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

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

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Answer 1 · 2016-09-20T11:32:18

$cos (\frac{θ}{2}) = - \sqrt{\frac{9}{10}}$ $cos(theta/2)=-sqrt(9/10)$

Explanation:

Using the identity $cos 2 A = 2 {cos}^{2} A - 1$ $cos2A=2cos^2A-1$ , we have

$cos θ = 2 {cos}^{2} (\frac{θ}{2}) - 1$ $costheta=2cos^2(theta/2)-1$ and hence $cos (\frac{θ}{2}) = \sqrt{\frac{1 + cos θ}{2}}$ $cos(theta/2)=sqrt((1+costheta)/2)$

As $cos θ = \frac{4}{5}$ $costheta=4/5$

$cos (\frac{θ}{2}) = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$ $cos(theta/2)=+-sqrt((1+4/5)/2)$

= $\pm \sqrt{\frac{\frac{9}{5}}{2}} = \pm \sqrt{\frac{9}{10}}$ $+-sqrt((9/5)/2)=+-sqrt(9/10)$

Now as $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2 < theta<2pi$ ,

$\frac{3 π}{4} < \frac{θ}{2} < π$ $(3pi)/4 < theta/2 < pi$ i.e. $\frac{θ}{2}$ $theta/2$ lies in second quadrant and

$cos (\frac{θ}{2}) = - \sqrt{\frac{9}{10}}$ $cos(theta/2)=-sqrt(9/10)$

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

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Explanation:

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If cosθ=45costheta=4/5 and 3π2<θ<2π(3pi)/2<theta<2pi, how do you find cos(θ2)cos (theta/2)?

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