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
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)