Determine Trigonometric Functions with Given Conditions

Linear Algebra Problems and Solutions

Problem 651

(a) Find a function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] such that $g(0) = g(\pi/2) = g(\pi) = 0$, where $a, b, c$ are constants.

(b) Find real numbers $a, b, c$ such that the function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] satisfies $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$.

 
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Solution.

(a) Condition: $g(0) = g(\pi/2) = g(\pi) = 0$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have the system of linear equations
\begin{align*}
g(0) &= a + b + c = 0 \\[6pt] g \left( \frac{\pi}{2} \right) &= -b = 0 \\[6pt] g(\pi) &= -a + b – c = 0.
\end{align*}


To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 1 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].
\end{align*}

The solution can now be read off: $a+c = 0$ and $b=0$.
Thus a general solution is of the form $g(\theta) = a \cos(\theta) – a \cos ( 3 \theta) $, where $a$ is any real number.

(b) Condition: $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have

\begin{align*}
g(0) &= a + b + c = 3\\
g(\pi/2) &= -b = 1\\
g(\pi) &= -a + b – c = -5.
\end{align*}


To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 3 \\ 0 & -1 & 0 & 1 \\ -1 & 1 & -1 & -5 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & -1 & -4 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{align*}

The solution can now be read off: $a+c = 4$ and $b=-1$. Thus a general solution is of the form
$$ g(\theta) = a \cos(\theta) – \cos(2 \theta) + (4 – a) \cos(3 \theta) , $$
where $a$ is any real number.


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