Section 5.3 Trigonometric Identities
The following trigonometric identities will be useful in calculus.
\begin{gather*}
\cos^2\theta + \sin^2\theta = 1 \\
\tan^2\theta + 1 = \sec^2\theta \\
\cot^2\theta + 1 = \csc^2\theta
\end{gather*}
\begin{gather*}
\sin(a+b) = \sin a \cos b +\cos a \sin b \\
\sin(a-b) = \sin a \cos b -\cos a \sin b \\
\cos(a+b) = \cos a \cos b -\sin a \sin b \\
\cos(a-b) = \cos a \cos b +\sin a \sin b
\end{gather*}
\begin{align*}
\sin 2\theta = \amp 2\sin\theta\cos\theta \\
\cos 2\theta = \amp \cos^2\theta - \sin^2\theta \\
=\amp 2\cos^2\theta - 1 \\
=\amp 1 - 2\sin^2\theta
\end{align*}
\begin{align*}
\cos^2\theta = \amp \frac{1+\cos 2\theta}{2}\\
\sin^2\theta = \amp \frac{1-\cos 2\theta}{2}
\end{align*}
\begin{gather*}
\cos\left(-\theta\right) =\cos\theta \\
\sin\left(-\theta\right) =-\sin\theta
\end{gather*}
It is not necessary to memorize all of the above identities! By knowing how an identity is derived, one can reduce the amount of memorization necessary.
The equation of the unit circle is \(x^2+y^2=1\text{.}\) The Pythagorean theorem also gives us the equation \(x^2+y^2=1\text{.}\) This equation gives us
\begin{equation*}
\cos^2\theta + \sin^2\theta =1\text{.}
\end{equation*}
Dividing this equation by \(\cos^2\theta\text{,}\) we obtain
\begin{equation*}
\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}
\text{ , or }
1 +\tan^2\theta=\sec^2\theta.
\end{equation*}
Dividing \(\cos^2\theta+\sin^2\theta=1\) by \(\sin^2\theta\text{,}\) we obtain
\begin{equation*}
\frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}
\text{ , or }
\cot^2\theta +1 =\csc^2\theta.
\end{equation*}
Using
\begin{equation*}
\sin(a+b) = \sin a \cos b + \cos a \sin b
\end{equation*}
and
\begin{equation*}
\cos(a+b) = \cos a\cos b - \sin a \sin b
\end{equation*}
with \(a=b=\theta\text{,}\) we obtain the formulas
\begin{equation*}
\sin 2\theta = 2\sin \theta \cos\theta \text{ and } \cos 2\theta = \cos^2\theta - \sin^2\theta.
\end{equation*}
Also
\begin{align*}
\cos 2\theta = \amp \cos^2\theta - \sin^2\theta \\
=\amp \cos^2\theta - (1-\cos^2\theta)\\
=\amp 2\cos^2\theta -1
\end{align*}
and
\begin{align*}
\cos 2\theta = \amp \cos^2\theta - \sin^2\theta \\
= \amp (1-\sin^2\theta) - \sin^2\theta \\
= \amp 1 - 2\sin^2\theta
\end{align*}
using the identity \(\cos^2\theta+\sin^2\theta=1\text{.}\) We can then solve for \(\cos^2\theta\) in \(\cos 2\theta=2\cos^2\theta-1\) to get
\begin{equation*}
\cos^2\theta=\frac{1+\cos2\theta}{2}.
\end{equation*}
Solving for \(\sin^2\theta\) in \(\cos2\theta = 1-2\sin^2\theta\) gives us
\begin{equation*}
\sin^2\theta = \frac{1-\cos2\theta}{2}.
\end{equation*}
