Question

If sin theta=3 divided by square root 15 and angle theta is in Quadrant I, what is the exact value of tan 2 theta insimplest radical form?

Answer 1

The given information is:

$\sin \theta=\frac{3}{\sqrt[]{15}}$

And angle theta is in quadrant I, then know that:

$\sin (x)=\frac{opposite}{\text{hypotenuse}}=\frac{y}{r}$

Then y=3 and r=square root (15)

By the Pythagorean theorem we can find x:

$\begin{gathered} r^2=x^2+y^2 \\ x^2=r^2-y^2 \\ x=\sqrt[]{r^2-y^2} \\ x=\sqrt[]{(\sqrt[]{15})^2-3^2} \\ x=\sqrt[]{15-9} \\ x=\sqrt[]{6} \end{gathered}$

And the tangent is:

$\begin{gathered} \tan (x)=\frac{y}{x} \\ \tan \theta=\frac{3}{\sqrt[]{6}} \end{gathered}$

Thus, tan 2theta:

$\begin{gathered} \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta} \\ \tan 2\theta=\frac{2\frac{3}{\sqrt[]{6}}}{1-(\frac{3}{\sqrt[]{6}})^2} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{1-\frac{9}{6}} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-\frac{3}{6}} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-\frac{1}{2}} \\ \tan 2\theta=\frac{6\cdot2}{-1\cdot\sqrt[]{6}} \\ \tan 2\theta=\frac{12}{-\sqrt[]{6}} \\ \tan 2\theta=-\frac{12}{\sqrt[]{6}} \end{gathered}$

Then the exact value of tan 2theta in simplest radical form is -12/square root(6)