# Thread: Proof of a trig relation using induction

1. ## Proof of a trig relation using induction

The number x and the real number $\theta$ are such that $x + \frac{1}{x}= 2 cos\theta$ . Use
mathematical induction to show that $x^n+ \frac{1}{x^n}= 2 cosn \theta$ for all positive integers n > 1.

Thank you

2. Hello differentiate
Originally Posted by differentiate
The number x and the real number $\theta$ are such that $x + \frac{1}{x}= 2 cos\theta$ . Use
mathematical induction to show that $x^n+ \frac{1}{x^n}= 2 cosn \theta$ for all positive integers n > 1.

Thank you
Hint:

$\left(x^k+\dfrac1{x^k}\right)\left(x+\dfrac1x\righ t)=x^{k+1}+\dfrac{1}{x^{k+1}}+x^{k-1}+\dfrac{1}{x^{k-1}}$

$\Rightarrow x^{k+1}+\dfrac{1}{x^{k+1}}=\left(x^k+\dfrac1{x^k}\ right)\left(x+\dfrac1x\right)-\left(x^{k-1}+\dfrac{1}{x^{k-1}\right)}$
Then assume that the result is true for all values up to $n = k$, and the RHS becomes $2(2\cos n\theta\cos\theta - \cos(n-1)\theta)$, which simplifies to $2\cos(n+1)\theta$.

Can you complete the proof?

3. Originally Posted by differentiate
The number x and the real number $\theta$ are such that $x + \frac{1}{x}= 2 cos\theta$ . Use
mathematical induction to show that $x^n+ \frac{1}{x^n}= 2 cosn \theta$ for all positive integers n > 1.

Thank you
There is also a solution without using induction...

Start of solution...

Let $z=rcis{\alpha}$, a complex number.

We will prove:

If $z + \frac{1}{z}= 2 cos\theta$

then: $x^n+ \frac{1}{z^n}= 2 cosn \theta$

Using de Moivre's formula, and $z=rcis{\alpha}$ on $z + \frac{1}{z}= 2 cos\theta$.

The result quickly proved...