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To solve for [tex]\(\cos(x + y)\)[/tex] 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 angles in the first quadrant, we will follow these steps:
1. Find [tex]\(\cos x\)[/tex]:
Since [tex]\(x\)[/tex] is in the first quadrant, both sine and cosine are positive. Using the Pythagorean identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex],
[tex]\[ \sin^2 x = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \][/tex]
[tex]\[ \cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
[tex]\[ \cos x = \sqrt{\frac{16}{25}} = \frac{4}{5} \][/tex]
Therefore, [tex]\(\cos x = \frac{4}{5}\)[/tex].
2. Find [tex]\(\sin y\)[/tex]:
Similarly, since [tex]\(y\)[/tex] is in the first quadrant, both sine and cosine are positive. Using the Pythagorean identity [tex]\(\cos^2 y + \sin^2 y = 1\)[/tex],
[tex]\[ \cos^2 y = \left(\frac{7}{25}\right)^2 = \frac{49}{625} \][/tex]
[tex]\[ \sin^2 y = 1 - \cos^2 y = 1 - \frac{49}{625} = \frac{625}{625} - \frac{49}{625} = \frac{576}{625} \][/tex]
[tex]\[ \sin y = \sqrt{\frac{576}{625}} = \frac{24}{25} \][/tex]
Therefore, [tex]\(\sin y = \frac{24}{25}\)[/tex].
3. Calculate [tex]\(\cos(x + y)\)[/tex]:
Using the angle addition formula for cosine: [tex]\(\cos(x + y) = \cos x \cos y - \sin x \sin y\)[/tex],
[tex]\[ \cos(x + y) = \left(\frac{4}{5} \cdot \frac{7}{25}\right) - \left(\frac{3}{5} \cdot \frac{24}{25}\right) \][/tex]
[tex]\[ \cos(x + y) = \frac{28}{125} - \frac{72}{125} = \frac{28 - 72}{125} = \frac{-44}{125} = -0.352 \][/tex]
Therefore, the values we have are:
[tex]\[ \cos x = \frac{4}{5} = 0.8 \][/tex]
[tex]\[ \sin y = \frac{24}{25} = 0.96 \][/tex]
[tex]\[ \cos(x + y) = -0.352 \][/tex]
So, when given [tex]\(\sin x = \frac{3}{5}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex] with angles in the first quadrant, the value of [tex]\(\cos(x + y)\)[/tex] is [tex]\(-0.352\)[/tex].
1. Find [tex]\(\cos x\)[/tex]:
Since [tex]\(x\)[/tex] is in the first quadrant, both sine and cosine are positive. Using the Pythagorean identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex],
[tex]\[ \sin^2 x = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \][/tex]
[tex]\[ \cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
[tex]\[ \cos x = \sqrt{\frac{16}{25}} = \frac{4}{5} \][/tex]
Therefore, [tex]\(\cos x = \frac{4}{5}\)[/tex].
2. Find [tex]\(\sin y\)[/tex]:
Similarly, since [tex]\(y\)[/tex] is in the first quadrant, both sine and cosine are positive. Using the Pythagorean identity [tex]\(\cos^2 y + \sin^2 y = 1\)[/tex],
[tex]\[ \cos^2 y = \left(\frac{7}{25}\right)^2 = \frac{49}{625} \][/tex]
[tex]\[ \sin^2 y = 1 - \cos^2 y = 1 - \frac{49}{625} = \frac{625}{625} - \frac{49}{625} = \frac{576}{625} \][/tex]
[tex]\[ \sin y = \sqrt{\frac{576}{625}} = \frac{24}{25} \][/tex]
Therefore, [tex]\(\sin y = \frac{24}{25}\)[/tex].
3. Calculate [tex]\(\cos(x + y)\)[/tex]:
Using the angle addition formula for cosine: [tex]\(\cos(x + y) = \cos x \cos y - \sin x \sin y\)[/tex],
[tex]\[ \cos(x + y) = \left(\frac{4}{5} \cdot \frac{7}{25}\right) - \left(\frac{3}{5} \cdot \frac{24}{25}\right) \][/tex]
[tex]\[ \cos(x + y) = \frac{28}{125} - \frac{72}{125} = \frac{28 - 72}{125} = \frac{-44}{125} = -0.352 \][/tex]
Therefore, the values we have are:
[tex]\[ \cos x = \frac{4}{5} = 0.8 \][/tex]
[tex]\[ \sin y = \frac{24}{25} = 0.96 \][/tex]
[tex]\[ \cos(x + y) = -0.352 \][/tex]
So, when given [tex]\(\sin x = \frac{3}{5}\)[/tex] and [tex]\(\cos y = \frac{7}{25}\)[/tex] with angles in the first quadrant, the value of [tex]\(\cos(x + y)\)[/tex] is [tex]\(-0.352\)[/tex].
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