# Good Question 16

I had an email last week from a teacher asking, how come I can use a substitution to find a power series for  $\cos \left( {2x} \right)$, and for  ${{e}^{{\left( {x-1} \right)}}}$, but not for  $\cos \left( {3x+\frac{\pi }{6}} \right)$?

The answer is that you can. Substituting (2x) in to the cosine’s series give you a Taylor series centered at x = 0, a Maclaurin Series. Substituting (x – 1) into the series for ex gives you a Taylor series centered at x = 1. And substituting $\left( {3x+\frac{\pi }{6}} \right)$ into the cosine series gives you a Taylor series centered at  $x=-\frac{\pi }{{18}}$. I suspect that she was hoping for or was asked to find a Maclaurin series, not one with such a strange center.

The center of a Taylor series is the value of x that makes its argument zero.

AP Exam Question 2004 BC 6(a)

This brought to mind the AP Exam question 2004 BC 6(a) where students were asked to write the third-degree Taylor polynomial about x = 0 for the function $f\left( x \right)=\sin \left( {5x+\frac{\pi }{4}} \right)$. The intended method was for students to find the first three derivative and substitute them into the general form for a Taylor series. That’s what students who got this correct did. This is the only time I can remember when students were expected to do that; usually they manipulate a given series or substitute into a known series.

A number of students tried to substitute $\left( {5x+\frac{\pi }{4}} \right)$ into the series for the sine. This gets a very nice Taylor series centered at  $x=-\frac{\pi }{{20}}$. This earned no credit since a Maclaurin series was required.

But there is an another way! (I originally wrote, “But there is an easier way!” but it’s only easier if you see how to do it.)

Trigonometry to the Rescue!

$\sin \left( {5x+\frac{\pi }{4}} \right)=\sin (5x)\cos \left( {\frac{\pi }{4}} \right)+\cos \left( {5x} \right)\sin \left( {\frac{\pi }{4}} \right)=\frac{{\sqrt{2}}}{2}\left( {\sin \left( {5x} \right)+\cos \left( {5x} \right)} \right)$

Then using the first two terms each from the series for sine and cosine you get the correct answer:

$\displaystyle \frac{{\sqrt{2}}}{2}\left( {\left( {5x-{{{\frac{{\left( {5x} \right)}}{{3!}}}}^{3}}} \right)+\left( {1-\frac{{{{{\left( {5x} \right)}}^{2}}}}{{2!}}} \right)} \right)=\frac{{\sqrt{2}}}{2}+\frac{{5\sqrt{2}}}{2}x-\frac{{25\sqrt{2}}}{{2\left( {2!} \right)}}{{x}^{2}}-\frac{{125\sqrt{2}}}{{2\left( {3!} \right)}}{{x}^{3}}$

This brings us to $\cos \left( {3x+\frac{\pi }{6}} \right)$, which can be approached the same way. Here is the entire Maclaurin series.

$\cos \left( {3x+\frac{\pi }{6}} \right)=\cos \left( {3x} \right)\cos \left( {\frac{\pi }{6}} \right)-\sin \left( {3x} \right)\sin \left( {\frac{\pi }{6}} \right)$

$\displaystyle =\frac{{\sqrt{3}}}{2}\cos \left( {3x} \right)-\frac{1}{2}\sin \left( {3x} \right)$

$\displaystyle =\frac{{\sqrt{3}}}{2}\sum\limits_{{n=0}}^{\infty }{{\frac{{{{{\left( {3x} \right)}}^{{2n}}}}}{{\left( {2n} \right)!}}}}-\frac{1}{2}\sum\limits_{{n=0}}^{\infty }{{\frac{{{{{\left( {3x} \right)}}^{{2n+1}}}}}{{\left( {2n+1} \right)!}}}}$

$\displaystyle =\sum\limits_{{n=0}}^{\infty }{{\left( {\frac{{\sqrt{3}\left( {{{3}^{{2n}}}} \right)}}{{2\left( {2n} \right)!}}{{x}^{{2n}}}-\frac{{1\left( {{{3}^{{2n+1}}}} \right)}}{{2\left( {2n+1} \right)!}}{{x}^{{2n+1}}}} \right)}}$

Moral: Trig can be very useful.

Here is a previous post, Geometric Series – Far Out , that shows a “mistake” you may find interesting.