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Trigonometry Problem | AIME I, 2015 | Question 13

Try this beautiful problem number 13 from the American Invitational Mathematics Examination, AIME, 2015 based on Trigonometry.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Trigonometry.

Trigonometry Problem – AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n), where (m) and (n) are integers greater than 1. Find (m+n).

• is 107
• is 91
• is 840
• cannot be determined from the given information

Key Concepts

Trigonometry

Sequence

Algebra

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x – \frac{1}{x}}{2i} = \frac{x^2 – 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 – 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ – z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 – 1| |x^6 – 1| |x^{10} – 1| \dots |x^{354} – 1| |x^{358} – 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z – x^2)(z – x^6)\dots (z – x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \dots |1 – x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.