Answer: [tex]sin\ 2x=-\dfrac{240}{289}\qquad cos\ 2x=\dfrac{161}{289}\qquad tan\ 2x=-\dfrac{240}{161}[/tex]
Step-by-step explanation:
[tex]cos\ x=\dfrac{15}{17}\quad and\quad csc\ x<0\implies sin\ x=-\dfrac{8}{17}\ and\ tan\ 2x=-\dfrac{8}{15}\\\\sin\ 2x=2(sin\ x\cdot cos\ x)\\\\.\qquad =2\bigg(\dfrac{-8}{17}\cdot \dfrac{15}{17}\bigg)\\\\\\.\qquad =2\bigg(\dfrac{-120}{289}\bigg)\\\\\\.\qquad=\large\boxed{-\dfrac{240}{289}}[/tex]
[tex]cos\ 2x=cos^2\ x-sin^2\ x\\\\.\qquad=\bigg(\dfrac{15}{17}\bigg)^2-\bigg(\dfrac{-8}{17}\bigg)^2\\\\\\.\qquad=\dfrac{225}{289}-\dfrac{64}{289}\\\\\\.\qquad=\large\boxed{\dfrac{161}{289}}[/tex]
[tex]tan\ 2x=\dfrac{sin\ 2x}{cos\ 2x}\\\\\\.\qquad=\large\boxed{-\dfrac{240}{161}}[/tex]
The trigonometry values are:[tex]\mathbf{sin(2x) = -\frac{240}{289}}[/tex], [tex]\mathbf{cos(2x) =\frac{161}{289}}[/tex] and [tex]\mathbf{tan(2x)= -\frac{240}{161}}\\[/tex]
The given parameter is:
[tex]\mathbf{cos(x) = \frac{15}{17}}[/tex]
If csc(x) is less than 0, then sin(x) is less than 0.
Using the following trigonometry ratio,
[tex]\mathbf{sin^2(x) +cos^2(x) = 1}[/tex]
Substitute [tex]\mathbf{cos(x) = \frac{15}{17}}[/tex]
[tex]\mathbf{sin^2(x) +(\frac{15}{17})^2 = 1}[/tex]
Expand
[tex]\mathbf{sin^2(x) +\frac{225}{289} = 1}[/tex]
Collect like terms
[tex]\mathbf{sin^2(x) = 1 -\frac{225}{289}}[/tex]
Take LCM
[tex]\mathbf{sin^2(x) = \frac{289-225}{289}}[/tex]
[tex]\mathbf{sin^2(x) = \frac{64}{289}}[/tex]
Take square roots
[tex]\mathbf{sin(x) = \pm\frac{8}{17}}[/tex]
Recall that, the sine of the angles is negative.
So, we have:
[tex]\mathbf{sin(x) = -\frac{8}{17}}[/tex]
sin(2x) is then calculated as:
[tex]\mathbf{sin(2x) = 2sin(x)cos(x)}[/tex]
This gives
[tex]\mathbf{sin(2x) = 2 \times \frac{-8}{17} \times \frac{15}{17}}[/tex]
[tex]\mathbf{sin(2x) = -\frac{240}{289}}[/tex]
cos(2x) is then calculated as:
[tex]\mathbf{cos(2x) =cos^2(x) - sin^2(x)}[/tex]
This gives
[tex]\mathbf{cos(2x) =(\frac{15}{17})^2 - \frac{64}{289}}[/tex]
[tex]\mathbf{cos(2x) =\frac{225}{289} - \frac{64}{289}}[/tex]
Take LCM
[tex]\mathbf{cos(2x) =\frac{225 - 64}{289}}[/tex]
[tex]\mathbf{cos(2x) =\frac{161}{289}}[/tex]
Lastly, tan(2x) is calculated using:
[tex]\mathbf{tan(2x)= \frac{sin(2x)}{cos(2x)}}[/tex]
So, we have:
[tex]\mathbf{tan(2x)= \frac{-240/289}{161/289}}[/tex]
[tex]\mathbf{tan(2x)= -\frac{240}{161}}\\[/tex]
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