Question

How do you use the unit circle to find the exact value for `cos ((7pi)/3)`?

Answer 1

`cos((7pi)/3)` is just `cos(2pi + pi/3)`. Since `cos(2pi) = cos 0`, `cos 2pi = 1`.

Go `pi/3` (`60^o`) past that, and you'll have `cos((7pi)/3) = cos(420^o) = cos(60^o)`. You'd go down two `30^o` steps from the value of 1, which becomes `1 -> sqrt3/2 -> 1/2`.

The pattern goes `1, sqrt3/2, 1/2, 0, -1/2, -sqrt3/2, -1, -sqrt3/2, -1/2, 0, 1/2, sqrt3/2, 1` at every `30^o`.

Or, you can use the additive identities of `cos`.

`\mathbf(cos(u + v) = cosucosv - sinusinv)`

Using `u + v = (7pi)/3`, we get:

`color(blue)(cos((7pi)/3))`

`cos((6pi)/3 + pi/3) = cos(2pi + pi/3)`

`= cos2picos(pi/3) - sin2pisin(pi/3)`

`= cos0cos(60^o) - sin0sin(60^o)`

`= 1*cos(60^o) - 0*sin(60^o)`

`= cos(60^o) = color(blue)(1/2)`

Answer 2

1/2

Explanation:

Another way.
cos ((7pi)/3) = cos (pi/3 + 2pi) = cos (pi/3), or cos 60^@
Call M the extremity of arc (pi/3), Call O the origin, and A the origin of all arcs. The triangle MAO is equilateral, since its 3 angles all equal to `60^@`.
`cos pi/3`, or cos 60^@, is equal to half of the radius OA = 1.
Therefor, `cos (pi/3) = 1/2`