The value of sum_{r=1}^{18} cos^2(5r)^circ,wh | vedclass

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(C) The given expression is $S = \sum_{r=1}^{18} \cos^2(5r)^\circ = \cos^2 5^\circ + \cos^2 10^\circ + \dots + \cos^2 85^\circ + \cos^2 90^\circ$.We k

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$\frac{17}{2}$

Vedclass · Answer

(C) The given expression is $S = \sum_{r=1}^{18} \cos^2(5r)^\circ = \cos^2 5^\circ + \cos^2 10^\circ + \dots + \cos^2 85^\circ + \cos^2 90^\circ$.We know that $\cos^2 90^\circ = 0$.For the remaining terms,we can pair them as $\cos^2 	heta + \cos^2(90^\circ - 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.There are $17$ terms excluding $\cos^2 90^\circ$. We can form $8$ pairs of $(\cos^2 5^\circ + \cos^2 85^\circ), (\cos^2 10^\circ + \cos^2 80^\circ), \dots, (\cos^2 40^\circ + \cos^2 50^\circ)$ and one middle term $\cos^2 45^\circ$.Each pair sums to $1$,so $8 	imes 1 = 8$.The middle term is $\cos^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.Thus,the total sum is $8 + \frac{1}{2} = \frac{17}{2}$.

The value of $\sum_{r=1}^{18} \cos^2(5r)^\circ$,where $x^\circ$ denotes the $x$ degree,is equal to

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