Analyzing the Setup
The expression we are evaluating is 2sin(22π)sin(223π)sin(225π)sin(227π)sin(229π). It is natural to feel a sense of unease when looking at this jumble of trigonometric functions.
However, in the world of JEE Advanced, chaos is often just order in disguise. Let us define P as the product of these five sine terms, such that our target expression is 2P.
The angles involved are 22π,223π,225π,227π,229π. Notice that they are all odd multiples of 22π, which serves as our first clue.
The Bridge to Cosine
The cosine function possesses an elegant identity for products of the form ∏k=1ncos(2n+1kπ). To utilize this, we apply the complementary angle identity: sin(θ)=cos(2π−θ).
We rewrite 2π as 2211π. For any angle 22kπ in our product, the sine term transforms as follows:
sin(22kπ)=cos(2211π−22kπ)=cos(22(11−k)π)
The Transformation
Applying this transformation to each term in our product P:
For k=1, we get cos(2210π)=cos(115π).
For k=3, we get cos(228π)=cos(114π).
For k=5, we get cos(226π)=cos(113π).
For k=7, we get cos(224π)=cos(112π).
For k=9, we get cos(222π)=cos(11π).
Rearranging these in increasing order, our product P becomes:
The Magical Identity
We utilize the standard trigonometric identity:
Setting 2n+1=11, we find 2n=10, which implies n=5. Since our product matches this form perfectly for n=5, we have:
The Final Calculation
The most common mistake in JEE is to stop at the value of P. The original question asks for the value of 2P.
We must multiply our result by 2:
The final value of the expression is 161.