proof of (cosh(x)+sinh(x))^n=cosh(xn)+sinh(xn)
A proof by Mathematical Induction that the statement
A(n): (cosh(x)+sinh(x))^n=cosh(nx)+sinh(nx) is true Ɐn
Base Case
n = 1
cosh(x)+sinh(x)=cosh(x)+sinh(x) – which is true.
Inductive step
n = k
Assume A(k) is true, i.e that (cosh(x)+sinh(x))^k=cosh(kx)+sinh(kx)
n = k+1
(cosh(x)+sinh(x))^(k+1) = (cosh(x)+sinh(x))^k *(cosh(x)+sinh(x))
(cosh(x)+sinh(x))^(k+1) = (cosh(kx)+sinh(kx)) *(cosh(x)+sinh(x)) (using A(k))
(cosh(x)+sinh(x))^(k+1) = cosh(kx).cosh(x)+sinh(kx).cosh(x) + cosh(kx).sinh(x) + sinh(kx).sinh(x)
(cosh(x)+sinh(x))^(k+1) = cosh(kx).cosh(x) + sinh(kx).sinh(x) + sinh(kx).cosh(x) + cosh(kx).sinh(x)
(cosh(x)+sinh(x))^(k+1) = cosh(kx + x) + sinh(kx + x)
(cosh(x)+sinh(x))^(k+1) = cosh((k+1)x) + sinh((k+1)x)
The last line above shows that if A(k) is true, then so also is A(k+1)
Hence, by mathematical induction A(n) is true Ɐn