Question

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

csc(x) = 8, tan(x) < 0

sin(2x) =

cos(2x) =

tan(2x) =

Explanation:

Step1: Determine the quadrant of \(x\)

Since \(\csc(x)=\frac{1}{\sin(x)} = 8>0\) and \(\tan(x)<0\), \(x\) is in the second quadrant.

Step2: Find \(\sin(x)\)

Given \(\csc(x) = 8\), then \(\sin(x)=\frac{1}{8}\)

Step3: Find \(\cos(x)\)

Using the identity \(\sin^{2}(x)+\cos^{2}(x) = 1\), we have \(\cos^{2}(x)=1-\sin^{2}(x)=1 - (\frac{1}{8})^{2}=1-\frac{1}{64}=\frac{63}{64}\).
Since \(x\) is in the second quadrant, \(\cos(x)=-\frac{3\sqrt{7}}{8}\)

Step4: Find \(\sin(2x)\)

Using the double - angle formula \(\sin(2x)=2\sin(x)\cos(x)\)
Substitute \(\sin(x)=\frac{1}{8}\) and \(\cos(x)=-\frac{3\sqrt{7}}{8}\)
\(\sin(2x)=2\times\frac{1}{8}\times(-\frac{3\sqrt{7}}{8})=-\frac{3\sqrt{7}}{32}\)

Step5: Find \(\cos(2x)\)

Using the double - angle formula \(\cos(2x)=1 - 2\sin^{2}(x)\)
Substitute \(\sin(x)=\frac{1}{8}\)
\(\cos(2x)=1-2\times(\frac{1}{8})^{2}=1-\frac{2}{64}=\frac{31}{32}\)

Step6: Find \(\tan(2x)\)

Using the formula \(\tan(2x)=\frac{\sin(2x)}{\cos(2x)}\)
Substitute \(\sin(2x)=-\frac{3\sqrt{7}}{32}\) and \(\cos(2x)=\frac{31}{32}\)
\(\tan(2x)=-\frac{3\sqrt{7}}{31}\)

Answer:

\(\sin(2x)=-\frac{3\sqrt{7}}{32}\)
\(\cos(2x)=\frac{31}{32}\)
\(\tan(2x)=-\frac{3\sqrt{7}}{31}\)