# Sine squared when obtaining a Maclaurin and taking it's limit as $x\to 0$

When determining the Maclaurin series for $\sin^2(x)$ we use the trigonometric identity $\frac{1-\cos(2x)}{2}$. But when taking the limit of e.g. the Maclaurin series for $\cos(\sin(x))$ as $x$ approaches zero, the Maclaurin series for e.g. the second term which is also $\sin^2(x)$ is simply the squared Maclaurin series of $\sin(x)$. How is this possible?

Edit.1: an example image

Edit.2: I think that when you substitute $sin(x)$ for $x$ in the Maclaurin series for $cos(x)$, you should insert the Maclaurin series for $sin(x)$ instead of just $sin(x)$ as I did in the third row in the image below. I think this is how they got the squared Maclaurin series of $sin(x)$ when $sin^2(x)$ instead of using the usual trigonometric identity. Please, if you think this conclusion is correct let me know. Or even if it is wrong!

• Can you clarify using some examples!? At this point I may just say that you are using different approximations in each case. It means using different $n$ for the remainder $O(x^n)$. – H. R. Jan 10 '16 at 8:30
• It bothers me because Maclaurin series obtained by squaring the Maclaurin series of $sin(x)$ and the Maclaurin series of $sin^2(x)$ obtained by using the trigonometric identity produce two different series. – Dick Armstrong Jan 10 '16 at 8:41
• @DickArmstrong. No ! If it does for you, it probably means that you did not take enough terms. – Claude Leibovici Jan 10 '16 at 8:42

After solving a few similar problems, I came to the following conclusion: Yes, when you have $f(x)=sin^2(x)$ you solve it by using the following trigonometric identity:
you can see e.g. the second term that you get after expanding the series, is $sin^2(x)$ but here you don't solve it using the trigonometric identity above, but instead you find the Maclaurin series for the $sin(x)$ then use it as a variable, like $(sin(x))^n$, where in this case $n=0,2,4,6, ...$