To find [tex]\(\cos x\)[/tex] and [tex]\(\sin y\)[/tex] given that [tex]\(\sin x = \frac{3}{5}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex], we can make use of the Pythagorean identities.
### Finding [tex]\(\cos x\)[/tex]:
Given that [tex]\(\sin x = \frac{3}{5}\)[/tex], we need to find [tex]\(\cos x\)[/tex]. We will use the Pythagorean identity for sine and cosine:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
Substitute the given value of [tex]\(\sin x\)[/tex]:
[tex]\[
\left(\frac{3}{5}\right)^2 + \cos^2 x = 1
\][/tex]
[tex]\[
\frac{9}{25} + \cos^2 x = 1
\][/tex]
[tex]\[
\cos^2 x = 1 - \frac{9}{25}
\][/tex]
[tex]\[
\cos^2 x = \frac{25}{25} - \frac{9}{25}
\][/tex]
[tex]\[
\cos^2 x = \frac{16}{25}
\][/tex]
Taking the positive square root (since [tex]\(x\)[/tex] is in the first quadrant where cosine values are positive):
[tex]\[
\cos x = \sqrt{\frac{16}{25}} = \frac{4}{5}
\][/tex]
So, [tex]\(\cos x = 0.8\)[/tex].
### Finding [tex]\(\sin y\)[/tex]:
Given that [tex]\(\cos y = \frac{7}{25}\)[/tex], we need to find [tex]\(\sin y\)[/tex]. Again, we use the Pythagorean identity:
[tex]\[
\sin^2 y + \cos^2 y = 1
\][/tex]
Substitute the given value of [tex]\(\cos y\)[/tex]:
[tex]\[
\sin^2 y + \left(\frac{7}{25}\right)^2 = 1
\][/tex]
[tex]\[
\sin^2 y + \frac{49}{625} = 1
\][/tex]
[tex]\[
\sin^2 y = 1 - \frac{49}{625}
\][/tex]
[tex]\[
\sin^2 y = \frac{625}{625} - \frac{49}{625}
\][/tex]
[tex]\[
\sin^2 y = \frac{576}{625}
\][/tex]
Taking the positive square root (since [tex]\(y\)[/tex] is in the first quadrant where sine values are positive):
[tex]\[
\sin y = \sqrt{\frac{576}{625}} = \frac{24}{25}
\][/tex]
So, [tex]\(\sin y = 0.96\)[/tex].
### Final Result:
[tex]\[
\cos x = 0.8 \quad \text{and} \quad \sin y = 0.96
\][/tex]
