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Question Number 143635 by SOMEDAVONG last updated on 16/Jun/21

	Answered by TheHoneyCat last updated on 16/Jun/21

	$let x \in C_{o r R i f y o u w a n t}$ $let s = \sin (x) and c = \cos (x)$ $s^{4} + c^{4} + s^{2} c^{2}$ $= {(s^{2})}^{2} + {(c^{2})}^{2} + 2 (s^{2}) (c^{2}) - {(s c)}^{2}$ $= (s^{2} + c^{2}) - {(s c)}^{2}$ $= 1 - {(s c)}^{2}$ $also,$ $c^{8} + s^{8}$ $= c^{2} c^{6} + s^{2} s^{6}$ $= (1 - s^{2}) c^{6} + (1 - c^{2}) s^{6}$ $= c^{6} + s^{6} - s^{2} c^{6} - c^{2} s^{6}$ $= (1 - s^{2}) c^{4} + (1 - c^{2}) s^{4} - c^{2} s^{2} (c^{4} + s^{4})$ $= c^{4} + s^{4} - s^{2} c^{4} - c^{2} s^{4} - c^{2} s^{2} (c^{4} + s^{4})$ $= (c^{4} + s^{4}) (1 - {(c s)}^{2}) - {(c s)}^{2} (c^{2} + s^{2})$ $= ((1 - s^{2}) c^{2} + (1 - c^{2}) s^{2}) (1 - {(c s)}^{2}) - {(c s)}^{2}$ $= (c^{2} + s^{2} - 2 {(c s)}^{2}) (1 - {(c s)}^{2}) - {(c s)}^{2}$ $= (1 - 2 {(c s)}^{2}) (1 - {(c s)}^{2}) - {(c s)}^{2}$ $= 1 - 4 {(c s)}^{2} + 2 {(c s)}^{4}$ $therefore :$ $N = 2 - 2 {(s c)}^{2} - 1 + 4 {(s c)}^{2} - 2 {(c s)}^{4}$ $= 1 + 2 {(s c)}^{2} - 2 {(s c)}^{4}$ $knowing that \sin . \cos (x) = \frac{1}{2} \sin^{2} (2 x)$ $N = 1 + \frac{1}{2} \sin^{2} (2 x) - \frac{1}{8} \sin^{4} (2 x) (✠)$ $= 1 + \frac{1}{2} \sin^{2} (2 x) - \frac{1}{8} (1 - \cos^{2} (2 x)) \sin^{2} (2 x)$ $= \frac{1}{8} (8 + {4 \sin}^{2} - \sin^{2} + {(\cos . \sin)}^{2}) (2 x)$ $= \frac{1}{8} (8 + {3 \sin}^{2} (2 x) + {(\frac{1}{2} \sin (4 x))}^{2})$ $= \frac{1}{64} (64 + {24 \sin}^{2} (2 x) + {2 \sin}^{2} (4 x))$ $= \frac{1}{64} (64 + 12 + 1 - 12 (1 - \sin^{2} (2 x)) - (1 - {2 \sin}^{2} (4 x)))$ $= \frac{1}{64} (77 - 12 \cos (4 x) - \cos (8 x))$ $Wich is the^{″} {simpliest}^{″} expression_{(i t i s i t s F o u r r i e r t r a n s f o r m a t i o n)}$ $thougth I do prefer (✠)$
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