Answer :
To solve the given problem, we need to express the given trigonometric expression as a single cosine term. The given expression is:
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) \][/tex]
First, recall the trigonometric identity:
[tex]\[ \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \][/tex]
Now, consider the expression [tex]\(\cos(A + B)\)[/tex] where [tex]\(A = \pi x\)[/tex] and [tex]\(B = \frac{x}{2}\)[/tex]. Using the identity provided:
[tex]\[ \cos(\pi x + \frac{x}{2}) = \cos(\pi x)\cos(\frac{x}{2}) - \sin(\pi x)\sin(\frac{x}{2}) \][/tex]
Notice that the given expression:
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) \][/tex]
is the negative of the expression derived from the trigonometric identity:
[tex]\[ - \left( \cos(\pi x)\cos(\frac{x}{2}) - \sin(\pi x)\sin(\frac{x}{2}) \right) \][/tex]
Therefore,
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) = -\cos\left(\pi x + \frac{x}{2}\right) = -\cos\left(\left(\pi + \frac{1}{2}\right) x\right) \][/tex]
So the expression simplifies to:
[tex]\[ -\cos\left(\left(\pi + \frac{1}{2}\right) x\right) \][/tex]
Comparing this with the multiple-choice answers:
1. [tex]\(\cos \left[\left(\pi+\frac{1}{2}\right) x\right]\)[/tex]
2. [tex]\(-\cos \left[\left(\pi-\frac{1}{2}\right) x\right]\)[/tex]
3. [tex]\(-\cos \left[\left(\pi+\frac{1}{2}\right) x\right]\)[/tex]
4. [tex]\(\cos \left[\left(\pi-\frac{1}{2}\right) x\right]\)[/tex]
5. [tex]\(-\cos \left(\frac{x}{2}\right)\)[/tex]
The correct answer is:
[tex]\[ \boxed{-\cos \left[\left(\pi+\frac{1}{2}\right) x\right]} \][/tex]
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) \][/tex]
First, recall the trigonometric identity:
[tex]\[ \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \][/tex]
Now, consider the expression [tex]\(\cos(A + B)\)[/tex] where [tex]\(A = \pi x\)[/tex] and [tex]\(B = \frac{x}{2}\)[/tex]. Using the identity provided:
[tex]\[ \cos(\pi x + \frac{x}{2}) = \cos(\pi x)\cos(\frac{x}{2}) - \sin(\pi x)\sin(\frac{x}{2}) \][/tex]
Notice that the given expression:
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) \][/tex]
is the negative of the expression derived from the trigonometric identity:
[tex]\[ - \left( \cos(\pi x)\cos(\frac{x}{2}) - \sin(\pi x)\sin(\frac{x}{2}) \right) \][/tex]
Therefore,
[tex]\[ \sin (\pi x) \sin \left(\frac{x}{2}\right) - \cos \left(\frac{x}{2}\right) \cos (\pi x) = -\cos\left(\pi x + \frac{x}{2}\right) = -\cos\left(\left(\pi + \frac{1}{2}\right) x\right) \][/tex]
So the expression simplifies to:
[tex]\[ -\cos\left(\left(\pi + \frac{1}{2}\right) x\right) \][/tex]
Comparing this with the multiple-choice answers:
1. [tex]\(\cos \left[\left(\pi+\frac{1}{2}\right) x\right]\)[/tex]
2. [tex]\(-\cos \left[\left(\pi-\frac{1}{2}\right) x\right]\)[/tex]
3. [tex]\(-\cos \left[\left(\pi+\frac{1}{2}\right) x\right]\)[/tex]
4. [tex]\(\cos \left[\left(\pi-\frac{1}{2}\right) x\right]\)[/tex]
5. [tex]\(-\cos \left(\frac{x}{2}\right)\)[/tex]
The correct answer is:
[tex]\[ \boxed{-\cos \left[\left(\pi+\frac{1}{2}\right) x\right]} \][/tex]
