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

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{{iTest box|year=2006|num-b=8|num-a=10|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
  
 
[[Category:Introductory Trigonometry Problems]]
 
[[Category:Introductory Trigonometry Problems]]

Revision as of 19:56, 28 November 2018

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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