Subsection B.2.4 Some More Simple Identities
¶Consider the figure below
The pair triangles on the left shows that there is a simple relationship between trigonometric functions evaluated at \(\theta\) and at \(-\theta\text{:}\)
\begin{align*}
\sin(-\theta)&=-\sin(\theta) & \cos(-\theta) &=\cos(\theta)
\end{align*}
That is — sine is an odd function, while cosine is even. Since the other trigonometric functions can be expressed in terms of sine and cosine we obtain
\begin{align*}
\tan(-\theta) &=-\tan(\theta) &
\csc(-\theta) &=-\csc(\theta) &
\sec(-\theta) &=\sec(\theta) &
\cot(-\theta) &=-\cot(\theta)
\end{align*}
Now consider the triangle on the right — if we consider the angle \(\frac{\pi}{2}-\theta\) the side-lengths of the triangle remain unchanged, but the roles of “opposite” and “adjacent” are swapped. Hence we have
\begin{align*}
\sin\left(\tfrac{\pi}{2}-\theta\right)&=\cos\theta &
\cos\left(\tfrac{\pi}{2}-\theta\right)&=\sin\theta
\end{align*}
Again these imply that
\begin{align*}
\tan\left(\tfrac{\pi}{2}-\theta\right)&=\cot\theta &
\csc\left(\tfrac{\pi}{2}-\theta\right)&=\sec\theta &
\sec\left(\tfrac{\pi}{2}-\theta\right)&=\csc\theta &
\cot\left(\tfrac{\pi}{2}-\theta\right)&=\tan\theta
\end{align*}
We can go further. Consider the following diagram:
This implies that
\begin{align*}
\sin(\pi-\theta)&=\sin(\theta) & \cos(\pi-\theta) &= -\cos(\theta)\\
\sin(\pi+\theta)&=-\sin(\theta) & \cos(\pi+\theta) &=-\cos(\theta)
\end{align*}
From which we can get the rules for the other four trigonometric functions.