find \\( \\sin ( 2 x ) , \\cos ( 2 x ) \\), and \\( \\tan ( 2 x ) \\) from the given information.\n\\…

find \\( \\sin ( 2 x ) , \\cos ( 2 x ) \\), and \\( \\tan ( 2 x ) \\) from the given information.\n\\ \n\\tan ( x ) = - \\frac { 12 } { 5 } , \\quad x \\text { in quadrant ii }\n\\\n\\ \n\\sin ( 2 x ) = - \\frac { 240 } { 289 }\n\\\n\\ \n\\cos ( 2 x ) = - \\frac { 161 } { 289 }\n\\\n\\ \n\\tan ( 2 x ) = \\frac { 240 } { 161 }\n\\\nresources

find \\( \\sin ( 2 x ) , \\cos ( 2 x ) \\), and \\( \\tan ( 2 x ) \\) from the given information.\n\\ \n\\tan ( x ) = - \\frac { 12 } { 5 } , \\quad x \\text { in quadrant ii }\n\\\n\\ \n\\sin ( 2 x ) = - \\frac { 240 } { 289 }\n\\\n\\ \n\\cos ( 2 x ) = - \\frac { 161 } { 289 }\n\\\n\\ \n\\tan ( 2 x ) = \\frac { 240 } { 161 }\n\\\nresources

Answer

Explanation:

Step1: Find $\sin x$ and $\cos x$

Since $\tan x =-\frac{12}{5}=\frac{\sin x}{\cos x}$, and $\sin^{2}x+\cos^{2}x = 1$. Also, $x$ is in Quadrant II, so $\sin x>0$ and $\cos x<0$. From $\tan x=\frac{\sin x}{\cos x}=-\frac{12}{5}$, we have $\sin x =-\frac{12}{5}\cos x$. Substitute into $\sin^{2}x+\cos^{2}x = 1$: $$\left(-\frac{12}{5}\cos x\right)^{2}+\cos^{2}x=1$$ $$\frac{144}{25}\cos^{2}x+\cos^{2}x = 1$$ $$\frac{144 + 25}{25}\cos^{2}x=1$$ $$\cos^{2}x=\frac{25}{169}$$ Since $\cos x<0$, $\cos x=-\frac{5}{13}$. Then $\sin x=-\frac{12}{5}\times\left(-\frac{5}{13}\right)=\frac{12}{13}$.

Step2: Use double - angle formulas

  • For $\sin(2x)$: The double - angle formula is $\sin(2x)=2\sin x\cos x$. Substitute $\sin x=\frac{12}{13}$ and $\cos x =-\frac{5}{13}$: $\sin(2x)=2\times\frac{12}{13}\times\left(-\frac{5}{13}\right)=-\frac{120}{169}$.
  • For $\cos(2x)$: Use the formula $\cos(2x)=\cos^{2}x-\sin^{2}x$. Substitute $\sin x=\frac{12}{13}$ and $\cos x =-\frac{5}{13}$: $\cos(2x)=\left(-\frac{5}{13}\right)^{2}-\left(\frac{12}{13}\right)^{2}=\frac{25 - 144}{169}=-\frac{119}{169}$.
  • For $\tan(2x)$: Use the formula $\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$. Substitute $\sin(2x)=-\frac{120}{169}$ and $\cos(2x)=-\frac{119}{169}$: $\tan(2x)=\frac{-\frac{120}{169}}{-\frac{119}{169}}=\frac{120}{119}$.

Answer:

$\sin(2x)=-\frac{120}{169}$, $\cos(2x)=-\frac{119}{169}$, $\tan(2x)=\frac{120}{119}$