Analyzing the Setup
Welcome, fellow traveler of the JEE path. Today, we are not just solving an equation; we are uncovering a hidden symmetry in the world of trigonometry.
The problem before us is:
cos2θcos2θ=cos3θcos29θ
At first glance, it looks like a chaotic mess of angles. But remember, in the realm of JEE Advanced, chaos is often just order in disguise.
The Transformation
We start with the equation cos2θcos2θ=cos3θcos29θ. When you see products of cosines, your mind should immediately race to the product-to-sum identities.
To use them, we need a factor of
2. We multiply both sides by
2:
2cos2θcos2θ=2cos3θcos29θ
Now, we apply the identity 2cosAcosB=cos(A+B)+cos(A−B).
On the left-hand side, let
A=2θ and
B=2θ. Adding them gives
25θ, and subtracting them gives
23θ. Thus, the left side becomes:
cos25θ+cos23θ
On the right-hand side, let
A=29θ and
B=3θ. Adding them gives
215θ, and subtracting them gives
23θ. Thus, the right side becomes:
cos215θ+cos23θ
The Elegant Cancellation
Look at what we have now:
cos25θ+cos23θ=cos215θ+cos23θ
The term
cos23θ is waiting on both sides like a silent observer. We subtract it away, and suddenly, the equation collapses into something beautiful:
cos215θ=cos25θ
This is the moment where the complexity vanishes, leaving us with a clean, solvable structure.
The General Solution
We know that the general solution for
cosX=cosα is
X=2kπ±α, where
k∈Z. Applying this to our equation, we get:
215θ=2kπ±25θ
This splits into two cases.
Case 1 (the positive sign):
215θ=2kπ+25θ⇒210θ=2kπ⇒5θ=2kπ⇒θ=52kπ
Case 2 (the negative sign):
215θ=2kπ−25θ⇒220θ=2kπ⇒10θ=2kπ⇒θ=5kπ
Since
52kπ is just a subset of
5kπ, our general solution is simply:
θ=5kπ
Final Calculation
We are restricted to the interval [−π,π]. Setting −π≤5kπ≤π, we find that k must be an integer in the range [−5,5].
The possible values for k are {−5,−4,−3,−2,−1,0,1,2,3,4,5}.
The positive values of θ occur for k∈{1,2,3,4,5}, so m=5. The negative values of θ occur for k∈{−1,−2,−3,−4,−5}, so n=5.
The product
mn is:
5×5=25
We have navigated the storm and arrived at the shore. The beauty of this problem lies not in the final number, but in the elegance of the path we took to find it.