# Subspace Spanned by Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$

## Problem 612

Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.

Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.

**(a)** Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.

**(b)** Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

Sponsored Links

Contents

## Solution.

### (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.

By definition of the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$, the we know that $B$ is a spanning set for $W$.

Thus, it remains to show that $B$ is linearly independent set.

Suppose that

\[c_1\sin^2(x)+c_2\cos^2(x)=0.\]
This equality is true for all $x\in [-2\pi, 2\pi]$.

In particular, evaluating at $x=0$, we see that $c_2=0$.

Also, plugging in $x=\pi/2$ yields $c_1=0$.

Therefore, $\sin^2(x)$ and $\cos^2(x)$ are linearly independent, that is, $B$ is linearly independent.

As $B$ is a linearly independent spanning set, it is a basis for $W$.

### Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

Note that $\sin^2(x)-\cos^2(x)$ and $1$ are both in $W$ since both functions are linear combination of $\sin^2(x)$ and $\cos^2(x)$. Here, we used the trigonometric identity $1=\sin^2(x)+\cos^(x)$.

By part (a), we see that $\dim(W)=2$. So if we show that the functions $\sin^2(x)-\cos^2(x)$ and $1$ are linearly independent, then they form a basis for $W$.

We consider the coordinate vectors of these functions with respect to the basis $B$.

We have

\begin{align*}

[\sin^2(x)-\cos^2(x)]_B=\begin{bmatrix}

1

\\ -1

\end{bmatrix}

\text{ and }\\

[1]_B=[\sin^2(x)+\cos^2(x)]_B=\begin{bmatrix}

1\\ 1

\end{bmatrix}.

\end{align*}

Since we have

\begin{align*}

\begin{bmatrix}

1& 1 \\

-1& 1

\end{bmatrix}

\xrightarrow{R_2+R_1}

\begin{bmatrix}

1& 1 \\

0& 2

\end{bmatrix}

\xrightarrow{\frac{1}{2}R_2}

\begin{bmatrix}

1& 1 \\

0& 1

\end{bmatrix}

\xrightarrow{R_1-R_1}

\begin{bmatrix}

1& 0 \\

0& 1

\end{bmatrix},

\end{align*}

the coordinate vectors are linearly independent, and hence $\sin^2(x)-\cos^2(x)$ and $1$ are linearly independent.

We conclude that $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

(Another reasoning is that since the coordinate vectors form a basis for $\R^2$, $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.)

### Comment

You may directly show that $\{\sin^2(x)-\cos^2(x), 1\}$ is linearly independent just like we did for part (a).

Add to solve later

Sponsored Links