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

Thus, tan 2theta:

`\begin{gathered} \tan 2\theta=(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-(9)/(6)} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-(3)/(6)} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-(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)