Question

Find exact value of each trigonometric function
cos 7π/3

Answer 1

\[\cos \frac{ 7 π}{ 3 }\]

Answer 2

\[\text{ two hints } \\ \text{hint 1:} \frac{2\pi}{3}+\frac{5\pi}{3}=\frac{7\pi}{3} \\ \\ \text{ hint 2: } \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]

Answer 3

\[\frac{ \sqrt{3}}{ 2 }*2\]

Answer 4

how did you get that?

Answer 5

\[\cos(\frac{7\pi}{3})= \cos(\frac{2\pi}{3}+\frac{5\pi}{3}) \\ =\cos(\frac{2\pi}{3})\cos(\frac{5\pi}{3})-\sin(\frac{2\pi}{3}) \sin(\frac{5\pi}{3}) \\ =?\]
use unit circle to evaluate cos(2pi/3) and cos(5pi/3) and sin(2pi/3) and sin(5pi/3)

Answer 6

because 2π=6π/3
then i added π/3

Answer 7

lol I should have thought about subtracting 2pi
you're so brilliant

Answer 8

that does make things easier

Answer 9

\[\cos(\frac{7\pi}{3})=\cos(\frac{7\pi}{3}-2\pi)=\cos(\frac{\pi}{3})\]

Answer 10

and cos(pi/3) should actually be ...

Answer 11

look at the x-coordinate at pi/3

Answer 12

would it be \[\sqrt{3}\]

Answer 13

cos(pi/3)
find pi/3 on the unit circle and use the x-coordinate since we are dealing with cosine
cos(pi/3)=1/2

Answer 14

so cos(7pi/3)=1/2

Answer 15

\[\frac{ \sqrt{3} }{ 2 } *2=\sqrt{3}\]

Answer 16

I'm confused. Are you not wanting to evaluate cos(7pi/3) anymore?

Answer 17

cos function has period 2pi
so cos(x)=cos(x-2pi)=cos(x+2pi)=cos(x+2pi*n)
where n can be any integer

cos(7pi/3)=cos(7pi/3-2pi)=cos(pi/3)=1/2
we just look at the x-coordinate at 7pi/3

Answer 18

i think i got it now
thanx for your help

Answer 19

*we just look at the x-coordinate at pi/3

Answer 20

another example:
\[\cos(\frac{19\pi}{6})=\cos(\frac{19\pi}{6}-2\pi)=\cos(\frac{7\pi}{6})=\frac{-\sqrt{3}}{2}\]