1. **State the problem:** Expand $\cos(x-h)$ in ascending powers of $h$ up to the term in $h^6$ using Taylor series, then use this to find $\cos 54.5^\circ$ correct to 1 decimal place. 2. **Recall the Taylor series formula:** For a function $f(x)$ expanded about $x=a$, $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$ 3. **Apply to $\cos(x-h)$ about $h=0$:** Let $f(h) = \cos(x-h)$, then expand about $h=0$: $$f(h) = f(0) + f'(0)h + \frac{f''(0)}{2!}h^2 + \frac{f'''(0)}{3!}h^3 + \frac{f^{(4)}(0)}{4!}h^4 + \frac{f^{(5)}(0)}{5!}h^5 + \frac{f^{(6)}(0)}{6!}h^6$$ 4. **Calculate derivatives:** - $f(h) = \cos(x-h)$ - $f'(h) = \sin(x-h)$ - $f''(h) = \cos(x-h)$ - $f'''(h) = -\sin(x-h)$ - $f^{(4)}(h) = -\cos(x-h)$ - $f^{(5)}(h) = -\sin(x-h)$ - $f^{(6)}(h) = \cos(x-h)$ Evaluate at $h=0$: - $f(0) = \cos x$ - $f'(0) = \sin x$ - $f''(0) = \cos x$ - $f'''(0) = -\sin x$ - $f^{(4)}(0) = -\cos x$ - $f^{(5)}(0) = -\sin x$ - $f^{(6)}(0) = \cos x$ 5. **Write the expansion:** $$\cos(x-h) = \cos x + \sin x \cdot h + \frac{\cos x}{2!} h^2 - \frac{\sin x}{3!} h^3 - \frac{\cos x}{4!} h^4 - \frac{\sin x}{5!} h^5 + \frac{\cos x}{6!} h^6$$ 6. **Substitute values:** We want to find $\cos 54.5^\circ$ using $x=55^\circ$ and $h=0.5^\circ$ (in radians). Convert degrees to radians: $$h = 0.5^\circ = 0.5 \times \frac{\pi}{180} = \frac{\pi}{360} \approx 0.00872665$$ $$x = 55^\circ = 55 \times \frac{\pi}{180} \approx 0.959931$$ Calculate $\cos x$ and $\sin x$: $$\cos 55^\circ \approx 0.573576$$ $$\sin 55^\circ \approx 0.819152$$ 7. **Calculate factorials:** $$2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720$$ 8. **Calculate powers of $h$:** $$h = 0.00872665$$ $$h^2 = 7.6154 \times 10^{-5}$$ $$h^3 = 6.646 \times 10^{-7}$$ $$h^4 = 5.8 \times 10^{-9}$$ $$h^5 = 5.07 \times 10^{-11}$$ $$h^6 = 4.43 \times 10^{-13}$$ 9. **Substitute all into expansion:** $$\cos(54.5^\circ) \approx 0.573576 + 0.819152 \times 0.00872665 + \frac{0.573576}{2} \times 7.6154 \times 10^{-5} - \frac{0.819152}{6} \times 6.646 \times 10^{-7} - \frac{0.573576}{24} \times 5.8 \times 10^{-9} - \frac{0.819152}{120} \times 5.07 \times 10^{-11} + \frac{0.573576}{720} \times 4.43 \times 10^{-13}$$ Calculate each term: - $0.819152 \times 0.00872665 \approx 0.007147$ - $\frac{0.573576}{2} \times 7.6154 \times 10^{-5} = 0.286788 \times 7.6154 \times 10^{-5} \approx 2.184 \times 10^{-5}$ - $\frac{0.819152}{6} \times 6.646 \times 10^{-7} = 0.136525 \times 6.646 \times 10^{-7} \approx 9.07 \times 10^{-8}$ - $\frac{0.573576}{24} \times 5.8 \times 10^{-9} = 0.023899 \times 5.8 \times 10^{-9} \approx 1.386 \times 10^{-10}$ - $\frac{0.819152}{120} \times 5.07 \times 10^{-11} = 0.006826 \times 5.07 \times 10^{-11} \approx 3.46 \times 10^{-13}$ - $\frac{0.573576}{720} \times 4.43 \times 10^{-13} = 0.000797 \times 4.43 \times 10^{-13} \approx 3.53 \times 10^{-16}$ 10. **Sum all terms:** $$0.573576 + 0.007147 + 0.00002184 - 0.0000000907 - 0.0000000001386 - 0.000000000000346 + 0.000000000000000353 \approx 0.580745$$ 11. **Final answer:** $$\cos 54.5^\circ \approx 0.5807$$ Rounded to 1 decimal place: $$0.6$$