Question

If cos(o) = 3/V24and angle o is in Quadrant I, what is the exact value of tan(2o) insimplest radical form?

Answer 1

In order to solve this question, we need to use the following properties of trigonometric functions:

`\begin{gathered} \tan \theta=(\sin\theta)/(\cos\theta) \\ \\ \sin 2\theta=2\sin ^{}\theta\cos \theta \\ \\ \cos 2\theta=\cos ^2\theta-\sin ^2\theta \\ \\ \sin ^2\theta=1-\cos ^2\theta\text{ }\Rightarrow\sin \theta=\sqrt[]{1-\cos ^2\theta} \end{gathered}`

Using those properties, we can write tan2θ in terms of cosθ, and then use the given information cosθ = 3/√24.

We have:

`\begin{gathered} \tan 2\theta=(\sin 2\theta)/(\cos 2\theta) \\ \\ =(2\sin \theta\cos \theta)/(\cos ^2\theta-\sin ^2\theta) \\ \\ =\frac{2\cos \theta\sqrt[]{1-\cos^2\theta}}{\cos ^2\theta-(1-\cos ^2\theta)} \\ \\ =\frac{2\cos \theta\sqrt[]{1-\cos^2\theta}}{2\cos ^2\theta-1} \\ \\ =\frac{2\cdot\frac{3}{\sqrt[]{24}}\sqrt[]{1-(9)/(24)}}{2\cdot(9)/(24)-1} \\ \\ =\frac{2\cdot\frac{3}{\sqrt[]{24}}\sqrt[]{(15)/(24)}}{(9)/(12)-1} \\ \\ =\frac{(6)/(24)\sqrt[]{15}}{-(3)/(12)} \\ \\ =(6)/(24)\cdot(-(12)/(3))\sqrt[]{15} \\ \\ =-\sqrt[]{15} \end{gathered}`

Therefore:

`\tan 2\theta=-\sqrt[]{15}`