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-

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} + \dots + {(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) + \dots + {(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 + \dots) +

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 + \dots)

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 - \dots

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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