1. **Problem Statement:** Construct the Fourier expansion for $y$ in terms of $x$ up to the first harmonic for the given data: $$\begin{array}{c|ccccccc} x^\circ & 0 & 60 & 120 & 180 & 240 & 300 & 360 \\ y & 10.5 & 26.4 & 27 & 12.5 & -19.2 & -15.8 & 10.5 \\\end{array}$$ 2. **Fourier Series Formula:** For a function with period $360^\circ$, the Fourier expansion up to the first harmonic is: $$y = a_0 + a_1 \cos\left(\frac{2\pi x}{360}\right) + b_1 \sin\left(\frac{2\pi x}{360}\right)$$ where $$a_0 = \frac{1}{N} \sum_{k=0}^{N-1} y_k$$ $$a_1 = \frac{2}{N} \sum_{k=0}^{N-1} y_k \cos\left(\frac{2\pi x_k}{360}\right)$$ $$b_1 = \frac{2}{N} \sum_{k=0}^{N-1} y_k \sin\left(\frac{2\pi x_k}{360}\right)$$ and $N=7$ data points. 3. **Calculate $a_0$:** $$a_0 = \frac{1}{7}(10.5 + 26.4 + 27 + 12.5 - 19.2 - 15.8 + 10.5) = \frac{1}{7}(51.9) = 7.4143$$ 4. **Calculate $a_1$:** Compute each term: - $\cos\left(\frac{2\pi \times 0}{360}\right) = \cos(0) = 1$ - $\cos\left(\frac{2\pi \times 60}{360}\right) = \cos\left(\frac{\pi}{3}\right) = 0.5$ - $\cos\left(\frac{2\pi \times 120}{360}\right) = \cos\left(\frac{2\pi}{3}\right) = -0.5$ - $\cos\left(\frac{2\pi \times 180}{360}\right) = \cos(\pi) = -1$ - $\cos\left(\frac{2\pi \times 240}{360}\right) = \cos\left(\frac{4\pi}{3}\right) = -0.5$ - $\cos\left(\frac{2\pi \times 300}{360}\right) = \cos\left(\frac{5\pi}{3}\right) = 0.5$ - $\cos\left(\frac{2\pi \times 360}{360}\right) = \cos(2\pi) = 1$ Calculate sum: $$\sum y_k \cos\left(\frac{2\pi x_k}{360}\right) = 10.5(1) + 26.4(0.5) + 27(-0.5) + 12.5(-1) + (-19.2)(-0.5) + (-15.8)(0.5) + 10.5(1)$$ $$= 10.5 + 13.2 - 13.5 - 12.5 + 9.6 - 7.9 + 10.5 = 9.9$$ Then $$a_1 = \frac{2}{7} \times 9.9 = 2.8286$$ 5. **Calculate $b_1$:** Compute each sine term: - $\sin\left(\frac{2\pi \times 0}{360}\right) = 0$ - $\sin\left(\frac{2\pi \times 60}{360}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.8660$ - $\sin\left(\frac{2\pi \times 120}{360}\right) = \sin\left(\frac{2\pi}{3}\right) = 0.8660$ - $\sin\left(\frac{2\pi \times 180}{360}\right) = \sin(\pi) = 0$ - $\sin\left(\frac{2\pi \times 240}{360}\right) = \sin\left(\frac{4\pi}{3}\right) = -0.8660$ - $\sin\left(\frac{2\pi \times 300}{360}\right) = \sin\left(\frac{5\pi}{3}\right) = -0.8660$ - $\sin\left(\frac{2\pi \times 360}{360}\right) = \sin(2\pi) = 0$ Calculate sum: $$\sum y_k \sin\left(\frac{2\pi x_k}{360}\right) = 10.5(0) + 26.4(0.8660) + 27(0.8660) + 12.5(0) + (-19.2)(-0.8660) + (-15.8)(-0.8660) + 10.5(0)$$ $$= 0 + 22.85 + 23.38 + 0 + 16.62 + 13.68 + 0 = 76.53$$ Then $$b_1 = \frac{2}{7} \times 76.53 = 21.866$$ 6. **Final Fourier expansion up to first harmonic:** $$\boxed{y = 7.4143 + 2.8286 \cos\left(\frac{\pi x}{180}\right) + 21.866 \sin\left(\frac{\pi x}{180}\right)}$$ This formula approximates the given data using the first harmonic of the Fourier series.