1. **State the problem:** We want to find an approximate root of the equation $$\sin(x) = x^2 - 1$$ using Newton's method. 2. **Rewrite the equation:** Define a function $$f(x) = \sin(x) - x^2 + 1$$. We want to find $$x$$ such that $$f(x) = 0$$. 3. **Newton's method formula:** The iterative formula is $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ where $$f'(x)$$ is the derivative of $$f(x)$$. 4. **Find the derivative:** $$ f'(x) = \cos(x) - 2x $$ 5. **Choose an initial guess:** Let's start with $$x_0 = 1$$ (a reasonable guess near where $$\sin(x)$$ and $$x^2 - 1$$ might intersect). 6. **Calculate iterations:** - Compute $$f(1) = \sin(1) - 1^2 + 1 = \sin(1) = 0.84147$$ - Compute $$f'(1) = \cos(1) - 2(1) = 0.54030 - 2 = -1.4597$$ Apply Newton's formula: $$ x_1 = 1 - \frac{0.84147}{-1.4597} = 1 + 0.5765 = 1.5765 $$ 7. **Second iteration:** - Compute $$f(1.5765) = \sin(1.5765) - (1.5765)^2 + 1 \approx 0.99999 - 2.4855 + 1 = -0.4855$$ - Compute $$f'(1.5765) = \cos(1.5765) - 2(1.5765) \approx -0.0063 - 3.153 = -3.1593$$ Apply Newton's formula: $$ x_2 = 1.5765 - \frac{-0.4855}{-3.1593} = 1.5765 - 0.1536 = 1.4229 $$ 8. **Third iteration:** - Compute $$f(1.4229) = \sin(1.4229) - (1.4229)^2 + 1 \approx 0.9899 - 2.0246 + 1 = -0.0347$$ - Compute $$f'(1.4229) = \cos(1.4229) - 2(1.4229) \approx 0.1467 - 2.8458 = -2.6991$$ Apply Newton's formula: $$ x_3 = 1.4229 - \frac{-0.0347}{-2.6991} = 1.4229 - 0.0129 = 1.4100 $$ 9. **Fourth iteration:** - Compute $$f(1.4100) = \sin(1.4100) - (1.4100)^2 + 1 \approx 0.9870 - 1.9881 + 1 = -0.0011$$ - Compute $$f'(1.4100) = \cos(1.4100) - 2(1.4100) \approx 0.1603 - 2.8200 = -2.6597$$ Apply Newton's formula: $$ x_4 = 1.4100 - \frac{-0.0011}{-2.6597} = 1.4100 - 0.0004 = 1.4096 $$ 10. **Conclusion:** The approximate root is $$x \approx 1.4096$$ after 4 iterations. This means $$\sin(x)$$ and $$x^2 - 1$$ are approximately equal at $$x = 1.4096$$.