SAT1000 - P925

By De Moivre's Theorem and then binomial expansion,

$\cos 5x + i \sin 5x$

$= (\cos x + i \sin x)^{5}$

$= {5 \choose 0} (\cos x)^{5} (i \sin x)^{0} + {5 \choose 1} (\cos x)^{4} (i \sin x)^{1} + {5 \choose 2} (\cos x)^{3} (i \sin x)^{2} + {5 \choose 3} (\cos x)^{2} (i \sin x)^{3} + {5 \choose 4} (\cos x)^{1} (i \sin x)^{4} + {5 \choose 5} (\cos x)^{0} (i \sin x)^{5}$

$= \cos^5 x + 5i \cos^4 x \sin x - 10 \cos^3 x \sin^2 x - 10i \cos^2 x \sin^3 x + 5 \cos x \sin^4 x + i \sin^5 x$

Equating real parts and then using the identity $\sin^2 x = 1 - \cos^2 x$ ,

$\cos 5x$

$= \cos^5 x - 10 \cos^3 x \sin^2 x + 5 \cos x \sin^4 x$

$= \cos^5 x - 10 \cos^3 x (1 - \cos^2 x) + 5 \cos x ((1 - \cos^2 x))^2$

$= 16\cos^5 x - 20 \cos^3 x + 5 \cos x$

Using the double angle formula $\cos 2\alpha = 2\cos^2 \alpha - 1$ ,

$\cos 10x$

$= \cos 2(5x)$

$= 2\cos^2 5x - 1$

$= 2(16\cos^5 x - 20 \cos^3 x + 5 \cos x)^2 - 1$

$= 512 \cos^{10} x - 1280 \cos^{8} x + 1120 \cos^{6} x - 400 \cos^{4} x + 50 \cos^{2} x - 1$

Therefore, $a_1 = 512$ , $a_4 = -400$ , $a_5 = 50$ , and $a_1 - a_4 + a_5 = \boxed{962}$ .

All expressions $\cos{nx}$ are related to $T_{n}(\cos{x}) = \cos{nx}$ , the $n$ th Chebyshev Polynomial of the First Kind. Here is a fabulous Brilliant article about these functions.

Importantly Chebyshev polynomials have the recurrence relation $T_{n+1}(x) = 2x T_{n}(x) - T_{n - 1}(x)$ where $T_{0}(x) = 1$ and $T_{1}(x) = x$ . Therefore, $T_{10}(x) = 512x^{10} - 1280x^8 + 1120x^6 - 400x^4 + 50x^2 - 1$ . Thus, $a_1 = 512, a_4 = -400, a_5 = 50$ and $a_1 - a_4 + a_5 = \boxed{962}$ .

$a_1=512,a_4=-400,a_5=50\implies a_1-a_4+a_5=\boxed {962}$ .