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1 Answer
- ?Lv 64 years agoFavorite Answer
By induction: when n=1 we have
sin(x + π/2)
= sin(x)·cos(π/2) + sin(π/2)·cos(x)
= sin(x)·0 + 1·cos(x)
= cos(x)
= (d/dx)(sin(x)),
establishing the base case. For the inductive hypothesis assume (dⁿ/dxⁿ)(sin(x)) = sin(x + nπ/2) for some positive integer n. Then
(dⁿ⁺¹/dxⁿ⁺¹)(sin(x))
= (d/dx)((dⁿ/dxⁿ)(sin(x)))
= (d/dx)(sin(x + nπ/2)) . . . . . . . . . . . . . . . . . . . . . . .(via the inductive hypothesis)
= (d(sin(x + nπ/2) / d(x + nπ/2))·(d(x + nπ/2) / dx) . (via the chain rule)
= sin((x+nπ/2) + π/2)·1 . . . . . . . . . . . . . . . . . . . . . (via the base case)
= sin(x + (n+1)π/2),
q.e.d.