Find $\sin(t+\pi)$ and $\cos(t+\pi)$ when $\sin(t)$ and $\cos(t)$ given I am learning precalculus and I do not know how to approach when solving this problem:

Which steps should i take?
 A: Hints:
For (a), use the fact that $\sin(t+π)=-\sin t$ and $\cos(t+π)=-\cos t$ as well.
For (b), you need to know that $\sin$ is odd and $\cos$ is even. Thus, $\sin(-t)=-\sin t$ and $\cos(-t)=\cos t.$
For (c), use the facts that $\cos(t+π/2)=-\sin t$ and $\sin(t+π/2)=\cos t.$
You should now be able to do (d) by applying the facts I suggested in (c) and (b) above.
A: Think of (cos(t), sin(t)) as (x,y) coordinates on the unit circle.
I would visualize the point on Quadrant 1, and see what happened next.
$$\begin{matrix}
t → t+\pi & (x,y) → (-x, -y) & \text{Quadrant 3}\cr
t → -t & (x,y) → (x, -y) & \text{Quadrant 4}\cr
t → t+\frac{\pi}{2} & (x,y) → (-y, x) & \text{Quadrant 2}\cr
t → -t+\frac{\pi}{2} & (x,y) → (y,x) & \text{Complementary angles}
\end{matrix}$$
Ignoring the signs, odd $\frac{\pi}{2}$ rotations will swap x and y's.
Then, from the Quadrants, add back the signs.
A: use addition/subtraction formulas:
$$\sin(a+b) = \sin(a) \cdot \cos(b)+ \cos (a)\cdot \sin(b)$$
Substitute $-b$ in the formula above, which yields:
\sin(a+(-b)) & = \sin(a-b)\ 
             & = \sin(a) \cdot \cos(-b)+ \cos (a)\cdot \sin(-b)\
             & = \sin(a) \cdot \cos(b) - \cos (a)\cdot \sin(b)
$$\cos(a+b) = \cos(a)\cdot \cos(b) - \sin(a)\cdot \sin(b)$$
Substitute $-b$ in the formula above, which yields:
\cos(a+(-b)) & = \cos(a-b)\
             & = \cos(a) \cdot \cos(-b) - \sin(a) \cdot \sin(-b)\ 
             & = \cos(a) \cdot \cos(b) + \sin(a) \cdot \sin(b)
Now rewrite each angle in terms of $a+b$, in terms of 2 quantities added together $ \\ $
i.e: $-t = 0-t = 0+(-t) $
if you used 0-t, you will use subtraction formula, for $a = 0, b = t \\ $
or you can use addition formula, $0+(-t)$, for $a = 0, b = -t$
