proof of (cosh(x)+sinh(x))^n=cosh(xn)+sinh(xn)

Question

proof that proof of (cosh(x)+sinh(x))^n=cosh(xn)+sinh(xn)

Answer 1

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