Answer:
sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...
Step-by-step explanation:
given that f(x) = sin x
we have to find the Taylor series for that
[tex]f(x) = sin x : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0[/tex]
and so on.
i.e. 2nd, 4th, 6th terms would be 0
and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative
Using the above we can write Taylor series as
[tex]f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...[/tex]
[tex]sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...[/tex]
This is valid for all real values of x.
x ∈[tex](-\infty, infty)[/tex]
