Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Trigonometry Simplification.
If \(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ} = tan x^\circ \)
Find \( x^\circ \) ?
Trigonometry
Geometry
Answer: 95
Singapore Mathematical Olympiad
Challenges and Thrill - Pre College Mathematics
If you really got stuck into this sum we can start from here
\(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ}\)
= \(\frac {cos 100^\circ}{1-2sin 50^\circ cos 50^\circ}\)
Now let's check with some basic values in trigonometry
\( Cos 2 A = cos^2 A - sin^2 A \) and
\(2 sin A cos A = sin 2 A\)
Now try the rest of the sum by using these two above mentioned values..................
Let's continue from the last hint :
\( cos 100^\circ = cos^2 50^\circ - sin^2 50^\circ \)
\( 2 sin 25^\circ cos 25^\circ = sin 50^\circ\)
\(\frac {cos^2 50^\circ - sin^2 50^\circ}{2sin 50^\circ cos 50^\circ}\)
\(\frac {cos^2 50^\circ - sin^2 50^\circ }{(cos 50^\circ - sin 50^\circ)^2}\)
Using \(a^2 - b^2 = (a+b) (a-b)\) formula
\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ - sin 50^\circ}\)
Do the rest of the steps .................
Starting from right after the last hint:
\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ - sin 50^\circ}\)
= \(\frac {1+ tan 50^\circ}{1-tan 50^\circ}\)
= \(\frac {tan 45^\circ + tan 50^\circ}{1-tan 45^\circ tan 50 ^\circ}\)
= \( tan 95^\circ\) - Answer
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