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

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Since <math>x</math> is in the third quadrant, <math>\cos(x)</math> is negative, so <math>\cos(x) = -\sqrt{1 - \tfrac{25}{169}} = -\tfrac{12}{13}</math>.  Using the half-angle identity (or the double angle cosine identity),
 
Since <math>x</math> is in the third quadrant, <math>\cos(x)</math> is negative, so <math>\cos(x) = -\sqrt{1 - \tfrac{25}{169}} = -\tfrac{12}{13}</math>.  Using the half-angle identity (or the double angle cosine identity),
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
|\cos(\frac{x}{2}) &= |\sqrt{\frac{\cos(x) + 1}{2}}| \\
+
|\cos(\frac{x}{2})| &= |\sqrt{\frac{\cos(x) + 1}{2}}| \\
 
&= \sqrt{\frac12 \cdot \frac{1}{13}} \\
 
&= \sqrt{\frac12 \cdot \frac{1}{13}} \\
 
&= \sqrt{\frac1{26}} \\
 
&= \sqrt{\frac1{26}} \\

Latest revision as of 11:33, 16 February 2021

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, Answer Key)
Preceded by:
Problem 8 		Followed by:
Problem 10
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