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How do you use the half angle identity to find exact value of sin(-15)?

1 Answer

Sep 6, 2015

$sin (- 15^{\circ}) = - \frac{\sqrt{2 - \sqrt{3}}}{2} \approx - 0.25882$ $sin(-15^@) = -sqrt(2-sqrt(3))/2 ~~ -0.25882$

Explanation:

Half angle formula for $sin (θ)$ $sin(theta)$
$XXXX sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - cos (θ)}{2}}$ $color(white)("XXXX")sin(theta/2) = +-sqrt((1-cos(theta))/2)$

$XXXX sin (- 15^{\circ}) = - \sqrt{(1 - \frac{cos (- 30)}{2})}$ $color(white)("XXXX")sin(-15^@) = -sqrt((1-cos(-30)/2)$
$XXXXXXXX$ $color(white)("XXXXXXXX")$ ...we can rule out the positive version as extraneous since $sin$ $sin$ is negative in Q IV

$cos (- 30^{\circ}) = \frac{\sqrt{3}}{2}$ $cos(-30^@) = sqrt(3)/2$
enter image source here

$sin (- 15^{\circ}) = - \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = - \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin(-15^@) = -sqrt((1-sqrt(3)/2)/2) = - sqrt(2-sqrt(3))/2$

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