$c o s 1^{0} + c o s 2^{0} + c o s 3^{0} + . . . . . . . + c o s 179^{0} =$

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Solution

The correct option is B $0$
$cos 1^{\circ} + cos 2^{\circ} + cos 3^{\circ} + \dots + cos 179^{\circ} = cos 1^{\circ} + cos 2^{\circ} + \dots + cos (180^{\circ} - 2^{\circ}) + cos (180^{\circ} - 1^{\circ})$
$= cos 1^{\circ} + cos 2^{\circ} + \dots + cos 89^{\circ} + cos 90^{\circ} - cos 81^{\circ} - \dots - cos 2^{\circ} - cos 1^{\circ}$
$= cos 90^{\circ}$
$= 0$

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Similar questions

c o s 10^{0} . c o s 20^{0} . c o s 30^{0} . . . . . . . . c o s 80^{0} . . . . . . . c o s 900^{0} . . . . . c o s 1790^{0}

	List I		List II
A.	$cos 20^{\circ} + cos 80^{\circ} - \sqrt{3} cos 50^{\circ}$	1.	$- 1$
B.	$cos 0^{\circ} + cos \frac{π}{7} + cos \frac{2 π}{7} + cos \frac{3 π}{7} + cos \frac{4 π}{7} + cos \frac{5 π}{7} + cos \frac{6 π}{7}$	2.	$- \frac{3}{4}$
C.	$cos 20^{\circ} + cos 40^{\circ} + cos 60^{\circ} + 4 cos 10^{\circ} cos 20^{\circ} cos 30^{\circ}$	3.	$1$
D.	$cos 20^{\circ} cos 100^{\circ} + cos 100^{\circ} cos 140^{\circ} - cos 140^{\circ} cos 200^{\circ}$	4.	$0$

c o s 1^{0} . c o s 2^{0} . . . . . . . . . c o s 90^{0} =

c o s 1^{o} \times c o s 2^{o} \times . . . . \times c o s 90^{o}

cos 40^{\circ} + cos 50^{\circ} + cos 70^{\circ} + cos 80^{\circ} = cos 20^{\circ} + cos 10^{\circ}

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