What is $sin (- 240^{\circ})$ $sin(-240^circ)$ ?

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What is $sin (- 240^{\circ})$ $sin(-240^circ)$ ?

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Answer 1 · 2016-12-30T12:53:46

$\frac{\sqrt{3}}{2}$ $sqrt3/2$

Explanation:

Use the unit circle and trig table -->
The co-terminal angle (reference angle) of $(- 240^{\circ})$ $(-240^@)$ is $(360^{\circ} - 240^{\circ} = 120^{\circ})$ $(360^@ - 240^@ = 120^@)$
$sin (- 240) = sin 120 = \frac{\sqrt{3}}{2}$ $sin (-240) = sin 120 = sqrt3/2$

Answer 2 · 2016-12-30T13:20:25

$sin (- 240^{\circ}) = \frac{\sqrt{3}}{2}$ $sin(-240^@)=sqrt(3)/2$
(see below for use of reference angle)

Explanation:

The reference angle is the angle between the angular arm and the X-axis when the angle is drawn in the standard position (i.e. with the base arm along the positive X-axis); as such any reference angle should be in $[0, \frac{π}{2}]$ $[0,pi/2]$

Negative angles are measured clockwise from the positive X-axis,
so for the angle $- 240^{\circ}$ $-240^@$ we have:
enter image source here
which gives us a reference angle of $60^{\circ}$ $60^@$ in Quadrant 2 where (according to CAST rules) the $sin$ $sin$ is positive.

$60^{\circ}$ $60^@$ is one of the standard angles with $sin (60^{\circ}) = \frac{\sqrt{3}}{2}$ $sin(60^@)=sqrt(3)/2$

Therefore $sin (- 240^{\circ}) = + sin (60^{\circ}) = \frac{\sqrt{3}}{2}$ $sin(-240^@)=+sin(60^@) =sqrt(3)/2$

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