Chapter 5 - Section 5.2 - Trigonometric Functions of Real Numbers - 5.2 Exercises - Page 418: 83

$sin(t+\pi) = -sin ~t$ $cos(t+\pi) = -cos ~t$ $tan(t+\pi) = tan ~t$

In the figure, we can see the following: $sin ~t = \frac{y}{1} = y$ $sin(t+\pi) = \frac{-y}{1} = -y = -sin ~t$ $cos ~t = \frac{x}{1} = x$ $cos(t+\pi) = \frac{-x}{1} = -x = -cos ~t$ $tan ~t = \frac{y}{x}$ $tan(t+\pi) = \frac{-y}{-x} = \frac{y}{x} = tan ~t$