the problem first…

Answer: 36

Source
Suggested Reading

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

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………

Second Hint

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 …………………..

Final Step

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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