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 12:33, 16 February 2021
Problem
If and is in the third quadrant, what is the absolute value of ?
Solution
Since is in the third quadrant, is negative, so . Using the half-angle identity (or the double angle cosine identity), The answer is .
See Also
2006 iTest (Problems) | ||
Preceded by: Problem 8 |
Followed by: Problem 10 | |
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