Question

cos(5pie/3)=
A. root 3/2
B. root 2/2
C. -root 2/2
D. 1/2

Answer 1

cos(5π/3) = cos(6π/3 - π/3) = cos(6π - π/3)

= cos(2π)cos(π/3) + sin(2π)sin(π/3)
= 1 • cos(π/3) + 0 • sin(2π/3)
= cos(π/3)

Answer 2

my cos says root 1/2

Answer 3

cos(5π/3) = cos(6π/3 - π/3) = cos(6π - π/3)

= cos(2π)cos(π/3) + sin(2π)sin(π/3)
= 1 • cos(π/3) + 0 • sin(π/3)
= cos(π/3)

Answer 4

so D?

Answer 5

hint:
\[\Large \begin{gathered}
\cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi + \frac{{2\pi }}{3}} \right) = \hfill \\
\hfill \\
= \cos \pi \cos \left( {\frac{{2\pi }}{3}} \right) - \sin \pi \sin \left( {\frac{{2\pi }}{3}} \right) = ... \hfill \\
\end{gathered} \]

Answer 6

is it D?

Answer 7

yes, D is right

Answer 8

hint:
\[\begin{gathered}
\cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi + \frac{{2\pi }}{3}} \right) = \hfill \\
\hfill \\
= \cos \pi \cos \left( {\frac{{2\pi }}{3}} \right) - \sin \pi \sin \left( {\frac{{2\pi }}{3}} \right) = ... \hfill \\
\end{gathered} \]

Answer 9

you can do it as (2π-π/3), it seems easier this way a bit.... (to me at least)

Answer 10

Thank you

Answer 11

yw

Answer 12

and I still had a typo, once agian

Answer 13

cos(5π/3) = cos(6π/3 - π/3) = cos(2π - π/3)

^
I said that was 6π (when it is 2π)

= cos(2π)cos(π/3) + sin(2π)sin(π/3)
= 1 • cos(π/3) + 0 • sin(2π/3)
= cos(π/3)

Answer 14

in any case, if you have questions ask, if not then good luck