To find [tex]\(\sin(a + b)\)[/tex] given [tex]\(\sin a = \frac{4}{9}\)[/tex] and [tex]\(\sin b = \frac{2}{5}\)[/tex], we can use the angle addition formula for sine:
[tex]\[
\sin(a + b) = \sin a \cos b + \cos a \sin b
\][/tex]
First, we need to find [tex]\(\cos a\)[/tex] and [tex]\(\cos b\)[/tex]. We can use the Pythagorean identity for this:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
For [tex]\(\cos a\)[/tex]:
[tex]\[
\cos^2 a = 1 - \sin^2 a = 1 - \left(\frac{4}{9}\right)^2 = 1 - \frac{16}{81} = \frac{81}{81} - \frac{16}{81} = \frac{65}{81}
\][/tex]
[tex]\[
\cos a = \sqrt{\frac{65}{81}} = \frac{\sqrt{65}}{9}
\][/tex]
For [tex]\(\cos b\)[/tex]:
[tex]\[
\cos^2 b = 1 - \sin^2 b = 1 - \left(\frac{2}{5}\right)^2 = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}
\][/tex]
[tex]\[
\cos b = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}
\][/tex]
Now we can use these values to find [tex]\(\sin(a + b)\)[/tex]:
[tex]\[
\sin(a + b) = \sin a \cos b + \cos a \sin b = \frac{4}{9} \cdot \frac{\sqrt{21}}{5} + \frac{\sqrt{65}}{9} \cdot \frac{2}{5}
\][/tex]
[tex]\[
\sin(a + b) = \frac{4 \sqrt{21}}{45} + \frac{2 \sqrt{65}}{45} = \frac{4 \sqrt{21} + 2 \sqrt{65}}{45}
\][/tex]
Thus,
[tex]\[
\sin(a + b) = \frac{2 \sqrt{65} + 4 \sqrt{21}}{45}
\][/tex]
Given the given choices:
A) [tex]\(\frac{2 \sqrt{65} - 4 \sqrt{21}}{45}\)[/tex]
B) [tex]\(\frac{2 \sqrt{65} + 4 \sqrt{21}}{45}\)[/tex]
C) [tex]\(\frac{4 \sqrt{65} - 2 \sqrt{21}}{45}\)[/tex]
D) [tex]\(\frac{4 \sqrt{65} + 2 \sqrt{21}}{45}\)[/tex]
The correct answer is:
B) [tex]\(\frac{2 \sqrt{65} + 4 \sqrt{21}}{45}\)[/tex]
