Difference between revisions of "005 Sample Final A, Question 13"

Revision as of 15:59, 19 May 2015

Question Give the exact value of the following if its defined, otherwise, write undefined.
$(a)\sin ^{-1}(2)\qquad \qquad (b)\sin \left({\frac {-32\pi }{3}}\right)\qquad \qquad (c)\sec \left({\frac {-17\pi }{6}}\right)$

Step 1:
For (a), we want an angle $\theta$ such that $\sin(\theta )=2$ . Since $-1\leq \sin(\theta )\leq 1$ , it is impossible
for $\sin(\theta )=2$ . So, $\sin ^{-1}(2)$ is undefined.

Step 2:
For (b), we need to find the reference angle for ${\frac {-32\pi }{3}}$ . If we add multiples of $2\pi$ to this angle, we get the
reference angle ${\frac {4\pi }{3}}$ . So, $\sin \left({\frac {-32\pi }{3}}\right)=\sin \left({\frac {4\pi }{3}}\right)={\frac {-{\sqrt {3}}}{2}}$ .

Step 3:
For (c), we need to find the reference angle for ${\frac {-17\pi }{6}}$ . If we add multiples of $2\pi$ to this angle, we get the
reference angle ${\frac {7\pi }{6}}$ . Since $\cos \left({\frac {7\pi }{6}}\right)={\frac {-{\sqrt {3}}}{2}}$ , we have
$\sec \left({\frac {-17\pi }{6}}\right)=\sec \left({\frac {7\pi }{6}}\right)={\frac {2}{-{\sqrt {3}}}}={\frac {-2{\sqrt {3}}}{3}}$ .

Final Answer:
a) undefined
b) ${\frac {-{\sqrt {3}}}{2}}$
c) ${\frac {-2{\sqrt {3}}}{3}}$

@@ Line 4: / Line 4: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Step 1:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+| For (a), we want an angle <math>\theta</math> such that <math>\sin(\theta)=2</math>. Since <math>-1\leq \sin (\theta)\leq 1</math>, it is impossible
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+|for <math>\sin(\theta)=2</math>. So, <math>\sin^{-1}(2)</math> is undefined.
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 2:
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+| For (b), we need to find the reference angle for <math>\frac{-32\pi}{3}</math>. If we add multiples of <math>2\pi</math> to this angle, we get the
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+|reference angle <math>\frac{4\pi}{3}</math>. So, <math>\sin\left(\frac{-32\pi}{3}\right)=\sin\left(\frac{4\pi}{3}\right)=\frac{-\sqrt{3}}{2}</math>.
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 3:
 |-
-|e) True.
+| For (c), we need to find the reference angle for <math>\frac{-17\pi}{6}</math>. If we add multiples of <math>2\pi</math> to this angle, we get the
 |-
-|f) False.
+|reference angle <math>\frac{7\pi}{6}</math>. Since <math>\cos\left(\frac{7\pi}{6}\right)=\frac{-\sqrt{3}}{2}</math>, we have
+|-
+|<math>\sec\left(\frac{-17\pi}{6}\right)=\sec\left(\frac{7\pi}{6}\right)=\frac{2}{-\sqrt{3}}=\frac{-2\sqrt{3}}{3}</math>.
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Final Answer:
+|-
+|a) undefined
+|-
+|b) <math>\frac{-\sqrt{3}}{2}</math>
+|-
+|c)<math>\frac{-2\sqrt{3}}{3}</math>
 |}

Difference between revisions of "005 Sample Final A, Question 13"

Revision as of 15:59, 19 May 2015

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