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Sagot :
Given that [tex]\(\tan x = -\frac{8}{15}\)[/tex] and [tex]\(x\)[/tex] terminates in quadrant II, we will find [tex]\(\sin 2x\)[/tex], [tex]\(\cos 2x\)[/tex], and [tex]\(\tan 2x\)[/tex].
First, we need to determine the values of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] based on the tangent value.
### Finding [tex]\(\cos x\)[/tex] and [tex]\(\sin x\)[/tex]:
Using the identity:
[tex]\[ 1 + \tan^2 x = \sec^2 x \][/tex]
Given [tex]\(\tan x = -\frac{8}{15}\)[/tex]:
[tex]\[ 1 + \left(-\frac{8}{15}\right)^2 = \sec^2 x \][/tex]
[tex]\[ 1 + \frac{64}{225} = \sec^2 x \][/tex]
[tex]\[ \sec^2 x = \frac{225}{225} + \frac{64}{225} = \frac{289}{225} \][/tex]
So, [tex]\(\sec x = \sqrt{\frac{289}{225}} = \frac{17}{15}\)[/tex].
Since [tex]\(\sec x = \frac{1}{\cos x}\)[/tex]:
[tex]\[ \cos x = \frac{15}{17} \][/tex]
However, since [tex]\(x\)[/tex] is in quadrant II, [tex]\(\cos x\)[/tex] is negative:
[tex]\[ \cos x = -\frac{15}{17} \][/tex]
Next, using the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Since we already have [tex]\(\cos x = -\frac{15}{17}\)[/tex]:
[tex]\[ \sin^2 x + \left(-\frac{15}{17}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 x + \frac{225}{289} = 1 \][/tex]
[tex]\[ \sin^2 x = 1 - \frac{225}{289} \][/tex]
[tex]\[ \sin^2 x = \frac{289}{289} - \frac{225}{289} \][/tex]
[tex]\[ \sin^2 x = \frac{64}{289} \][/tex]
Thus:
[tex]\[ \sin x = \sqrt{\frac{64}{289}} = \frac{8}{17} \][/tex]
Since [tex]\(x\)[/tex] terminates in quadrant II, [tex]\(\sin x\)[/tex] is positive:
[tex]\[ \sin x = \frac{8}{17} \][/tex]
### Finding [tex]\(\sin 2x\)[/tex] and [tex]\(\cos 2x\)[/tex]:
Using the double angle identities:
[tex]\[ \sin 2x = 2 \sin x \cos x \][/tex]
[tex]\[ \sin 2x = 2 \left(\frac{8}{17}\right) \left(-\frac{15}{17}\right) \][/tex]
[tex]\[ \sin 2x = 2 \left(-\frac{120}{289}\right) \][/tex]
[tex]\[ \sin 2x = -\frac{240}{289} \][/tex]
Next, for [tex]\(\cos 2x\)[/tex]:
[tex]\[ \cos 2x = \cos^2 x - \sin^2 x \][/tex]
[tex]\[ \cos 2x = \left(-\frac{15}{17}\right)^2 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \cos 2x = \frac{225}{289} - \frac{64}{289} \][/tex]
[tex]\[ \cos 2x = \frac{225 - 64}{289} \][/tex]
[tex]\[ \cos 2x = \frac{161}{289} \][/tex]
### Finding [tex]\(\tan 2x\)[/tex]:
Finally, using the identity:
[tex]\[ \tan 2x = \frac{\sin 2x}{\cos 2x} \][/tex]
[tex]\[ \tan 2x = \frac{-\frac{240}{289}}{\frac{161}{289}} \][/tex]
[tex]\[ \tan 2x = \frac{-240}{161} \][/tex]
Rewriting it as a simplified fraction:
[tex]\[ \tan 2x = -\frac{240}{161} \][/tex]
Thus, the values are:
[tex]\[ \sin 2x = -\frac{240}{289} \][/tex]
[tex]\[ \cos 2x = \frac{161}{289} \][/tex]
[tex]\[ \tan 2x = -\frac{240}{161} \][/tex]
First, we need to determine the values of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] based on the tangent value.
### Finding [tex]\(\cos x\)[/tex] and [tex]\(\sin x\)[/tex]:
Using the identity:
[tex]\[ 1 + \tan^2 x = \sec^2 x \][/tex]
Given [tex]\(\tan x = -\frac{8}{15}\)[/tex]:
[tex]\[ 1 + \left(-\frac{8}{15}\right)^2 = \sec^2 x \][/tex]
[tex]\[ 1 + \frac{64}{225} = \sec^2 x \][/tex]
[tex]\[ \sec^2 x = \frac{225}{225} + \frac{64}{225} = \frac{289}{225} \][/tex]
So, [tex]\(\sec x = \sqrt{\frac{289}{225}} = \frac{17}{15}\)[/tex].
Since [tex]\(\sec x = \frac{1}{\cos x}\)[/tex]:
[tex]\[ \cos x = \frac{15}{17} \][/tex]
However, since [tex]\(x\)[/tex] is in quadrant II, [tex]\(\cos x\)[/tex] is negative:
[tex]\[ \cos x = -\frac{15}{17} \][/tex]
Next, using the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Since we already have [tex]\(\cos x = -\frac{15}{17}\)[/tex]:
[tex]\[ \sin^2 x + \left(-\frac{15}{17}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 x + \frac{225}{289} = 1 \][/tex]
[tex]\[ \sin^2 x = 1 - \frac{225}{289} \][/tex]
[tex]\[ \sin^2 x = \frac{289}{289} - \frac{225}{289} \][/tex]
[tex]\[ \sin^2 x = \frac{64}{289} \][/tex]
Thus:
[tex]\[ \sin x = \sqrt{\frac{64}{289}} = \frac{8}{17} \][/tex]
Since [tex]\(x\)[/tex] terminates in quadrant II, [tex]\(\sin x\)[/tex] is positive:
[tex]\[ \sin x = \frac{8}{17} \][/tex]
### Finding [tex]\(\sin 2x\)[/tex] and [tex]\(\cos 2x\)[/tex]:
Using the double angle identities:
[tex]\[ \sin 2x = 2 \sin x \cos x \][/tex]
[tex]\[ \sin 2x = 2 \left(\frac{8}{17}\right) \left(-\frac{15}{17}\right) \][/tex]
[tex]\[ \sin 2x = 2 \left(-\frac{120}{289}\right) \][/tex]
[tex]\[ \sin 2x = -\frac{240}{289} \][/tex]
Next, for [tex]\(\cos 2x\)[/tex]:
[tex]\[ \cos 2x = \cos^2 x - \sin^2 x \][/tex]
[tex]\[ \cos 2x = \left(-\frac{15}{17}\right)^2 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \cos 2x = \frac{225}{289} - \frac{64}{289} \][/tex]
[tex]\[ \cos 2x = \frac{225 - 64}{289} \][/tex]
[tex]\[ \cos 2x = \frac{161}{289} \][/tex]
### Finding [tex]\(\tan 2x\)[/tex]:
Finally, using the identity:
[tex]\[ \tan 2x = \frac{\sin 2x}{\cos 2x} \][/tex]
[tex]\[ \tan 2x = \frac{-\frac{240}{289}}{\frac{161}{289}} \][/tex]
[tex]\[ \tan 2x = \frac{-240}{161} \][/tex]
Rewriting it as a simplified fraction:
[tex]\[ \tan 2x = -\frac{240}{161} \][/tex]
Thus, the values are:
[tex]\[ \sin 2x = -\frac{240}{289} \][/tex]
[tex]\[ \cos 2x = \frac{161}{289} \][/tex]
[tex]\[ \tan 2x = -\frac{240}{161} \][/tex]
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