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

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

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Answer 1 · 2015-06-12T16:57:48

$cos((7pi)/3)$ 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 · 2016-03-05T06:01:16

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, $cos (pi/3) = 1/2$

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

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