Respuesta :
Given:
[tex]\sin x=-\dfrac{15}{17}[/tex]
x lies in the III quadrant.
To find:
The values of [tex]\sin 2x, \cos 2x, \tan 2x [/tex].
Solution:
It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.
We know that,
[tex]\sin^2 x+\cos^2 x=1[/tex]
[tex](-\dfrac{15}{17})^2+\cos^2 x=1[/tex]
[tex]\cos^2 x=1-\dfrac{225}{289}[/tex]
[tex]\cos x=\pm\sqrt{\dfrac{289-225}{289}}[/tex]
x lies in the III quadrant. So,
[tex]\cos x=-\sqrt{\dfrac{64}{289}}[/tex]
[tex]\cos x=-\dfrac{8}{17}[/tex]
Now,
[tex]\sin 2x=2\sin x\cos x[/tex]
[tex]\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})[/tex]
[tex]\sin 2x=-\dfrac{240}{289}[/tex]
We know that,
[tex]\cos 2x=1-2\sin^2x[/tex]
[tex]\cos 2x=1-2(-\dfrac{15}{17})^2[/tex]
[tex]\cos 2x=1-2(\dfrac{225}{289})[/tex]
[tex]\cos 2x=\dfrac{289-450}{289}[/tex]
[tex]\cos 2x=-\dfrac{161}{289}[/tex]
Using the trigonometric ratios, we get
[tex]\tan 2x=\dfrac{\sin 2x}{\cos 2x}[/tex]
[tex]\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}[/tex]
[tex]\tan 2x=\dfrac{240}{161}[/tex]
Hence, the required values are [tex]\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}[/tex].
