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Given that [tex]\(\sin x = \frac{3}{5}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex], with [tex]\(x\)[/tex] and [tex]\(y\)[/tex] both being Quadrant I angles, find [tex]\(\sin 2y\)[/tex].

Sagot :

To solve the problem, we need to find the value of [tex]\(\sin 2y\)[/tex], given that [tex]\(\sin x = \frac{3}{5}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex], with both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] being angles in the first quadrant.

### Step-by-Step Solution:

1. Determine [tex]\(\cos x\)[/tex]:
Since [tex]\(x\)[/tex] is in the first quadrant, both [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] are positive. Using the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Substitute [tex]\(\sin x = \frac{3}{5}\)[/tex]:
[tex]\[ \left(\frac{3}{5}\right)^2 + \cos^2 x = 1 \][/tex]
[tex]\[ \frac{9}{25} + \cos^2 x = 1 \][/tex]
Simplifying, we find:
[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]
Therefore:
[tex]\[ \cos x = \sqrt{\frac{16}{25}} = \frac{4}{5} \][/tex]

2. Determine [tex]\(\sin y\)[/tex]:
Since [tex]\(y\)[/tex] is in the first quadrant, both [tex]\(\cos y\)[/tex] and [tex]\(\sin y\)[/tex] are positive. Again, using the Pythagorean identity:
[tex]\[ \cos^2 y + \sin^2 y = 1 \][/tex]
Substitute [tex]\(\cos y = \frac{7}{25}\)[/tex]:
[tex]\[ \left(\frac{7}{25}\right)^2 + \sin^2 y = 1 \][/tex]
[tex]\[ \frac{49}{625} + \sin^2 y = 1 \][/tex]
Simplifying, we find:
[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]
Therefore:
[tex]\[ \sin y = \sqrt{\frac{576}{625}} = \frac{24}{25} \][/tex]

3. Calculate [tex]\(\sin 2y\)[/tex]:
The double-angle formula for sine is:
[tex]\[ \sin 2y = 2 \sin y \cos y \][/tex]
Substituting [tex]\(\sin y = \frac{24}{25}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex]:
[tex]\[ \sin 2y = 2 \times \frac{24}{25} \times \frac{7}{25} \][/tex]
Simplifying the product:
[tex]\[ \sin 2y = 2 \times \frac{168}{625} \][/tex]
[tex]\[ \sin 2y = \frac{336}{625} \][/tex]
[tex]\[ \sin 2y \approx 0.5376 \][/tex]

### Final Answer:
[tex]\[ \sin 2y = 0.5376 \][/tex]

In summary, [tex]\(\sin 2y \approx 0.5376\)[/tex], [tex]\(\cos x = 0.8\)[/tex], and [tex]\(\sin y = 0.96\)[/tex].