2 Answers

Fred Pym · Answer 1 · 2023-12-21T11:09:15+0000

Answer:

The general solution to
$\cos(x - 1) = -\cos(x)$ is
$x = (k + (1/2))\, \pi + (1/2)$ for all integer
$k \in \mathbb{Z}$ .

Explanation:

Given:

$\cos(x - 1) = -\cos(x)$ .

Rearrange to obtain:

$\cos(x - 1) + \cos(x) = 0$ .

By sum-of-angle identity for cosines:

$\cos(a + b) = \cos(a) \,\cos(b) - \sin(a) \, \sin(b)$ .

Since
$\sin(-b) = -\sin(b)$ , the following is also an identity:

$\begin{aligned}\cos(a - b) &= \cos(a) \,\cos(b) - \sin(a) \, \sin(-b) \\ &= \cos(a)\, \cos(b) - \sin(a) \, (-\sin(b)) \\ &= \cos(a) \, \cos(b) + \sin(a)\, \sin(b)\end{aligned}$ .

Add these two identities together to obtain the sum-to-product formula:

$\begin{aligned}& \cos(a + b) \\ &+ \cos(a - b)\\ &= \cos(a) \, \cos(b) - \sin(a)\, \sin(b) \\ &\quad + \cos(a)\, \cos(b) + \sin(a)\, \sin(b)\end{aligned}$ .

Simplify to obtain the formula:

$\cos(a + b) + \cos(a - b) = 2\, \cos(a)\, \cos(b)$ .

Before applying this sum-to-product formula to
$\cos(x - 1) + \cos(x) = 0$ , it would be necessary to find the
and
such that:

.
.

Solve this system of equations for
and
to obtain
a = x - (1/2) and
b = -(1/2) .

Apply the sum-to-product formula:

$\begin{aligned}& 2\, \cos(x - (1/2))\, \cos(-(1/2)) \\ =\; & 2\, \cos(a)\, \cos(b) \\ =\; & \cos(a + b) + \cos(a - b) && (\text{sum-to-product}) \\ =\; & \cos(x - (1/2) + (-1/2)) \\ &+ \cos(x - (1/2) - (-1/2)) \\ =\; & \cos(x - 1) + \cos(x) \\ =\; & 0\end{aligned}$ .

In other words, the original equation
$\cos(x - 1) + \cos(x) = 0$ is equivalent to
$2\, \cos(x - (1/2))\, \cos(-(1/2)) = 0$ . Solve this new equation for
: