prove

**perash** · December 16th 2007, 02:16 PM

$\sin 47° + \sin 61° - \sin 11° - \sin 25° = \cos 7°$ .

**TwistedOne151** · December 16th 2007, 05:23 PM

First, note that 47-11=61-25=36, so rearrange the left hand side to get $(\sin 47^\circ - \sin 11^\circ) + (\sin 61^\circ - \sin 25^\circ)$ . From this, we use the sum-to-product formulas, specifically the formula

$\sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right ) \sin \left ( \frac{A-B}{2} \right )$ .

Thus we get
$\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = (\sin 47^\circ - \sin 11^\circ) + (\sin 61^\circ - \sin 25^\circ)$
$\,\,= 2 \cos 29^\circ \sin 18^\circ + 2 \cos 43^\circ \sin 18^\circ$
$\,\,= 2(\cos 29^\circ + \cos 43^\circ)\sin 18^\circ$

We use the sum-to-product formulas again, this time using $\cos A + \cos B = 2\cos \left ( \frac{A+B}{2} \right ) \cos \left ( \frac{A-B}{2} \right )$ , so
$\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2(\cos 29^\circ + \cos 43^\circ)\sin 18^\circ$
$\,\,= 2(2\cos 36^\circ \cos (-7^\circ))\sin 18^\circ$
$\,\,= 4\cos 7^\circ \cos 36^\circ \sin 18^\circ$ , as $\cos(-x)=\cos x$ .

Working from the geometry of the pentagon, one can find the sine and cosine for angles that are multiples of 18 degrees, and thus find that:
$\cos 36^\circ = \frac{1 + \sqrt{5}}{4}$ , $\sin 18^\circ = \frac{-1 + \sqrt{5}}{4}$ , and $\cos 36^\circ \sin 18^\circ = \frac{1 + \sqrt{5}}{4}\frac{-1 + \sqrt{5}}{4}=\frac{5-1}{16}=\frac{1}{4}$

So we get:
$\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \cos 36^\circ \sin 18^\circ$
$\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \cdot \frac{1}{4}=\cos 7^\circ$

--Kevin C.

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