Question

How do you show that `sin(x + pi) = -sin(x)` ?

Answer 1

Use the identity:

`sin(A+B) = sin(A)cos(B)+cos(A)sin(B)`

Substitute x for A and `pi` for B:

`sin(x+pi) = sin(x)cos(pi)+cos(x)sin(pi)`

The second term disappears, because `sin(pi) = 0`

`sin(x+pi) = sin(x)cos(pi)`

The fact that `cos(pi) = -1` finishes the proof:

`sin(x+pi) = -sin(x)`

Q.E.D.

Answer 2

You can graph it and show it.

`sin(x+pi)` is just `sinx` shifted left by `pi` units. Since `sinx` has a period of `2pi`, that offsets `sinx` by half a period leftwards. Since `sinx` is odd about its half-period, we've transformed the function into `-sinx`.

`sinx`:

graph{sinx [-4.385, 4.384, -2.187, 2.198]}

`sin(x + pi)`:

graph{sin(x + pi) [-4.385, 4.384, -2.187, 2.198]}

`-sinx`:

graph{-sinx [-4.385, 4.384, -2.187, 2.198]}

Answer 3

A geometric argument...

Explanation:

The points on the unit circle have coordinates `(cos theta, sin theta)` where `theta` is the counterclockwise angle from the `x` axis.

So the point `(cos (x+pi), sin (x+pi))` is the point on the unit circle at angle `pi` from `(cos x, sin x)`.

This point is directly opposite `(cos x, sin x)` across the centre `(0, 0)` of the unit circle, so its coordinates can also be expressed as `(-cos x, -sin x)`.

Equating coordinates, we find:

`{ (cos (x+pi) = -cos x), (sin (x+pi) = -sin x) :}`