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Towards Maclaurin

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

You are asked in part (5) about generalising this method. What formulae would you expect to derive this way? This method shows a sequence of polynomial approximations to the trig. functions. You are not being asked to give a rigorous proof that the formulae hold in general; that requires a little more work.

Here is another route to the same result. It is not a solution to the problem because the problem specified another method. This should help in seeing the bigger picture more clearly.

We know $\cos x \leq 1$ for all $x$.
The function $f_1(x) = 1 - \cos x$ is positive for all x and so the integral of this function from 0 to $x$ is positive for all $x$. Hence $$\int_0^x f_1 (x) dx =x -\sin x $$ is positive so $\sin x \leq x$.

Integrating again, where $f_2(x) = x - \sin x$ $$\int_0^x f_ 2(x) dx ={x^2\over 2} + \cos x - 1$$ is positive so $\cos x \geq 1 - {x^2\over 2}$.

Integrating again, where $f_3(x)= \cos x - \left(1 - {x^2\over 2}\right)$ $$\int_0^x f_ 3(x) dx =\sin x - \left(x - {x^3\over 3!}\right)$$ is positive so $\sin x \geq x- {x^3\over 3!}$.

Integrating again, where $f_4(x)= \sin x - \left(x - {x^3\over 3!}\right)$ $$\int_0^x f_ 4(x) dx =-\cos x - \left({x^2\over 2!} - {x^4\over 4!}\right)+ 1$$ is positive so $\cos x \leq 1 - {x^2\over 2!} + {x^4\over 4!}$

and so on ...