probably, cos nx=2^(n−1) cos^n x−n2^(n−3) cos^(n−2) +(((n−3)n)/2)2^(n−5) cos^(n−4) x… wow


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Question Number 43268 by Rauny last updated on 09/Sep/18

$probably, \cos n x = 2^{n - 1} \cos^{n} x - n 2^{n - 3} \cos^{n - 2}$ $+ \frac{(n - 3) n}{2} 2^{n - 5} \cos^{n - 4} x \dots$ $wow$
Commented by Rauny last updated on 09/Sep/18

$oh thx!$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18

$c o s x + i s i n x = e^{i x}$ ${(c o s x + i s i n x)}^{n} = {(e^{i x})}^{n}$ $e^{i n x} = {(c + i s)}^{n} c = c o s x s = s i n x$ $c o s n x + i s i n n x = c^{n} + {n C}_{1} c^{n - 1} (i s) + {n C}_{2} c^{n - 2} {(i s)}^{2} +$ ${n C}_{3} c^{n - 3} {(i s)}^{3} + . . . + {(i s)}^{n}$ $= {c o s}^{n} x + i ({n C}_{1} {c o s}^{n - 1} x . s i n x) + {n C}_{2} {c o s}^{n - 2} x . (- {s i n}^{2} x) +$ ${n C}_{3} {c o s}^{n - 3} x (- {i s i n}^{3} x) + . . . + {(i s i n x)}^{n}$ $n o w a r r a n g i n g r e a l p a r t a n d i m a g i n a r y p a r t i n r i g h t$ $h a n d s i d e$ $c o s n x + i s i n n x = ({c o s}^{n} x - {n C}_{2} {c o s}^{n - 2} {s i n}^{2} x + {n C}_{4} {c o s}^{n - 4} {x s i n}^{4} x + . . .) +$ $i ({n C}_{1} {c o s}^{n - 1} x s i n x - {n C}_{3} {c o s}^{n - 3} {x s i n}^{3} x + . . .)$ $s o c o s n x = {c o s}^{n} x - {n C}_{2} {c o s}^{n - 2} {s i n}^{2} x + {n C}_{4} {c o s}^{n - 4} {x s i n}^{4} x - . . .$
Commented by malwaan last updated on 09/Sep/18

$how ?$
Commented by Rauny last updated on 09/Sep/18

$If you calculate it yourself,$ $you will know the reason .$
Commented by malwaan last updated on 20/Oct/18

$thank you sir$
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