Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
