Question

Prove that :$(\sin {55}^{\circ }-\sin {19}^{\circ })+(\sin {53}^{\circ }-\sin {17}^{\circ })=\cos 1^{\circ }$

Answer 1

Lets take LHS and then equate it to RHS

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

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

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

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

$=2\cos 1^{\circ }\left[2\cos {36}^{\circ }\sin {18}^{\circ }\right]$ [$\because \sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$]

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

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

$=\cos 1^{\circ }$

$=$ RHS

$\therefore (\sin {55}^{\circ }-\sin {19}^{\circ })+(\sin {53}^{\circ }-\sin {17}^{\circ })=\cos 1^{\circ }$