To solve for [tex]\( A \)[/tex] in the equation [tex]\( 2 \sin 52.5^\circ \sin 97.5^\circ = \frac{\sqrt{2} + \sqrt{A}}{2} \)[/tex], we should use the product-to-sum formula for sine functions. The product-to-sum formula states:
[tex]\[
2 \sin(x) \sin(y) = \cos(x - y) - \cos(x + y)
\][/tex]
Let [tex]\( x = 52.5^\circ \)[/tex] and [tex]\( y = 97.5^\circ \)[/tex]. Applying the product-to-sum formula:
[tex]\[
2 \sin(52.5^\circ) \sin(97.5^\circ) = \cos(52.5^\circ - 97.5^\circ) - \cos(52.5^\circ + 97.5^\circ)
\][/tex]
First, simplify the arguments in the cosine functions:
[tex]\[
52.5^\circ - 97.5^\circ = -45^\circ
\][/tex]
[tex]\[
52.5^\circ + 97.5^\circ = 150^\circ
\][/tex]
Then, substitute these values back into the equation:
[tex]\[
2 \sin(52.5^\circ) \sin(97.5^\circ) = \cos(-45^\circ) - \cos(150^\circ)
\][/tex]
Next, use the known values for the cosine function:
[tex]\[
\cos(-45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}
\][/tex]
[tex]\[
\cos(150^\circ) = -\frac{\sqrt{3}}{2}
\][/tex]
Then, substitute these values back into the equation:
[tex]\[
2 \sin(52.5^\circ) \sin(97.5^\circ) = \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}
\][/tex]
Thus:
[tex]\[
2 \sin(52.5^\circ) \sin(97.5^\circ) = \frac{\sqrt{2} + \sqrt{3}}{2}
\][/tex]
Given the original equation:
[tex]\[
2 \sin 52.5^\circ \sin 97.5^\circ = \frac{\sqrt{2} + \sqrt{A}}{2}
\][/tex]
By comparing both expressions:
[tex]\[
\frac{\sqrt{2} + \sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{A}}{2}
\][/tex]
Since the expressions on both sides of the equation must be equal, we equate the radicands:
[tex]\[
\sqrt{A} = \sqrt{3}
\][/tex]
By squaring both sides:
[tex]\[
A = 3
\][/tex]
Therefore, [tex]\( A = 3 \)[/tex].
