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Sagot :
To find [tex]\(\cos(\theta)\)[/tex] given that [tex]\(\sin(\theta) = \frac{3}{8}\)[/tex], we can use the Pythagorean identity, which states:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
First, we need to compute [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) = \left( \frac{3}{8} \right)^2 = \frac{9}{64} \][/tex]
Using the Pythagorean identity, solve for [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{9}{64} = \frac{64}{64} - \frac{9}{64} = \frac{55}{64} \][/tex]
Next, take the square root of both sides to solve for [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(\theta) = \sqrt{\frac{55}{64}} \][/tex]
Simplifying the square root:
[tex]\[ \cos(\theta) = \frac{\sqrt{55}}{8} \][/tex]
So the correct answer is:
B. [tex]\(\frac{\sqrt{55}}{8}\)[/tex]
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
First, we need to compute [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) = \left( \frac{3}{8} \right)^2 = \frac{9}{64} \][/tex]
Using the Pythagorean identity, solve for [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{9}{64} = \frac{64}{64} - \frac{9}{64} = \frac{55}{64} \][/tex]
Next, take the square root of both sides to solve for [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(\theta) = \sqrt{\frac{55}{64}} \][/tex]
Simplifying the square root:
[tex]\[ \cos(\theta) = \frac{\sqrt{55}}{8} \][/tex]
So the correct answer is:
B. [tex]\(\frac{\sqrt{55}}{8}\)[/tex]
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