To determine whether the given trigonometric identity is true or false, let's follow a step-by-step verification process.
Given identity:
[tex]$
\cos x \cos y = \frac{1}{2}[\sin (x+y) + \sin (x-y)]
$[/tex]
We need to verify whether this equation holds for various values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Consider [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to be any arbitrary angles.
1. First, let's rewrite the given equation for better visualization:
[tex]$
\cos x \cos y \stackrel{?}{=} \frac{1}{2}[\sin (x+y) + \sin (x-y)]
$[/tex]
2. According to angle addition and subtraction formulas in trigonometry:
- [tex]\(\sin (x+y) = \sin x \cos y + \cos x \sin y\)[/tex]
- [tex]\(\sin (x-y) = \sin x \cos y - \cos x \sin y\)[/tex]
3. If we substitute these formulas:
[tex]$
\frac{1}{2}[\sin (x+y) + \sin (x-y)] = \frac{1}{2}[(\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)]
$[/tex]
4. Simplifying the right-hand side, we get:
[tex]$
\frac{1}{2}[(\sin x \cos y + \sin x \cos y) + (\cos x \sin y - \cos x \sin y)] = \frac{1}{2}[2 \sin x \cos y]
$[/tex]
5. Therefore:
[tex]$
\frac{1}{2}[2 \sin x \cos y] = \sin x \cos y
$[/tex]
6. Now comparing the two expressions:
- Left side: [tex]\(\cos x \cos y\)[/tex]
- Right side we obtained after simplification: [tex]\(\sin x \cos y\)[/tex]
Clearly, [tex]\(\cos x \cos y \neq \sin x \cos y\)[/tex].
Hence, the equation [tex]\(\cos x \cos y = \frac{1}{2}[\sin (x+y) + \sin (x-y)]\)[/tex] does not hold true for all values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
Therefore, the statement is:
B. False
