Question

I really appreciate the help!


How do I find sin(11pi/3)? I might have a pop quiz tomorrow and I really need help to figure this out. Thanks!

Answer 1

@ganeshie8 @Michele_Laino @zepdrix @Luigi0210 @Compassionate @Whitemonsterbunny17 @adrynicoleb @Xmoses1

Answer 2

\[\sin \frac{ 11 \times 3.14 }{ 3 }\]

Answer 3

I wish i would have paid attention in high school...

Answer 4

@Compassionate but how do I calculate it? I know the answer is sqrt(2)/2.

Answer 5

@Compassionate I think there is a graph method, but I don't get it.

Answer 6

Someone help please ;_;

Answer 7

hint:
\[\Large \frac{{11\pi }}{3} = 3\pi + \frac{{2\pi }}{3}\]
then we have to use the formula of addition for "sin" function

Answer 8

what is that?

Answer 9

Is it sinacosb+sinbcosa?

Answer 10

yes!
\[\Large \begin{gathered}
\sin \left( {\frac{{11\pi }}{3}} \right) = \sin \left( {3\pi + \frac{{2\pi }}{3}} \right) = \hfill \\
\hfill \\
= \sin \left( {3\pi } \right)\cos \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {3\pi } \right)\sin \left( {\frac{{2\pi }}{3}} \right) = ...? \hfill \\
\end{gathered} \]

Answer 11

I don't really understand where to go from there.

Answer 12

it is simple, we have these values:
\[\Large \begin{gathered}
\sin \left( {3\pi } \right) = 0 \hfill \\
\cos \left( {\frac{{2\pi }}{3}} \right) = - \frac{1}{2} \hfill \\
\cos \left( {3\pi } \right) = - 1 \hfill \\
\sin \left( {\frac{{2\pi }}{3}} \right) = \frac{{\sqrt 3 }}{2} \hfill \\
\end{gathered} \]
please substitute them into my formula above

Answer 13

-sqrt(3)/2

Answer 14

hint:
\[\Large \begin{gathered}
\sin \left( {\frac{{11\pi }}{3}} \right) = \sin \left( {3\pi + \frac{{2\pi }}{3}} \right) = \hfill \\
\hfill \\
= \sin \left( {3\pi } \right)\cos \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {3\pi } \right)\sin \left( {\frac{{2\pi }}{3}} \right) = \hfill \\
\hfill \\
= 0 \cdot \left( { - \frac{1}{2}} \right) + \left( { - 1} \right) \cdot \left( {\frac{{\sqrt 3 }}{2}} \right) = ...? \hfill \\
\end{gathered} \]

Answer 15

yup, thanks for the help!

Answer 16

:)

Answer 17

Hopefully I can ace the pop quiz if it is tomorrow! :D Now to sleep.

Answer 18

\[\sin(n\pi+\theta)=(-1)^n.\sin(\theta) \forall n \in \mathbb{Z}^+\mathbb{Z}^-\]