find the sum of sin^2 1°+sin^2 2°+...+sin^2 60°=?


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Question Number 209706 by mr W last updated on 19/Jul/24

$f i n d t h e s u m o f$ $\sin^{2} 1 ° + \sin^{2} 2 ° + . . . + \sin^{2} 60 ° = ?$
	Answered by BaliramKumar last updated on 19/Jul/24

	$\underset{x = 1}{\sum^{60}} \sin^{2} x = \underset{x = 1}{\sum^{29}} \sin^{2} x + 15.5$
	Commented by mr W last updated on 19/Jul/24

	$a n d \underset{x = 1}{\sum^{29}} \sin^{2} x = ?$
	Answered by MM42 last updated on 19/Jul/24

	${s i n}^{2} a = \frac{1}{2} (1 - c o s 2 a)$ $\Rightarrow S = 30 - \frac{1}{2} (c o s 2 + c o s 4 + . . . + c o s 120)$ $s = c o s 2 + c o s 4 + . . . + c o s 120$ $\Rightarrow 2 s i n 1 \times s = 2 s i n 1 c o s 2 + 2 s i n 1 c o s 4 + . . . + 2 s i n 1 c o s 120$ $= (s i n 3 - s i n 1) + (s i n 5 - s i n 3) + . . . + (s i n 121 - s i n 119)$ $= s i n 121 - s i n 1 = 2 s i n 60 c o s 61$ $\Rightarrow s = \frac{\sqrt{3}}{2} \times \frac{c o s 61}{s i n 1} \sim 24.06$ $\Rightarrow S \sim 30 - 12.03 = 17.7 ✓$
	Commented by mr W last updated on 19/Jul/24

	$t h a n k s s i r!$ $p l e a s e r e c h e c k .$ $S = 30 - \frac{\sqrt{3} \cot 1 ° - 3}{8} \approx 17.97$
	Commented by MM42 last updated on 19/Jul/24

	$c o s 61 = c o s 60 c o s 1 - s i n 1 s i n 60$ $= \frac{1}{2} (c o s 1 - \sqrt{3} s i n 1)$ $\div 2 s i n 1 = \frac{1}{4} (c o t 1 - \sqrt{3}) ★$
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