### SubsectionB.2.4Some 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.