Question

The value of $\sin {55}^o-\sin {19}^o+\sin {53}^o-\sin {17}^o$ is always equal to

• A. $\frac{\cos 1^o}{2}$
• B. $\sin 1^o$
• C. $\tan 1^o$
• D. $\cot 1^o$

Answer 1

The correct option is A $\frac{\cos 1^o}{2}$

$\sin {55}^{\circ }-\sin {19}^{\circ }+\sin {53}^{\circ }-\sin {17}^{\circ }$

$=\sin {55}^{\circ }+\sin {53}^{\circ }-(\sin {19}^{\circ }+\sin {17}^{\circ })$

$2\sin \frac{(55+53)}{2}\cos \left(\frac{55-53}{2}\right)=2\left(\sin \right(\frac{19+17}{2}\left)\right)cos\left(\frac{19-17}{2}\right)$

$=2\sin {54}^{\circ }\cos 1^{\circ }-2\sin {18}^{\circ }\cos 1^{\circ }$

$=2\cos 1^{\circ }(\sin {54}^{\circ }-\sin {18}^{\circ })$

$=2\cos 1^{\circ }\left(2sin\right(\frac{54-18}{2}\left)\right)\cos \left(\frac{54+18}{2}\right)$

$=2\cos 1^{\circ }sin{18}^{\circ }cos{36}^{\circ }$

$=2\cos 1^{\circ }\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)$

$=2\cos 1^{\circ }\left(\frac{4}{16}\right)$

$=\frac{cos{1}^{\circ }}{2}$.