How would you find the exact value of the six trigonometric function of −π/3?

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Answer 1 · 2017-01-29T03:05:43

Use the identities:
$cos(x)=cos(-x)$
$sin(x)=+-sqrt(1-cos^2(x))$
$tan(x)=sin(x)/cos(x)$
$cot(x)=1/tan(x)$
$sec(x)=1/cos(x)$
$csc(x)=1/sin(x)$

It is well known that $cos(pi/3) = 1/2$

Using the identity, $cos(x) = cos(-x)$ , then:

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

Using the identity $sin(x) = +-sqrt(1-cos^2(x))$ :

$sin(-pi/3) = +-sqrt(1-cos^2(-pi/3))$

Substitute $(1/2)^2$ for $cos^2(-pi/3)$

$sin(-pi/3) = +-sqrt(1-(1/2)^2)$

$sin(-pi/3) = +-sqrt(3)/2$

We know that the sine function is negative in this quadrant, therefore, drop the +:

$sin(-pi/3) = -sqrt(3)/2$

Use the identity $tan(x) = sin(x)/cos(x)$

$tan(-pi/3) = (-sqrt(3)/2)/(1/2)$

$tan(-pi/3) = -sqrt(3)$

Use the identity $cot(x) = 1/tan(x)$ :

$cot(x) = 1/-sqrt(3) = -sqrt3/3$

Use the identity $sec(x) = 1/cos(x)$

$sec(-pi/3) = 1/cos(pi/3) = 1/(1/2) = 2$

Use the identity $csc(x) = 1/sin(x)$

$csc(-pi/3) = 1/sin(pi/3) = 1/(-sqrt(3)/2) = -2sqrt(3)/3$