cos ^{2} 5^{circ}+cos ^{2} 10^{circ}+cos ^{2} | vedclass

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Question

(A) The given expression is $S = \cos ^{2} 5^{\circ}+\cos ^{2} 10^{\circ}+\cos ^{2} 15^{\circ}+\cdots+\cos ^{2} 90^{\circ}$.We know that $\cos(90^{\ci

Vedclass · Accepted Answer

$8 \frac{1}{2}$

Vedclass · Answer

(A) The given expression is $S = \cos ^{2} 5^{\circ}+\cos ^{2} 10^{\circ}+\cos ^{2} 15^{\circ}+\cdots+\cos ^{2} 90^{\circ}$.We know that $\cos(90^{\circ} - 	heta) = \sin 	heta$,so $\cos^{2}(90^{\circ} - 	heta) = \sin^{2} 	heta$.Pairing the terms: $(\cos ^{2} 5^{\circ}+\cos ^{2} 85^{\circ}) + (\cos ^{2} 10^{\circ}+\cos ^{2} 80^{\circ}) + \cdots + (\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}) + \cos ^{2} 45^{\circ} + \cos ^{2} 90^{\circ}$.Each pair $(\cos ^{2} 	heta + \cos ^{2} (90^{\circ} - 	heta)) = (\cos ^{2} 	heta + \sin ^{2} 	heta) = 1$.There are $8$ such pairs (for angles $5^{\circ}, 10^{\circ}, \dots, 40^{\circ}$).Thus,the sum is $(8 	imes 1) + \cos ^{2} 45^{\circ} + \cos ^{2} 90^{\circ}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$\cos^{2} 45^{\circ} = \frac{1}{2}$. Also,$\cos 90^{\circ} = 0$.Total sum $= 8 + \frac{1}{2} + 0 = 8 \frac{1}{2}$.

$\cos ^{2} 5^{\circ}+\cos ^{2} 10^{\circ}+\cos ^{2} 15^{\circ}+\cdots+\cos ^{2} 90^{\circ}=$

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