Question

Find the Maclaurian series for sin^2x. (Hint: Use the identity sin^2x=1/2(1-cos2x)

Answer 1

The Maclaurin series is just the special case for the Taylor series centered around $a=0$ .

So, $f(x)= f(0) + \frac{f'(0)}{1} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + ....$

Given $g(x) = sin^2 x = \frac{1}{2} (1-cos(2x))$.

Assume $f(x)=cos(2x) \implies f(0)=cos(0)= 1$

$f'(x) = -2sin(2x) \implies f'(0) = 0$

$f''(x)=-4cos(2x) \implies f''(0) = -4$

$f'''(x)=8sin(2x) \implies f'''(0) = 0$

$f''''(x)=16cos(2x) \implies f''''(0) = 16$ and so on.

So, $cos(2x)= 1 + \frac{-4}{2!} x^2 + \frac{16}{4!} x^4 + .... = 1 - 2 x^2 + \frac{2}{3} x^4 + ....$

So, $sin^2 x = \frac{1}{2} [1 - (1-2x^2 + \frac{2}{3} x^4 + ... ) ] = x^2 - \frac{1}{3} x^4 + ....$