Difference between revisions of "008A Sample Final A, Question 18"

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(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...")
 
 
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'''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>
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'''Question: ''' Compute <math>\cos(\arctan\frac{5}{3})</math>
  
 
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!Foundations
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!Foundations: &nbsp;
 
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|1) How is secant related to either sine or cosine?
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|1) Arctan can be thought of as referencing an angle in a triangle. What are the side lengths of the triangle?
 
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|2) What quadrant is each angle in? What is the reference angle for each?
 
 
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|Answer:
 
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|1) <math> sec(x) = \frac{1}{cos(x)}</math>
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|1) Since tangent is opposite/adjacent, the side lengths of the triangle are <math>3, 5\text{, and } \sqrt{34}</math>  
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|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>
 
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Solution:
 
Solution:
  
 
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! Final Answer A:
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! Step 1: &nbsp;
 
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|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>
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|<math>\arctan\left(\frac{5}{3}\right)</math> &nbsp; is the measure of an angle in the triangle with side lengths <math>3, 5\text{, and } \sqrt{34}</math>. The angle that corresponds to <math>\arctan\left(\frac{5}{3}\right)</math> is the one between the side of length 3 and the side of length &nbsp; <math> \sqrt{34}</math>
 
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! Final Answer B:
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!Step 2: &nbsp;
 
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|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>
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|Now we just have to take <math>\cos</math>&nbsp; of the angle referred to in step 1.
 
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! Final Answer C:
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!Final Answer: &nbsp;
 
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|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>
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|<math>\cos\left(\arctan\frac{5}{3}\right) = \frac{5}{\sqrt{34}}</math>
 
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[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]
 
[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

Latest revision as of 23:04, 25 May 2015

Question: Compute

Foundations:  
1) Arctan can be thought of as referencing an angle in a triangle. What are the side lengths of the triangle?
Answer:
1) Since tangent is opposite/adjacent, the side lengths of the triangle are

Solution:

Step 1:  
  is the measure of an angle in the triangle with side lengths . The angle that corresponds to is the one between the side of length 3 and the side of length  
Step 2:  
Now we just have to take   of the angle referred to in step 1.
Final Answer:  

Return to Sample Exam

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