Proof that $\sin^2(kx)+\cos^2(kx)=1$ with k being any integer

Question

I know that $\sin^2(x)+cos^2(x)=1$. But how comes $\sin^2(2x)+cos^2(2x)=1$ or $\sin^2(4x)+cos^2(4x)=1$?

Is there any geometrical proof for all these expressions?

Answer 1

Use induction to prove: $\sin^2(nx)+\cos^2(nx)=1$.

Base case: $n=1$: $\sin^2x+\cos^2x=1$.

Assume $\sin^2(nx)+\cos^2(nx)=1$ is true and prove: $$\sin^2((n+1)x)+\cos^2((n+1)x)=\\ (\sin nx\cos x+\cos nx\sin x)^2+(\cos nx\cos x-\sin nx\sin x)^2=\\ \color{red}{\sin^2nx\cos^2x}+\color{blue}{\cos^2nx\sin^2x}+\color{red}{\cos^2nx\cos^2x}+\color{blue}{\sin^2nx\sin^2x}=\\ \color{red}{(\sin^2nx+\cos^2nx)\cos^2x}+\color{blue}{(\sin^2nx+\cos^2nx)\sin^2x}=\\ \sin^2(nx)+\cos^2(nx)=1.$$

Answer 2

Hint:

$$\cos^2\left(\frac{\sqrt[\pi]{\log(t^2+h)}}{z\csc(\theta)}\right)+\sin^2\left(\frac{\sqrt[\pi]{\log(t^2+h)}}{z\csc(\theta)}\right)=1.$$

Answer 3

It might be worthwhile to point out the underlying logical basis for answering this question. Namely, this is an example of the substitution principle: if you know that an equation is true for all $x$ in some set, then by substituting any expression for an element of that set in place of $x$, you get another true equation.

So, $\sin^2(x)+\cos^2(x)=1$ is true for all real numbers $x$. Therefore, if you substitute any expression for a real number in place of $x$, you get a true equation.

To take an example, if $x$ is a real number then $4x$ is also a real number, because real numbers are closed under multiplication. So, if I substitute $4x$ in place of $x$, I get a true equation: $$\sin^2(4x) + \cos^2(4x)=1 $$

Here's one way I try to make this clear when teaching precalculus. If the $x$ is confusing you, replace it by anything else. For example, replace it by FNORK: $$\sin^2(\text{FNORK}) + \cos^2(\text{FNORK}) = 1 $$ This is a true equation, where we take FNORK as an expression representing any real number. Now feel free to replace FNORK by any other expression representing a real number. For example replace FNORK by $4x$; as explained earlier, $4x$ is a real number if $x$ is a real number.

And, of course, it turns out that you can replace FNORK by any complex number, as hinted in the other answer.

Answer 4

You were looking for a geometrical proof.

Think of a plane and vectors in it transformed by some operator. In particular consider linear operators in a cartesian coordinate systems. The coordinate of a vector are transformed by the multiplication by a matrix representing a given linear operator.

In this setting $$\cos^2\theta+\sin^2\theta$$ can be interpreted as how much of the first coordinate of a vector goes into the first coordinate of the transformed vector, and at the same time how much of the second coordinate of a vector goes into the second coordinate of the transformed vector in the case, for instance, that a vector is subjected to two tranformations: a rotation of $\theta$ radians and a subsequent rotation of $-\theta$ (that collectively amouts to no transformation at all).

Indeed, the matrix representation of this global (neutral) transformation is then given by the multiplication of these two matrix:

\begin{equation} \begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\phantom{-}\cos\theta \end{bmatrix} \cdot \begin{bmatrix} \phantom{-}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \end{equation} and then \begin{equation} \begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} \cos^2\theta+\sin^2\theta&0\\ 0&\cos^2\theta+\sin^2\theta \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \end{equation}

But we know that this global transformation is a no transformation so it must be that: \begin{equation} \begin{bmatrix} \cos^2\theta+\sin^2\theta&0\\ 0&\cos^2\theta+\sin^2\theta \end{bmatrix} = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \end{equation} and so it must be $$\cos^2\theta+\sin^2\theta = 1$$.

Now let's see what happens to $$\cos^2\theta'+\sin^2\theta'$$ when $\theta' = k\theta$, for whatever real $k$. Now if we apply the same geometrical reasoning as before for $\theta'=k\theta$ (two rotations of the same radians but of opposite orientation) we will again have a no transformation and again we will conclude that $$\cos^2\theta'+\sin^2\theta' = 1$$ therefore it is also $$\cos^2 k\theta+\sin^2 k\theta = 1$$