Find the solution of the equation f(x) = sin(x) − cos(1 − x^2) −
1 using Newton-Raphson

method with an initial guess of xi = 1.2. Perform two iterations
and calculate percentage

relative error after the second iteration.

Angles should be in radians

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f(x) = sin(x) - cos(1-x2) - 1 Find solution using Newton - Raphson Method . Newton - Raphson Method is x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} f^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x} f(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left(\sin (x)-\cos \left(1-x^{2}\right)-1\right) f^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x}(\sin (x))-\frac{\mathrm{d}}{\mathrm{d} x}\left(\cos \left(1-x^{2}\right)\right)-\frac{\mathrm{d}}{\mathrm{d} x}(1) f^{\prime}(x)=\cos (x)+\sin \left(1-x^{2}\right) \frac{\mathrm{d}}{\mathrm{d} x}\left(1-x^{2}\right)-0        f^{\prime}(x)=\cos (x)+\sin \left(1-x^{2}\right)(-2 x) f^{\prime}(x)=\cos (x)-2 x \sin \left(1-x^{2}\right) Initial guess x0 = 1.2 [Given] Ist Iteration : Put n = 0 in above formula-  x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} f(1.2)=\sin (1.2)-\cos \left[1-(1.2)^{2}\right]-1 f(1.2)=\sin (1.2)-\cos [1-1.44]-1 f(1.2)=\sin (1.2)-\cos (-0.44)-1 f(1.2)=0.932039086-0.9047516632-1 f(1.2) = -0.9727 f^{\prime}(1.2)=\cos (1.2)-2 \times 1.2 \times \sin \left(1-(1.2)^{2}\right) f^{\prime}(1.2)=\cos (1.2)-2.4 \times \sin (1-1.44) f^{\prime}(1.2)=\cos (1.2)-2.4 \times \sin (-0.44) f^{\prime}(1.2)=0.3624-2.4 \times(-0.4259) f^{\prime}(1.2)=0.3624+1.0223 f^{\prime}(1.2)=1.3847 Put in formula x_{1}=1.2-\frac{-0.9727}{1.3847} x_{1}=1.2+0.7025 x_{1}=1.9025 2nd Iteration : x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)} f(1.9025)=\sin (1.9025)-\cos \left[1-(1.9025)^{2}\right]-1 f(1.9025)=\sin (1.9025)-\cos [1-3.6195]-1 f(1.9025)=\sin (1.9025)-\cos (-2.6195)-1 f(1.9025)=0.9455+0.8668-1 f(1.9025) = 0.8123 f^{\prime}(1.9025)=\cos (1.9025)-2 \times 1.9025 \times \sin \left(1-(1.9025)^{2}\right) f^{\prime}(1.9025)=\cos (1.9025)-3.805 \times \sin (1-3.6195) f^{\prime}(1.9025)=\cos (1.9025)-3.805 \times \sin (-2.6195) f^{\prime}(1.9025)=-0.3257+1.8975 f^{\prime}(1.9025)=1.5718 Put in formula x_{2}=1.9025-\frac{0.8123}{1.5718} x_{2}=1.9025-0.5168 x_{2}=1.3857 Percentage relative error is -         =\left|\frac{x_{2}-x_{1}}{x_{2}}\right| \times 100         =\left|\frac{1.3857-1.9025}{1.3857}\right| \times 100         =\left|\frac{-0.5168}{1.3857}\right| \times 100         =0.3729 \times 100          = 37.29% THANKS. ...