What is the Sum of Cosine Products in POTW #141?

[anemone]

Evaluate $\cos 1^{\circ}\cos 2^{\circ}+\cos 2^{\circ}\cos 3^{\circ}+\cdots+\cos 88^{\circ}\cos 89^{\circ}$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. greg1313
2. MarkFL
3. lfdahl
3. kaliprasad

Solution from MarkFL:

Spoiler

Using the product to sum identity:

$$\cos(\alpha)\cos(\beta)=\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}$$

And the negative angle identity for cosine (cosine is an even function):

$$\cos(-x)=\cos(x)$$

And then factoring, the sum becomes:

$$\frac{\cos\left(1^{\circ}\right)}{2}\cdot\sum_{k=1}^{88}(1)+\frac{1}{2}\left(\cos\left(3^{\circ}\right)+\cos\left(5^{\circ}\right)+\cdots+\cos\left(177^{\circ}\right)\right)$$

Next, use $$\sum_{k=1}^n(1)=n$$ and then group as follows:

$$44\cos\left(1^{\circ}\right)+\left(\left(\cos\left(3^{\circ}\right)+\cos\left(177^{\circ}\right)\right)+\left(\cos\left(5^{\circ}\right)+\cos\left(175^{\circ}\right)\right)+\cdots+\left(\cos\left(89^{\circ}\right)+\cos\left(91^{\circ}\right)\right)\right)$$

Now, applying the sum to product identity:

$$\cos(\alpha)+\cos(\beta)=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$$

And using the fact that:

$$\cos\left(90^{\circ}\right)=0$$

We obtain:

$$44\cos\left(1^{\circ}\right)+\left(0+0+\cdots+0\right)=44\cos\left(1^{\circ}\right)$$

Solution from kaliprasad:

Spoiler

we have $2 \cos(x) \cos (y) = \cos (x+y) + \cos (y-x)$

so $2 (\cos\,1^\circ \cos\,2^\circ ) = \cos\,3^\circ + \cos\,1^\circ$

$2 (\cos\,2^\circ \cos\,3^\circ ) = \cos\,5^\circ + \cos\,1^\circ$

so on till

$2 (\cos\,44^\circ \cos\,45^\circ ) = \cos\,89^\circ + \cos\,1^\circ$

$2 (\cos\,45^\circ\cos\,46^\circ ) = \cos\,91^\circ + \cos\,1^\circ$

or $2 (\cos\,45^\circ \cos\,46^\circ ) = - \cos\,89^\circ + \cos\,1^\circ$

so on till

$2 (\cos\,88^\circ \cos\,89^\circ ) = - \cos\,3^\circ + \cos\,1^\circ$

on adding above for each positive term in 1st half for the first term there is a -ve term for the second half and we are left wth 88 times $\cos\,1^\circ$

so $2( \cos\,1^\circ \cos\,2^\circ + \cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ) = 88 \cos\,1^\circ$

or $ \cos\,1^\circ \cos\,2^\circ + \cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ = 44 \cos\,1^\circ$

hence the given expression is $44 \cos\,1^\circ$