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Sagot :
To find [tex]\(\sin \theta\)[/tex] given that [tex]\(\cos \theta = \frac{15}{17}\)[/tex], we can use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Given [tex]\(\cos \theta = \frac{15}{17}\)[/tex], let's denote this value for clarity:
[tex]\[ \cos \theta = \frac{15}{17} \][/tex]
First, let's square [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos^2 \theta = \left(\frac{15}{17}\right)^2 = \frac{225}{289} \][/tex]
Next, substitute [tex]\(\cos^2 \theta\)[/tex] into the Pythagorean identity:
[tex]\[ \sin^2 \theta + \frac{225}{289} = 1 \][/tex]
Now, solve for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = 1 - \frac{225}{289} \][/tex]
To proceed, we need to subtract [tex]\(\frac{225}{289}\)[/tex] from 1. We can convert 1 to a fraction with the same denominator:
[tex]\[ 1 = \frac{289}{289} \][/tex]
Thus:
[tex]\[ \sin^2 \theta = \frac{289}{289} - \frac{225}{289} = \frac{64}{289} \][/tex]
Now, taking the square root of both sides to find [tex]\(\sin \theta\)[/tex]:
[tex]\[ \sin \theta = \sqrt{\frac{64}{289}} = \frac{\sqrt{64}}{\sqrt{289}} = \frac{8}{17} \][/tex]
Given that [tex]\(\cos \theta\)[/tex] is positive and considering standard trigonometric functions in different quadrants, [tex]\(\sin \theta\)[/tex] should also be positive in this case.
Therefore, the value of [tex]\(\sin \theta\)[/tex] is:
[tex]\[ \boxed{\frac{8}{17}} \][/tex]
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Given [tex]\(\cos \theta = \frac{15}{17}\)[/tex], let's denote this value for clarity:
[tex]\[ \cos \theta = \frac{15}{17} \][/tex]
First, let's square [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos^2 \theta = \left(\frac{15}{17}\right)^2 = \frac{225}{289} \][/tex]
Next, substitute [tex]\(\cos^2 \theta\)[/tex] into the Pythagorean identity:
[tex]\[ \sin^2 \theta + \frac{225}{289} = 1 \][/tex]
Now, solve for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = 1 - \frac{225}{289} \][/tex]
To proceed, we need to subtract [tex]\(\frac{225}{289}\)[/tex] from 1. We can convert 1 to a fraction with the same denominator:
[tex]\[ 1 = \frac{289}{289} \][/tex]
Thus:
[tex]\[ \sin^2 \theta = \frac{289}{289} - \frac{225}{289} = \frac{64}{289} \][/tex]
Now, taking the square root of both sides to find [tex]\(\sin \theta\)[/tex]:
[tex]\[ \sin \theta = \sqrt{\frac{64}{289}} = \frac{\sqrt{64}}{\sqrt{289}} = \frac{8}{17} \][/tex]
Given that [tex]\(\cos \theta\)[/tex] is positive and considering standard trigonometric functions in different quadrants, [tex]\(\sin \theta\)[/tex] should also be positive in this case.
Therefore, the value of [tex]\(\sin \theta\)[/tex] is:
[tex]\[ \boxed{\frac{8}{17}} \][/tex]
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