To determine why [tex]\(\sin 30^\circ \sin 60^\circ + \cos 30^\circ \cos 60^\circ = 2(\sin 30^\circ \sin 60^\circ) = 2(\cos 30^\circ \cos 60^\circ)\)[/tex] is true, let's break down the trigonometric identities and relationships involved.
First, recall the exact values of the sines and cosines for the given angles:
- [tex]\(\sin 30^\circ = \frac{1}{2}\)[/tex]
- [tex]\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\cos 60^\circ = \frac{1}{2}\)[/tex]
Now, let's evaluate the left-hand side (LHS) and the right-hand side (RHS) of the given equation separately.
### Evaluating the LHS
[tex]\[
\text{LHS} = \sin 30^\circ \sin 60^\circ + \cos 30^\circ \cos 60^\circ
\][/tex]
Substitute the known values:
[tex]\[
\text{LHS} = \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)
\][/tex]
[tex]\[
\text{LHS} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}
\][/tex]
[tex]\[
\text{LHS} = \frac{2\sqrt{3}}{4}
\][/tex]
[tex]\[
\text{LHS} = \frac{\sqrt{3}}{2}
\][/tex]
### Evaluating the RHS
[tex]\[
\text{RHS} = 2(\sin 30^\circ \sin 60^\circ)
\][/tex]
Substitute the known values:
[tex]\[
\text{RHS} = 2 \left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right)
\][/tex]
[tex]\[
\text{RHS} = 2 \left(\frac{\sqrt{3}}{4}\right)
\][/tex]
[tex]\[
\text{RHS} = \frac{2\sqrt{3}}{4}
\][/tex]
[tex]\[
\text{RHS} = \frac{\sqrt{3}}{2}
\][/tex]
Both LHS and RHS have the same value:
[tex]\[
\frac{\sqrt{3}}{2}
\][/tex]
Therefore, [tex]\(\sin 30^\circ \sin 60^\circ + \cos 30^\circ \cos 60^\circ\)[/tex] is indeed equal to [tex]\(2(\sin 30^\circ \sin 60^\circ)\)[/tex] which simplifies to [tex]\(2(\cos 30^\circ \cos 60^\circ)\)[/tex].
The correct reasoning for why this identity holds can be inferred from:
b. The sine of an angle is equal to the cosine of its complement.
In the case of the angles 30° and 60°, these angles are complements of each other (since 30° + 60° = 90°). The identity used in the problem is a specific instance of the angle sum identity for cosine, i.e., [tex]\(\cos (a - b) = \cos a \cos b + \sin a \sin b\)[/tex], where [tex]\(a = 60^\circ\)[/tex] and [tex]\(b = 30^\circ\)[/tex], hence [tex]\(\cos (60^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}\)[/tex].
Thus, option b is the most appropriate reason.
