Question

Evaluate sine, cosine, and tangent at the following value. Use the reference angle, θ′, and write your answer in exact form: 41π/6

Answer 1

Let us find the reference angle:

`\begin{gathered} (41)/(6)\pi\Rightarrow(41)/(6)\cdot(2)/(2)\pi=(41)/(12)\cdot2\pi=(3+(5)/(12))2\pi \\ (41)/(6)\pi\Rightarrow(5)/(6)\pi \end{gathered}`

Now, let us to calculate the sine and cosine of this angle:

`\begin{gathered} \sin ((5)/(6)\pi)=\sin (150\degree)=0.5 \\ \cos ((5)/(6)\pi)=\cos (150\degree)=\sqrt[]{1-\sin ^2((5)/(6)\pi)}=\sqrt[]{1-0.5^2}=\sqrt[]{(3)/(4)}=\frac{\sqrt[]{3}}{2} \\ \end{gathered}`

From this, we have:

`\begin{gathered} \sin ((41\pi)/(6))=(1)/(2) \\ \cos ((41\pi)/(6))=\frac{\sqrt[]{3}}{2} \end{gathered}`

And because:

`\tan (\theta)=(\sin (\theta))/(\cos (\theta))`

we can calculate:

`\tan ((41\pi)/(6))=\frac{(1)/(2)}{\frac{\sqrt[]{3}}{2}}=(1)/(2)\cdot\frac{2}{\sqrt[]{3}}=\frac{\sqrt[]{3}}{3}`

From the solution developed above, we are able to summarize the solution as:

`\begin{gathered} \sin ((41\pi)/(6))=(1)/(2) \\ \\ \cos ((41\pi)/(6))=\frac{\sqrt[]{3}}{2} \\ \\ \tan ((41\pi)/(6))=\frac{\sqrt[]{3}}{3} \end{gathered}`