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\title{Simple Trigonometric Formulas}

\author{Math 126: Calculus II}
\date{Fall Semester 1999}

\begin{document}

\maketitle
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\section*{Fundamental Identities}
The functions $\cos(\theta)$ and $\sin(\theta)$ are defined to be the $x$
and $y$~coordinates of the point at an angle of $\theta$ on the unit circle.
Therefore,
$\sin(-\theta) = -\sin(\theta)$, $\cos(-\theta) = \cos(\theta)$, and
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\] 
The other trigonometric functions are defined in terms of sine and cosine:
\[
\begin{array}{rclrcl}
\tan(\theta) &=& \sin(\theta)/\cos(\theta)\quad & \quad
\cot(\theta) &=& \cos(\theta)/\sin(\theta) = 1/\tan(\theta)\\
\sec(\theta) &=& 1/\cos(\theta)\quad & \quad
\csc(\theta) &=& 1/\sin(\theta)
\end{array}
\]
Dividing $\sin^2(\theta) + \cos^2(\theta) = 1$ by
$\cos^2(\theta)$ or $\sin^2(\theta)$ gives
\begin{eqnarray*}
\tan^2(\theta) + 1 &=& \sec^2(\theta)\\
1 + \cot^2(\theta) &=& \csc^2(\theta)
\end{eqnarray*}

\section*{Addition Formulas}
The following two formulas are somewhat difficult to derive and should
be memorized.
\begin{eqnarray*}
\sin(\alpha+\beta) &=& \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \\
\cos(\alpha+\beta) &=& \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)
\end{eqnarray*}
These formulas can be used to prove simple identities like
$\sin(\pi/2 - \theta) = \sin(\pi/2)\cos(\theta) + \cos(\pi/2)\sin(\theta) =
\cos(\theta)$, or $\cos(x-\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) =
-\cos(x)$.  If we set $\alpha = \beta$ in the addition formulas we get the 
double-angle formulas:
\begin{eqnarray*}
\sin(2\alpha) &=&  2\sin(\alpha)\cos(\alpha) \\
\cos(2\alpha) &=& \cos^2(\alpha) - \sin^2(\alpha)
\end{eqnarray*}
The formula for $\cos(2\alpha)$ is often rewritten by replacing
$\cos^2(\alpha)$ with $1-\sin^2(\alpha)$ or replacing $\sin^2(\alpha)$
with $1-\cos^2(\alpha)$ to get
\begin{eqnarray*}
\cos(2\alpha) &=& 1 - 2\sin^2(\alpha) \\
\cos(2\alpha) &=& 2\cos^2(\alpha) - 1
\end{eqnarray*}
Solving for $\sin^2(\alpha)$ and $\cos^2(\alpha)$ yields identities
important for integration:
\begin{eqnarray*}
\sin^2(\alpha) &=& \frac{1}{2}(1 - \cos(2\alpha)) \\
\cos^2(\alpha) &=& \frac{1}{2}(1 + \cos(2\alpha))
\end{eqnarray*}

\section*{Derivatives and Integrals}
\[
\begin{array}{rclrcl}
\frac{d}{dx}\sin(x) &=& \cos(x)\quad &\quad
\frac{d}{dx}\sec(x) &=& \sec(x)\tan(x)\\
\frac{d}{dx}\cos(x) &=& -\sin(x)\quad &\quad
\frac{d}{dx}\csc(x) &=& -\csc(x)\cot(x) \\
\frac{d}{dx}\tan(x) &=& \sec^2(x)\quad &\quad
\frac{d}{dx}\cot(x) &=& -\csc^2(x) \\
\\
\int\sin(x)\,dx &=& -\cos(x)+C\quad &\quad
\int\sec(x)\,dx &=& \ln|\sec(x)+\tan(x)|+C\\
\int\cos(x)\,dx &=& \sin(x)+C\quad &\quad
\int\csc(x)\,dx &=& -\ln|\csc(x)+\cot(x)|+C \\
\int\tan(x)\,dx &=& \ln|\sec(x)|+C\quad &\quad 
\int\cot(x)\,dx &=& -\ln|\csc(x)|+C
\end{array}
\]

%An addition formula for tangent can also be derived from the
%ones for sine and cosine. Dividing the first equation by the second gives
%\[
%\tan(\alpha+\beta) = \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}
%                 {\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)}
%\]
%Now divide the numerator and denominator by $\cos(\alpha)\cos(\beta)$ to get
%\[
%\tan(\alpha+\beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 -
%\tan(\alpha)\tan(\beta)}
%\]
%\[
%\tan(2\alpha) &=& \frac{2\tan(\alpha)}{1-\tan^2(\alpha)} \\
%\]

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