If #sin x + "cosec" x = 2#, then find #sin^n x + "cosec" ^n x# ?

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Answer 1 · 2016-09-03T10:46:07

#sin^nx + 1/(sin^n x) = 2#

Solving

#sin x + 1/sin x = 2 -> sin^2 x - 2 sinx + 1 = 0#

we have

#sin x = 1# then

#sin^nx + 1/(sin^n x) = 2#

This can be also demonstrated by finite induction.

1) First it is true for #n=1# (with sin x = 1 of course)
2) Supposing that it is true for #n# then
3) Prove that it is true for #n+1#

It is quite easy and it is left as an exercise.