If-sin-2-x-cos-2-x-1-then-what-the-value-of-sin-11-x-cos-11-x-

Question Number 117257 by bemath last updated on 10/Oct/20

I f \sin^{2} (x) + \cos^{2} (x) = 1 t h e n

w h a t t h e v a l u e o f \sin^{11} (x) + \cos^{11} (x) = ?

Commented by bemath last updated on 10/Oct/20

nice question

Answered by mathmax by abdo last updated on 10/Oct/20

generally let find \sin^{2 n + 1} x + \cos^{2 n + 1} x = u_{n} (x)

u_{n} (x) = {(\frac{e^{ix} - e^{- ix}}{2 i})}^{2 n + 1} + {(\frac{e^{ix} + e^{- ix}}{2})}^{2 n + 1}

= \frac{1}{{(2 i)}^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} {(e^{ix})}^{k} {(- e^{- ix})}^{2 n + 1 - k}

+ \frac{1}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} {(e^{ix})}^{k} {(e^{- ix})}^{2 n + 1 - k}

= \frac{1}{{(2 i)}^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{ikx} {(- 1)}^{2 n + 1 - k} e^{- i (2 n + 1 - k) x}

+ \frac{1}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{ikx} e^{- i (2 n + 1 - k) x}

= - \frac{{(- 1)}^{n}}{2^{2 n + 1} i} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{i (k - (2 n + 1) + k) x}

+ \frac{1}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{i (k + k - (2 n + 1)) x}

= i \frac{{(- 1)}^{n}}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{i (2 k - (2 n + 1)) x} + \frac{1}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{i (2 k - (2 n + 1)) x}

= (1 + i {(- 1)}^{n}) \times \frac{1}{2^{2 n + 1}} \sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} e^{i (2 k - (2 n + 1)) x}

n = 5 \Rightarrow \sin^{11} x + \cos^{11} x = \frac{1 - i}{2^{11}} \sum_{k = 0}^{11} C_{11}^{k} e^{i (2 k - 11) x}

= \frac{1 - i}{2^{11}} {C_{11}^{0} e^{- 11 ix} + C_{11}^{1} e^{i (- 9) x} + C_{11}^{2} e^{i (- 7) x} + \dots . C_{11}^{11} e^{i 11 x}}