$sin 24^{\circ} + sin 32^{\circ} + sin 204^{\circ} + sin 212^{\circ} =$

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0$
$sin 24 ° + sin 32 ° + sin 204 ° + sin 212 °$
$= sin 24 ° + sin 32 ° + sin (180 ° + 24 °) + sin (180 ° + 32 °)$
$= sin 24 ° + sin 32 ° - sin 24 ° + sin 32 °$
$= 0$
Hence, the answer is $0.$

Suggest Corrections

Similar questions

{cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ} = \frac{\sqrt{5} + 1}{8}

Prove that $({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ}) = \frac{(\sqrt{5} + 1)}{8}$

sin 20 (4+sec20 ) is

{sin}^{2} 12^{\circ} + {sin}^{2} 21^{\circ} + {sin}^{2} 39^{\circ} + {sin}^{2} 48^{\circ} - {sin}^{2} 9^{\circ} - {sin}^{2} 18^{\circ} = 1

cos 6^{\circ} sin 24^{\circ} cos 72^{\circ}

\frac{3}{16}

cos 10^{\circ} cos 30^{\circ} cos 50^{\circ} cos 70^{\circ}

\frac{1}{16}

sin 2^{\circ} sin 24^{\circ} sin 48^{\circ} sin 84^{\circ}

\frac{3}{16}

sin 20^{\circ} sin 40^{\circ} sin 60^{\circ} sin 80^{\circ}

\frac{1}{8}

Join BYJU'S Learning Program