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How would I prove the following result?

$$\cos A+\cos(A+B)+\ldots +\cos(A+ (n-1)B) = \frac{\sin\left(\frac{nB}{2}\right) \cos\left[A+\frac{(n-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$

Caleb Stanford
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user93343
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3 Answers3

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Hint: Use $\sum_{k = 1}^{n} \text{cos}(A+kB) = \text{Re}(\sum_{k=1}^{n}e^{i(A+kB)})$ and proceed with the sum of a geometric series.

dayar
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  • Yep. You have to be careful about the case where $b=2\pi\ell$ for some integer $\ell$ when summing the power series. – Thomas Andrews Sep 05 '13 at 23:21
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    $k$ should start at $0$ and end at $n-1$ (the OP started with $\cos(A)$, which is $k = 0$, and then said there were $n$ terms total). Otherwise, this is a good answer. – Caleb Stanford Sep 05 '13 at 23:28
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Let $$ C=Cos(A)+Cos(A+B)+\cdots+Cos(A+(n-1)B) $$ and $$ S=Sin(A)+Sin(A+B)+\cdots+Sin(A+(n-1)B) $$ if $$i=\sqrt{-1}$$ Consider

$$C+iS=e^{iA}+e^{i(A+B)}+\cdots+e^{i(A+(n-1)B)}$$ $\implies$

$$C+iS=e^{iA}\left(1+e^{iB}+e^{i2B}+\cdots+e^{i(n-1)B}\right)$$ $\implies$ $$C+iS=e^{iA}\left(\frac{e^{inB}-1}{e^{iB}-1}\right)$$ $\implies$ $$C+iS=e^{i\left(A+(n-1)\frac{B}{2}\right)} \frac{Sin(\frac{nB}{2})}{Sin(\frac{B}{2})}$$ From this we get Real and Imaginary parts as the Required Results.

Ekaveera Gouribhatla
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Induction on $n$.

Checking for $n=1$:

$$\cos A =\frac{\sin(\frac{B}{2})}{\sin(\frac{B}{2})}\cos A =\cos A $$

Induction hypothesis:

Suppose that the equality holds for $n=k$, thus

$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$

Induction step:

Proving for $n=k+1$

$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$

Now, we try to simplify the left right side:

$$\frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$

We have

$$\frac{\cos(A+kB)\sin (\frac{B}{2})+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$

Using the identity:

$$\cos x \sin x = \frac{1}{2}[\sin(x+y)-\sin(x-y)]$$

$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+kB+\frac{B}{2})-\sin(A+kB-\frac{B}{2})]$$

Or:

$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]$$

Plugging it:

$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$

Using the identity:

$$ \sin x\cos x = \frac{1}{2}[\sin(x+y)+\sin(x-y)]$$

We will get:

$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\frac{1}{2}[\sin(A+\frac{B(2k-1)}{2})+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$

After simplifying more:

$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})]+\frac{1}{2}[+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$

Using the identity:

$$\sin(x\pm y)=\sin x\cos y \pm \cos x \sin y $$

We will have:

$$\frac{\frac{1}{2}[\sin A \cos(\frac{B(2k+1)}{2})+\cos A\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2})\cos(A)-\cos(\frac{B}{2})\sin(A)]}{\sin \left(\frac{B}{2}\right)}$$

Which equals to:

$$\frac{\frac{1}{2}[\sin( A)(\cos(\frac{B(2k+1)}{2})-\cos(\frac{B}{2}))+\cos(A)(\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2}))}{\sin \left(\frac{B}{2}\right)}$$

Using the identities:

$$\cos x-\cos y=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$$

$$\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$$

We will get:

$$\frac{\frac{1}{2}[-2\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+2\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$

After simplifying:

$$\frac{-\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$

and

$$\frac{\sin(\frac{B(k+1)}{2})[\cos(A)\cos(\frac{Bk}{2})-\sin(A)\sin(\frac{B(k+1)}{2})]}{\sin \left(\frac{B}{2}\right)}$$

Using the identity:

$$\cos x \cos y - \sin x \sin y=\cos(x+y)$$

We will get:

$$\frac{\sin(\frac{B(k+1)}{2})(\cos(A+Bk))}{\sin \left(\frac{B}{2}\right)}$$

Therefore

$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$

And we prove for $n=k+1$, thus the induction step completed, and the identity holds for any $n$.

Salech Alhasov
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