https://gradwiki.math.ucr.edu/index.php?action=history&feed=atom&title=8A_F11_Q17
8A F11 Q17 - Revision history
2026-04-20T18:52:33Z
Revision history for this page on the wiki
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https://gradwiki.math.ucr.edu/index.php?title=8A_F11_Q17&diff=335&oldid=prev
Matt Lee at 23:01, 6 April 2015
2015-04-06T23:01:12Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 23:01, 6 April 2015</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|Sin(-120) = - sin(120). So you can either compute sin(120) or sin(-120) = sin(240). Since the reference angle is 60 degrees, or <math>\frac{\pi}{3}</math>, So <math> sin(-120) = \frac{\sqrt{3}}{2}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|Sin(-120) = - sin(120). So you can either compute sin(120) or sin(-120) = sin(240). Since the reference angle is 60 degrees, or <math>\frac{\pi}{3}</math>, So <math> sin(-120) = \frac{\sqrt{3}}{2}</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|}</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
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Matt Lee
https://gradwiki.math.ucr.edu/index.php?title=8A_F11_Q17&diff=254&oldid=prev
Matt Lee at 18:58, 29 March 2015
2015-03-29T18:58:12Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 18:58, 29 March 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|2) a) Quadrant 2, b) Quadrant 4, c) Quadrant 3. The reference angles are: <math> \frac{\pi}{4}, \frac{\pi}{6}</math>, and 60 degrees or <math>\frac{\pi}{3}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|2) a) Quadrant 2, b) Quadrant 4, c) Quadrant 3. The reference angles are: <math> \frac{\pi}{4}, \frac{\pi}{6}</math>, and 60 degrees or <math>\frac{\pi}{3}</math></div></td></tr>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">|}</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Solution:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Solution:</div></td></tr>
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Matt Lee
https://gradwiki.math.ucr.edu/index.php?title=8A_F11_Q17&diff=253&oldid=prev
Matt Lee: Created page with "'''Question: ''' Compute the following trig ratios: a) <math> \sec \frac{3\pi}{4}</math> b) <math> \tan \frac{11\pi}{6}</math> c) <m..."
2015-03-29T18:57:49Z
<p>Created page with "'''Question: ''' Compute the following trig ratios: a) <math> \sec \frac{3\pi}{4}</math> b) <math> \tan \frac{11\pi}{6}</math> c) <m..."</p>
<p><b>New page</b></p><div>'''Question: ''' Compute the following trig ratios: a) <math> \sec \frac{3\pi}{4}</math> &nbsp; &nbsp; &nbsp; b) <math> \tan \frac{11\pi}{6}</math> &nbsp; &nbsp; &nbsp; c) <math>\sin(-120)</math><br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Foundations<br />
|-<br />
|1) How is secant related to either sine or cosine?<br />
|-<br />
|2) What quadrant is each angle in? What is the reference angle for each?<br />
|Answer:<br />
|-<br />
|1) <math> sec(x) = \frac{1}{cos(x)}</math> <br />
|-<br />
|2) a) Quadrant 2, b) Quadrant 4, c) Quadrant 3. The reference angles are: <math> \frac{\pi}{4}, \frac{\pi}{6}</math>, and 60 degrees or <math>\frac{\pi}{3}</math><br />
<br />
Solution:<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
! Final Answer A:<br />
|-<br />
|Since <math> \sec(x) = \frac{1}{\cos(x)} </math>, and the angle is in quadrant 2, <math> \sec(\frac{3\pi}{4}) = \frac{1}{\cos(\frac{3\pi}{4})} = \frac{1}{\frac{-1}{\sqrt{2}}} = -\sqrt{2}</math><br />
|}<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
! Final Answer B:<br />
|-<br />
|The reference angle is <math> \frac{\pi}{6} </math> and is in the fourth quadrant. So tangent will be negative. Since the angle is 30 degees, using the 30-60-90 right triangle, we can conclude that <math>\tan(\frac{11\pi}{6}) = -\frac{\sqrt{3}}{3}</math><br />
|}<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
! Final Answer C:<br />
|-<br />
|Sin(-120) = - sin(120). So you can either compute sin(120) or sin(-120) = sin(240). Since the reference angle is 60 degrees, or <math>\frac{\pi}{3}</math>, So <math> sin(-120) = \frac{\sqrt{3}}{2}</math><br />
|}</div>
Matt Lee