Find: 1−(sin30°)^2 + (sin30°)^4 − (sin30°)^6 + ...


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Question Number 202356 by hardmath last updated on 25/Dec/23

$Find :$ $1 - {(\sin 30 °)}^{2} + {(\sin 30 °)}^{4} - {(\sin 30 °)}^{6} + . . .$
	Answered by Rasheed.Sindhi last updated on 25/Dec/23
	$1−(sin30°)^2 + (sin30°)^4 − (sin30°)^6 + ... (1+(sin30°)^4 +(sin30°)^8 +...)−{(sin30°)^2 +(sin30°)^6 +(sin30°)^(10) ...} ={1+((1/2))^4 +((1/2))^8 +...} −{((1/2))^2 +((1/2))^6 +((1/2))^(10) +...} In 1st series: a=1,r=((1/2))^4 In 2nd series: a=((1/2))^2 ,r=((1/2))^4 =(1/(1−(1/(16))))−((1/4)/(1−(1/(16))))=(((1−(1/4)))/((1−(1/4))(1+(1/4)))) =(1/(5/4))=(4/5) ✓ Formula used: determinant (((a+ar+ar^2 +...=(a/(1−r)) ; ∣r∣<1)))$
	$1 - {(\sin 30 °)}^{2} + {(\sin 30 °)}^{4} - {(\sin 30 °)}^{6} + . . .$ $(1 + {(\sin 30 °)}^{4} + {(\sin 30 °)}^{8} + . . .) - {{(\sin 30 °)}^{2} + {(\sin 30 °)}^{6} + {(\sin 30 °)}^{10} . . .}$ $= {1 + {(\frac{1}{2})}^{4} + {(\frac{1}{2})}^{8} + . . .}$ $- {{(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{6} + {(\frac{1}{2})}^{10} + . . .}$ $I n 1 st series : a = 1, r = {(\frac{1}{2})}^{4}$ $I n 2 nd series : a = {(\frac{1}{2})}^{2}, r = {(\frac{1}{2})}^{4}$ $= \frac{1}{1 - \frac{1}{16}} - \frac{\frac{1}{4}}{1 - \frac{1}{16}} = \frac{(1 - \frac{1}{4})}{(1 - \frac{1}{4}) (1 + \frac{1}{4})}$ $= \frac{1}{5 / 4} = \frac{4}{5} ✓$ $F o r m u l a u s e d :$ $\begin{matrix} a + a r + {a r}^{2} + . . . = \frac{a}{1 - r}; ∣ r ∣< 1 \end{matrix}$
	Commented by hardmath last updated on 25/Dec/23

	Thank you dear professor, what are the values of a and r in the exercise...
	Commented by Rasheed.Sindhi last updated on 25/Dec/23

	$H a v e a d d e d t w o r e d l i n e s t o e x p l a i n .$
	Answered by Sutrisno last updated on 25/Dec/23

	$s = \frac{1}{1 - (- {(s i n 30 °)}^{2})}$ $s = \frac{1}{1 - (- \frac{1}{4})}$ $s = \frac{4}{5}$
	Answered by Rajpurohith last updated on 26/Dec/23

	$C a l l x = s i n (30 °)$ $t h e n t h e e x p r e s s i o n$ $= 1 - x^{2} + x^{4} - x^{6} + . . . = (\frac{1}{1 + x^{2}}) = \frac{1}{1 + {s i n}^{2} (30 °)}$ $= \frac{1}{1 + \frac{1}{4}} = \frac{4}{5}$
	Answered by MathematicalUser2357 last updated on 29/Dec/23

	$= \underset{x = 0}{\sum^{\infty}} ({(- 1)}^{n} \sin^{2 n} \frac{π}{6}) = \frac{4}{5}$
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