Question

How do I do this trig question?

Answer 1

\[\cos (\frac{ -13\pi }{ 6 }) + \sin \frac{ 13\pi }{ 6 }\]

Answer 2

hmmm I'd guess you simply get their values and add them up :)

Answer 3

since it's not really an equation anyway

Answer 4

you can use
\[13\pi/6=2\pi +\pi/6\]
use then the fact that sin and cos are 2pi periodic

Answer 5

yah i guess too

Answer 6

\[\frac{ \sqrt{3} + 1 }{ 2 }?\]

Answer 7

you have cos(-x)=cos(x)
so you just get rid of - sign for cos (-13pi/6)

Answer 8

hmm not quite

Answer 9

what is sin pi/6

Answer 10

and cos pi/6

Answer 11

\(\bf cos \left( -\cfrac{ 13\pi }{ 6 } \right) + sin \left( \cfrac{ 13\pi }{ 6 } \right)
\\ \quad \\
cos(-\theta)=cos(\theta)\qquad thus
\\ \quad \\
cos \left( -\cfrac{ 13\pi }{ 6 } \right) + sin \left( \cfrac{ 13\pi }{ 6 } \right)\iff cos \left( \cfrac{ 13\pi }{ 6 } \right) + sin \left( \cfrac{ 13\pi }{ 6 } \right)\)

Answer 12

eh hold on i guess you are correct my bad

Answer 13

Haha lol. I was panicking for a moment there.

Answer 14

root 3+1 /2 is good

Answer 15

Thanks guys. :) I appreciate the help.

Answer 16

yw

Answer 17

yw