47^ + 61^ - 11^ - 25^ = 7^

Question

$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}=\cos 7^{\circ}$

✓

Answer

$\begin{aligned}
& \text { L.H.S. }=\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ} \\
& =\left(\sin 47^{\circ}-\sin 25^{\circ}\right)+\left(\sin 61^{\circ}-\sin 11^{\circ}\right) \\
& =2 \cos \left(\frac{47^{\circ}+25^{\circ}}{2}\right) \sin \left(\frac{47^{\circ}-25^{\circ}}{2}\right) \\
& \quad+2 \cos \left(\frac{61^{\circ}+11^{\circ}}{2}\right) \sin \left(\frac{61^{\circ}-11^{\circ}}{2}\right) \\
& =2 \cos 36^{\circ} \sin 11^{\circ}+2 \cos 36^{\circ} \sin 25^{\circ} \\
& =2 \cos 36^{\circ}\left(\sin 11^{\circ}+\sin 25^{\circ}\right) \\
& =2 \cos 36^{\circ}\left[2 \sin \left(\frac{25^{\circ}+11^{\circ}}{2}\right) \cos \left(\frac{25^{\circ}-11^{\circ}}{2}\right)\right] \\
& =2 \cos 36^{\circ}\left(2 \sin 18^{\circ} \cos 7^{\circ}\right) \\
& =4\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right) \cos 7^{\circ} \\
& =\frac{4(5-1)}{16} \cos 7^{\circ} \\
& =\cos 7^{\circ} \\
& =\text { R.H.S. }
\end{aligned}$

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