# Difference between revisions of "2006 iTest Problems/Problem 9"

## Problem

If $\sin(x) = -\frac{5}{13}$ and $x$ is in the third quadrant, what is the absolute value of $\cos(\frac{x}{2})$?

$\mathrm{(A)}\,\frac{\sqrt{3}}{3}\quad\mathrm{(B)}\,\frac{2\sqrt{3}}{3}\quad\mathrm{(C)}\,\frac{6}{13}\quad\mathrm{(D)}\,\frac{5}{13}\quad\mathrm{(E)}\,-\frac{5}{13} \\ \quad\mathrm{(F)}\,\frac{\sqrt{26}}{26}\quad\mathrm{(G)}\,-\frac{\sqrt{26}}{26}\quad\mathrm{(H)}\,\frac{\sqrt{2}}{2}\quad\mathrm{(I)}\,\text{none of the above}$

## Solution

Since $x$ is in the third quadrant, $\cos(x)$ is negative, so $\cos(x) = -\sqrt{1 - \tfrac{25}{169}} = -\tfrac{12}{13}$. Using the half-angle identity (or the double angle cosine identity), \begin{align*} |\cos(\frac{x}{2}) &= |\sqrt{\frac{\cos(x) + 1}{2}}| \\ &= \sqrt{\frac12 \cdot \frac{1}{13}} \\ &= \sqrt{\frac1{26}} \\ &= \frac{\sqrt{26}}{26} \end{align*} The answer is $\boxed{\textbf{(F)}}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 8 Followed by:Problem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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