Sure, let's solve the trigonometric expression [tex]\(\sin(264^\circ) \cos(6^\circ) + \cos(264^\circ) \sin(6^\circ)\)[/tex].
First, we should recognize the trigonometric identity for the sine of a sum:
[tex]\[
\sin (a + b) = \sin a \cos b + \cos a \sin b
\][/tex]
In this problem, we can let [tex]\(a = 264^\circ\)[/tex] and [tex]\(b = 6^\circ\)[/tex]. Applying the given identity:
[tex]\[
\sin(264^\circ) \cos(6^\circ) + \cos(264^\circ) \sin(6^\circ) = \sin(264^\circ + 6^\circ)
\][/tex]
Next, we add the angles inside the sine function:
[tex]\[
264^\circ + 6^\circ = 270^\circ
\][/tex]
Now we need to find [tex]\(\sin(270^\circ)\)[/tex].
The sine of [tex]\(270^\circ\)[/tex] can be determined from the unit circle. [tex]\(270^\circ\)[/tex] corresponds to the point [tex]\((0, -1)\)[/tex] on the unit circle. Therefore:
[tex]\[
\sin(270^\circ) = -1
\][/tex]
Thus, the value of [tex]\(\sin(264^\circ) \cos(6^\circ) + \cos(264^\circ) \sin(6^\circ)\)[/tex] is:
[tex]\[
-1
\][/tex]
So, the exact value is:
[tex]\[
\boxed{-1}
\][/tex]
