Would someone mind explaining to me why I got this question wrong? Thank you so much for your help!

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Would someone mind explaining to me why I got this question wrong? Thank you so much for your help!

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Answer 1 · 2018-06-23T18:43:14

$- \frac{\sqrt{5}}{5}$ $-sqrt5/5$

Explanation:

$using the half angle identity for cos$ $"using the "color(blue)"half angle identity for cos"$

$∙ x cos (\frac{θ}{2}) = \pm \sqrt{\frac{1 + cos θ}{2}}$ $•color(white)(x)cos(theta/2)=+-sqrt((1+costheta)/2)$

$given π < θ < \frac{3 π}{2} then$ $"given "pi< theta <(3pi)/2" then"$

$\frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4} \leftarrow second quadrant$ $pi/2< theta/2< (3pi)/4larrcolor(blue)"second quadrant"$

$where cos (\frac{θ}{2}) < 0$ $"where "cos(theta/2)<0$

$cos (\frac{θ}{2}) = - \sqrt{\frac{1 - \frac{3}{5}}{2}}$ $cos(theta/2)=-sqrt((1-3/5)/2)$

$\times \times \times = - \sqrt{\frac{1}{5}}$ $color(white)(xxxxxx)=-sqrt(1/5)$

$\times \times \times = - \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = - \frac{\sqrt{5}}{5}$ $color(white)(xxxxxx)=-1/sqrt5xxsqrt5/sqrt5=-sqrt5/5$

Answer 2 · 2018-06-23T18:51:16

See explanation.

Explanation:

Think about it logically. $π < θ < \frac{3 π}{2}$ $pi < theta < (3pi)/2$ , so $θ$ $theta$ lies in the 3rd quadrant. To get to the 3rd quadrant, you take the basic acute angle and add $π$ $pi$ or $180^{\circ}$ $180^@$ to it. If you halve an angle between $π$ $pi$ and $\frac{3 π}{2}$ $(3pi)/2$ , it will always be greater than $\frac{π}{2}$ $pi/2$ because the original angle was greater than $π$ $pi$ to begin with. Now, realize that since $\frac{θ}{2}$ $theta/2$ is in the 2nd quadrant, $cos (\frac{θ}{2})$ $cos(theta/2)$ will always be negative because $cos θ$ $costheta$ is negative in the 2nd and 3rd quadrant. This is why the correct answer should be $- \frac{\sqrt{5}}{5}$ $-sqrt5/5$ .

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Would someone mind explaining to me why I got this question wrong? Thank you so much for your help!

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