Given $cos (x) = - \frac{3}{- 7}$ $cos(x) = -3/-7$ and $\frac{π}{2} < x < π$ $pi/2 < x < pi$ , how do you find $cos (\frac{x}{2})$ $cos(x/2)$ ?

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Given $cos (x) = - \frac{3}{- 7}$ $cos(x) = -3/-7$ and $\frac{π}{2} < x < π$ $pi/2 < x < pi$ , how do you find $cos (\frac{x}{2})$ $cos(x/2)$ ?

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Answer 1 · 2018-07-27T10:52:04

$x = 1.13$ $x=1.13$

Explanation:

$\frac{3}{7} = cos x$ $3/7=cosx$

$= cos (2 \times \frac{x}{2})$ $=cos(2timesx/2)$

$= {cos}^{2} (\frac{x}{2}) - {sin}^{2} (\frac{x}{2})$ $=cos^2(x/2)-sin^2(x/2)$

$= {cos}^{2} (\frac{x}{2}) - (1 - {cos}^{2} (\frac{x}{2}))$ $=cos^2(x/2)-(1-cos^2(x/2))$

$= 2 {cos}^{2} (\frac{x}{2}) - 1$ $=2cos^2(x/2)-1$

$\frac{3}{7} = 2 {cos}^{2} (\frac{x}{2}) - 1$ $3/7=2cos^2(x/2)-1$

$2 {cos}^{2} (\frac{x}{2}) = \frac{10}{7}$ $2cos^2(x/2)=10/7$

${cos}^{2} (\frac{x}{2}) = \frac{5}{7}$ $cos^2(x/2)=5/7$

$cos (\frac{x}{2}) = \pm \sqrt{\frac{5}{7}}$ $cos(x/2)=+-sqrt(5/7)$

The domain was originally $\frac{π}{2} < x < π$ $pi/2 < x < pi$ but since it is now $\frac{x}{2}$ $x/2$ , the domain has changed to $\frac{π}{4} < \frac{x}{2} < \frac{π}{2}$ $pi/4 < x/2 < pi/2$ ie first quadrant only

Therefore, $cos (\frac{x}{2}) = \sqrt{\frac{5}{7}}$ $cos(x/2)=sqrt(5/7)$ only as cosine is positive in the first and fourth quadrant and negative in the second and third quadrant

$\frac{x}{2} = 0.56$ $x/2=0.56$

$x = 1.13$ $x=1.13$

Answer 2 · 2018-07-27T12:57:04

$cos x = - \frac{3}{- 7} = \frac{3}{7} \Rightarrow x \in Q_{1}$ $cos x = -3 / ( - 7 ) = 3/7 rArr x in Q_1$ or $Q_{4}$ $Q_4$ .

So, $\frac{π}{2} < x < π$ $pi/2 < x < pi$ is incompatible.

If $cos x = - \frac{3}{7}, x \in Q_{2}$ $cos x = - 3/7, x in Q_2$ or $Q_{3} and \frac{x}{2} \in Q_{1}$ $Q_3 and x/2 in Q_1$ or Q_2#.

$cos ( x/2 ) = sqrt (1/2(1+ cos x )) = sqrt (2/7), if x in Q_2$ , and
$= - sqrt (2/7), x in Q_3$ .

Additional examples:

If $x = 2/3pi, cos x = -1/2$ and $cos (x/2) = cos (pi/3) = 1/2$

If $x = 4/3pi, cos x = -1/2$ and

$cos (x/2 ) = cos (2/3pi) = - 1/2$

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Given $cos (x) = - \frac{3}{- 7}$ $cos(x) = -3/-7$ and $\frac{π}{2} < x < π$ $pi/2 < x < pi$ , how do you find $cos (\frac{x}{2})$ $cos(x/2)$ ?

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Given cos(x) = -3/-7cos(x)=−3−7cos(x) = -3/-7 and pi/2 < x < piπ2<x<πpi/2 < x < pi, how do you find cos(x/2)cos(x2)cos(x/2)?

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Given $cos (x) = - \frac{3}{- 7}$ $cos(x) = -3/-7$ and $\frac{π}{2} < x < π$ $pi/2 < x < pi$ , how do you find $cos (\frac{x}{2})$ $cos(x/2)$ ?