If $cos θ = - \frac{4}{5}$ $costheta=-4/5$ and $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you find $sin (\frac{θ}{2})$ $sin (theta/2)$ ?

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If $cos θ = - \frac{4}{5}$ $costheta=-4/5$ and $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you find $sin (\frac{θ}{2})$ $sin (theta/2)$ ?

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Answer 1 · 2016-11-21T13:55:11

$\frac{3 \sqrt{10}}{10}$ $(3sqrt10)/10$

Explanation:

Use the trig identity:
$1 - cos 2 a = 2 {sin}^{2} a$ $1 - cos 2a = 2sin^2 a$
Replacing a by $sin (\frac{t}{2})$ $sin (t/2)$ , we get:
$2 {sin}^{2} (\frac{t}{2}) = 1 - cos t = 1 + \frac{4}{5} = \frac{9}{5}$ $2sin^2 (t/2) = 1 - cos t = 1 + 4/5 = 9/5$
${sin}^{2} (\frac{t}{2}) = \frac{9}{10}$ $sin^2 (t/2) = 9/10$
$sin (\frac{t}{2}) = \pm \frac{3}{\sqrt{10}} = \pm \frac{3 \sqrt{10}}{10}$ $sin (t/2) = +- 3/sqrt10 = +- (3sqrt10)/ 10$
Since t is Quadrant III $(π < t < \frac{3 π}{2})$ $(pi < t < (3pi)/2)$ , then $\frac{t}{2}$ $t/2$ is in Quadrant II $(\frac{π}{2} < \frac{t}{2} < π)$ $(pi/2 < t/2 < pi)$ . There for, $sin (\frac{t}{2})$ $sin (t/2)$ is positive.
$sin t = \frac{3 \sqrt{10}}{10}$ $sin t = (3sqrt10)/10$

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If $cos θ = - \frac{4}{5}$ $costheta=-4/5$ and $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you find $sin (\frac{θ}{2})$ $sin (theta/2)$ ?

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If $cos θ = - \frac{4}{5}$ $costheta=-4/5$ and $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you find $sin (\frac{θ}{2})$ $sin (theta/2)$ ?