How would I go about showing $\cos(x+\frac{\pi}{2})=-\sin(x)$?

Question

I want to look at the summation for both cos,sin..

$cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...= \sum\nolimits_{i=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$ so, $cos(x+\frac{\pi}{2})=\sum\nolimits_{i=0}^\infty \frac{(-1)^n(x+\frac{\pi}{2})^{2n}}{(2n)!}$. So how I need to expand $(x+\frac{\pi}{2})^{2n}$ using the binomial formula maybe?

I want to look at manipulating the POWER series specifically for proving this identity.

Answer 1

Start from $$\cos(x+\pi/2) = \sum_{n\ge 0} (-1)^n \frac{(x+\pi/2)^{2n}}{(2n)!}.$$ Now ask yourself about the coefficient of $x^k$ in this series, it is given by $$\sum_{2n\ge k} \frac{(-1)^n}{(2n)!} \binom{2n}{k} (\pi/2)^{2n-k} = \frac{1}{k!} \sum_{2n\ge k} \frac{(-1)^n}{(2n-k)!} (\pi/2)^{2n-k}.$$ In this last series the $2n-k$ term starts at zero when $k$ is even and is given by $$\frac{1}{k!} \sum_{q\ge 0} \frac{(-1)^{q+k/2}}{(2q)!} (\pi/2)^{2q} = \frac{(-1)^{k/2}}{k!} \sum_{q\ge 0} \frac{(-1)^q}{(2q)!} (\pi/2)^{2q} = \frac{(-1)^{k/2}}{k!} \cos(\pi/2) = 0.$$ When $k$ is odd the $2n-k$ term starts at one and is given by $$\frac{1}{k!} \sum_{q\ge 0} \frac{(-1)^{q+(k+1)/2}}{(2q+1)!} (\pi/2)^{2q+1} \\= \frac{(-1)^{(k+1)/2}}{k!} \sum_{q\ge 0} \frac{(-1)^q}{(2q+1)!} (\pi/2)^{2q+1} = \frac{(-1)^{(k+1)/2}}{k!} \sin(\pi/2) = \frac{(-1)^{(k+1)/2}}{k!}.$$ It follows that $$\cos(x+\pi/2) = \sum_{m\ge 0} \frac{(-1)^{m+1}}{(2m+1)!} x^{2m+1} = -\sin(x).$$

Answer 2

What you sketch there is definitely the hard way to do this. When you expand the binomials, each of them will contribute terms to every term of lower degree in the resulting series, and then you have to sum all of these to find the coefficients.

It's much easier (not to mention more instructive) to follow the programs outlined in answers to How to prove periodicity of sin(x) or cos(x) starting from the Taylor series expansion?.

Answer 3

As an example, one can use the statement that

PROP If $f(0)=f'(0)=0$ and $f''+f=0$, then $f=0$.

P The ODE gives $f'f''+ff'=0$, so that $(f')^2+f^2=K$. But then $K=0$ from the initial conditions. This gives that $f=0$.

COR If $f(0)=a,f'(0)=b$ and $f''+f=0$ then $f=a\cos+b\sin$.

P Set $g=a\cos +b\sin $. Then $h=f-g$. It follows $h''+h=0$ and $h(0)=0,h'(0)=0$ so $h\equiv 0$ by the lemma.

(Giveaway: let $f(x)=\cos (x+y)-\cos x\cos y+\sin x\sin y$ in the first result, or $f(x)=\cos (x+y)$ in the corrollary)

Then you can prove that $\cos(x+y)=\cos x\cos y-\sin x\sin y$ using the above, letting $y$ be a constant.

Alternatively, let $f=\sin x+\cos \left(x+\dfrac \pi 2\right)$ above. You will need to use the special values $\sin \dfrac \pi 2=1$ and $\cos \dfrac\pi 2=0$, though. One can in fact define $\dfrac \pi 2$ to be the unique number determined by $$\inf\{x>0:\cos x=0\}$$

Answer 4

$$ \begin{align} \cos\left(x+\frac\pi2\right) &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(x+\frac\pi2\right)^{2k}\\ &=\sum_{k=0}^\infty\sum_{j=0}^{2k}\frac{(-1)^k}{(2k)!}\binom{2k}{j}x^j(\pi/2)^{2k-j}\\ &=\sum_{j=0}^\infty\sum_{k=\lceil j/2\rceil}^\infty\frac{(-1)^k}{(2k)!}\binom{2k}{j}x^j(\pi/2)^{2k-j}\\ &=\sum_{j=0}^\infty\sum_{k=\lceil j/2\rceil}^\infty\frac{(-1)^k}{(2k-j)!j!}x^j(\pi/2)^{2k-j}\\ &=\sum_{j=0}^\infty\frac{x^j}{j!}\sum_{k=\lceil j/2\rceil}^\infty\frac{(-1)^k}{(2k-j)!}(\pi/2)^{2k-j}\tag{1} \end{align} $$ Notice that when $j$ is even, $$ \begin{align} \sum_{k=\lceil j/2\rceil}^\infty\frac{(-1)^k}{(2k-j)!}(\pi/2)^{2k-j} &=\sum_{k=0}^\infty\frac{(-1)^{k+j/2}}{(2k)!}(\pi/2)^{2k}\\ &=(-1)^{j/2}\cos(\pi/2)\\[12pt] &=0\tag{2} \end{align} $$ Notice that when $j$ is odd, $$ \begin{align} \sum_{k=\lceil j/2\rceil}^\infty\frac{(-1)^k}{(2k-j)!}(\pi/2)^{2k-j} &=\sum_{k=0}^\infty\frac{(-1)^{k+(j+1)/2}}{(2k+1)!}(\pi/2)^{2k+1}\\ &=(-1)^{(j+1)/2}\sin(\pi/2)\\[12pt] &=(-1)^{(j+1)/2}\tag{3} \end{align} $$ Plugging $(2)$ and $(3)$ back into $(1)$, removing the even terms and reindexing the odd terms, yields $$ \begin{align} \cos\left(x+\frac\pi2\right) &=-\sum_{j=0}^\infty\frac{(-1)^j}{(2j+1)!}x^{2j+1}\\ &=-\sin(x)\tag{4} \end{align} $$