In Exercises useidsforvaluesfirst01 - useidsforvalueslast01, use the Reciprocal and Quotient Identities (Theorem recipquotidfull) along with the Pythagorean Identities (Theorem pythids), to find the value of the circular function requested below. (Find the exact value unless otherwise indicated.)
If \(\sin (\theta ) = \frac {\sqrt {5}}{5}\), find \(\csc (\theta )\).

\(\csc (\theta ) = \sqrt {5}\)
If \(\sec (\theta ) = - 4\), find \(\cos (\theta )\).

\(\cos (\theta ) = \answer {-\frac {1}{4}}\)

If \(\tan (t) = 3\), find \(\cot (t)\).

\(\cot (t) = \frac {1}{3}\)
If \(\theta \) is a Quadrant IV angle with \(\cos (\theta ) = \frac {5}{13}\), find \(\sin (\theta )\).

\(\sin (\theta ) = \answer {-\frac {12}{13}}\)

If \(\theta \) is a Quadrant III angle with \(\tan (\theta ) = 2\), find \(\sec (\theta )\).

\(\sec (\theta ) = -\sqrt {5}\).
If \(\frac {\pi }{2} < t < \pi \) with \(\cot (t) = -2\), find \(\csc (t)\).

\(\csc (t) = \answer {\sqrt {5}}\).

If \(\sec (\theta ) = 3\) and \(\sin (\theta ) < 0\), find \(\tan (\theta )\).

\(\tan (\theta ) = -2\sqrt {2}\).
If \(\sin (\theta ) = -\frac {2}{3}\) but \(\tan (\theta ) > 0\), find \(\cos (\theta )\).

\(\cos (\theta ) = \answer {-\frac {\sqrt {5}}{3}}\).

If \(0 < t < \frac {\pi }{2}\) and \(\sin (t) = 0.42\), find \(\cos (t)\), rounded to four decimal places.

\(\cos (t) \approx 0.9075\).
If \(\theta \) is Quadrant IV angle with \(\sec (\theta ) = 1.17\), find \(\tan (\theta )\), rounded to four decimal places.

\(\tan (\theta ) \approx \answer {- 0.6074}\).

If \(\pi < t < \frac {3\pi }{2}\) with \(\cot (t) = 4.2\), find \(\csc (t)\), rounded to four decimal places.

\(\csc (t) \approx -4.079\).
In Exercises useidsforvaluesfirst02 - useidsforvalueslast02, use the Reciprocal and Quotient Identities (Theorem recipquotidfull) along with the Pythagorean Identities (Theorem pythids), to find the exact values of the remaining circular functions. (Compare your methods with how you solved Exercises findothercircfirst - findothercirclast in Section TheOtherCircularFunctions.)
\(\sin (\theta ) = \frac {3}{5}\) with \(\theta \) in Quadrant II

\(\cos (\theta ) = \)

\(\frac {4}{5}\) \(-\frac {4}{5}\) \(\frac {3}{4}\) \(-\frac {3}{4}\)

\(\tan (\theta ) = \)

\(\frac {4}{5}\) \(-\frac {4}{5}\) \(\frac {3}{4}\) \(-\frac {3}{4}\)

\(\csc (\theta ) = \)

\(\frac {5}{3}\) \(\frac {5}{4}\) \(-\frac {5}{4}\) \(-\frac {4}{3}\)

\(\sec (\theta ) = \)

\(\frac {5}{3}\) \(\frac {5}{4}\) \(-\frac {5}{4}\) \(-\frac {4}{3}\)

\(\cot (\theta ) = \)

\(\frac {5}{3}\) \(\frac {5}{4}\) \(-\frac {5}{4}\) \(-\frac {4}{3}\)
\(\tan (\theta ) = \frac {12}{5}\) with \(\theta \) in Quadrant III

\(\sin (\theta ) = -\frac {12}{13}, \cos (\theta ) = -\frac {5}{13}, \csc (\theta ) = -\frac {13}{12}, \sec (\theta ) = -\frac {13}{5}, \cot (\theta ) = \frac {5}{12}\)
\(\csc (\theta ) = \frac {25}{24}\) with \(\theta \) in Quadrant I

\(\sin (\theta ) = \answer {\frac {24}{25}}\)

\(\cos (\theta ) = \answer {\frac {7}{25}}\)

\(\tan (\theta ) = \answer {\frac {24}{7}}\)

\(\sec (\theta ) = \answer {\frac {25}{7}}\)

\(\cot (\theta ) = \answer {\frac {7}{24}}\)

\(\sec (\theta ) = 7\) with \(\theta \) in Quadrant IV

\(\sin (\theta ) = \frac {-4\sqrt {3}}{7}, \cos (\theta ) = \frac {1}{7}, \tan (\theta ) = -4\sqrt {3}, \csc (\theta ) = -\frac {7\sqrt {3}}{12}, \sec (\theta ) = 7, \cot (\theta ) = -\frac {\sqrt {3}}{12}\)
\(\csc (\theta ) = -\frac {10\sqrt {91}}{91}\) with \(\theta \) in Quadrant III

\(\sin (\theta ) = \answer {-\frac {\sqrt {91}}{10}}\)

\(\cos (\theta ) = \answer {-\frac {3}{10}}\)

\(\tan (\theta ) = \answer {\frac {\sqrt {91}}{3}}\)

\(\csc (\theta ) = \answer {-\frac {10\sqrt {91}}{91}}\)

\(\sec (\theta ) = \answer {-\frac {10}{3}}\)

\(\cot (\theta ) = \answer {\frac {3\sqrt {91}}{91}}\)

\(\cot (\theta ) = -23\) with \(\theta \) in Quadrant II

\(\sin (\theta ) = \frac {\sqrt {530}}{530}, \cos (\theta ) = -\frac {23\sqrt {530}}{530}, \tan (\theta ) = -\frac {1}{23}, \csc (\theta ) = \sqrt {530}, \sec (\theta ) = -\frac {\sqrt {530}}{23}, \cot (\theta ) = -23\)
\(\tan (\theta ) = -2\) with \(\theta \) in Quadrant IV.

\(\sin (\theta ) = \answer {-\frac {2\sqrt {5}}{5}}\)

\(\cos (\theta ) = \answer {\frac {\sqrt {5}}{5}}\)

\(\tan (\theta ) = \answer {-2}\)

\(\csc (\theta ) = \answer {-\frac {\sqrt {5}}{2}}\)

\(\sec (\theta ) = \answer {\sqrt {5}}\)

\(\cot (\theta ) = \answer {-\frac {1}{2}}\)

\(\sec (\theta ) = -4\) with \(\theta \) in Quadrant II.

\(\sin (\theta ) = \frac {\sqrt {15}}{4}, \cos (\theta ) = -\frac {1}{4}, \tan (\theta ) = -\sqrt {15}, \csc (\theta ) = \frac {4\sqrt {15}}{15}, \sec (\theta ) = -4, \cot (\theta ) = -\frac {\sqrt {15}}{15}\)
\(\cot (\theta ) = \sqrt {5}\) with \(\theta \) in Quadrant III.

\(\sin (\theta ) = \answer {-\frac {\sqrt {6}}{6}}\)

\(\cos (\theta ) = \answer {-\frac {\sqrt {30}}{6}}\)

\(\tan (\theta ) = \answer {\frac {\sqrt {5}}{5}}\)

\(\csc (\theta ) = \answer {-\sqrt {6}}\)

\(\sec (\theta ) = \answer {-\frac {\sqrt {30}}{5}}\)

\(\cot (\theta ) = \answer {\sqrt {5}}\)

\(\cos (\theta ) = \frac {1}{3}\) with \(\theta \) in Quadrant I.

\(\sin (\theta ) = \frac {2\sqrt {2}}{3}, \cos (\theta ) = \frac {1}{3}, \tan (\theta ) = 2\sqrt {2}, \csc (\theta ) = \frac {3\sqrt {2}}{4}, \sec (\theta ) = 3, \cot (\theta ) = \frac {\sqrt {2}}{4}\)
\(\cot (t) = 2\) with \(0 < t < \frac {\pi }{2}\).

\(\sin (t) = \answer {\frac {\sqrt {5}}{5}}\)

\(\cos (t) = \answer {\frac {2\sqrt {5}}{5}}\)

\(\tan (t) = \answer {\frac {1}{2}}\)

\(\csc (t) = \answer {\sqrt {5}}\)

\(\sec (t) = \answer {\frac {\sqrt {5}}{2}}\)

\(\cot (t) = \answer {2}\)

\(\csc (t) = 5\) with \(\frac {\pi }{2} < t < \pi \).

\(\sin (t) = \frac {1}{5}, \cos (t) = -\frac {2\sqrt {6}}{5}, \tan (t) = -\frac {\sqrt {6}}{12}, \csc (t) = 5, \sec (t) = -\frac {5\sqrt {6}}{12}, \cot (t) = -2\sqrt {6}\)
\(\tan (t) = \sqrt {10}\) with \(\pi < t < \frac {3\pi }{2}\).

\(\sin (t) = \answer {-\frac {\sqrt {110}}{11}}\)

\(\cos (t) = \answer {-\frac {\sqrt {11}}{11}}\)

\(\tan (t) = \answer {\sqrt {10}}\)

\(\csc (t) = \answer {-\frac {\sqrt {110}}{10}}\)

\(\sec (t) = \answer {-\sqrt {11}}\)

\(\cot (t) = \answer {\frac {\sqrt {10}}{10}}\)

\(\sec (t) = 2\sqrt {5}\) with \(\frac {3\pi }{2} < t < 2\pi \).

\(\sin (t) = -\frac {\sqrt {95}}{10}, \cos (t) = \frac {\sqrt {5}}{10}, \tan (t) = -\sqrt {19}, \csc (t) = -\frac {2\sqrt {95}}{19}, \sec (t) = 2\sqrt {5}, \cot (t) = -\frac {\sqrt {19}}{19}\)
Skippy claims \(\cos (\theta ) + \sin (\theta ) = 1\) is an identity because when \(\theta = 0\), the equation is true. Is Skippy correct? Explain.

No, Skippy is not correct. In order to be an identity, an equation must hold for all applicable angles. For example, \(\cos (\theta ) + \sin (\theta ) = 1\) does not hold when \(\theta = \pi \).
In Exercises firstcirciden - lastcirciden, verify the identity. Assume that all quantities are defined.
\(\cos (\theta ) \sec (\theta ) = 1\)
\(\tan (t)\cos (t) = \sin (t)\)
\(\sin (\theta ) \csc (\theta ) = 1\)
\(\tan (t) \cot (t) = 1\)
\(\csc (x) \cos (x) = \cot (x)\)
\(\frac {\sin (t)}{\cos ^{2}(t)} = \sec (t) \tan (t)\)
\(\frac {\cos (\theta )}{\sin ^{2}(\theta )} = \csc (\theta ) \cot (\theta )\)
\(\frac {1+ \sin (x)}{\cos (x)} = \sec (x) + \tan (x)\)
\(\frac {1 - \cos (\theta )}{\sin (\theta )} = \csc (\theta ) - \cot (\theta )\)
\(\frac {\cos (t)}{1 - \sin ^{2}(t)} = \sec (t)\)
\(\frac {\sin (x)}{1 - \cos ^{2}(x)} = \csc (x)\)
\(\frac {\sec (t)}{1 + \tan ^{2}(t)} = \cos (t)\)
\(\frac {\csc (\theta )}{1 + \cot ^{2}(\theta )} = \sin (\theta )\)
\(\frac {\tan (x)}{\sec ^{2}(x) - 1} = \cot (x)\)
\(\frac {\cot (t)}{\csc ^{2}(t) - 1} = \tan (t)\)
\(4 \cos ^{2}(\theta ) + 4 \sin ^{2}(\theta ) = 4\)
\(9 - \cos ^{2}(t) - \sin ^{2}(t) = 8\)
\(\tan ^{3}(t) = \tan (t)\sec ^{2}(t) - \tan (t)\)
\(\sin ^{5}(x) = \left (1-\cos ^{2}(x)\right )^{2} \sin (x)\)
\(\sec ^{10}(t) = \left (1 + \tan ^{2}(t)\right )^4 \sec ^{2}(t)\)
\(\cos ^{2}(x)\tan ^{3}(x) = \tan (x) - \sin (x)\cos (x)\)
\(\sec ^{4}(t) - \sec ^{2}(t) = \tan ^{2}(t) + \tan ^{4}(t)\)
\(\frac {\cos (\theta ) + 1}{\cos (\theta ) - 1} = \frac {1 + \sec (\theta )}{1 - \sec (\theta )}\)
\(\frac {\sin (t) + 1}{\sin (t) - 1} = \frac {1 + \csc (t)}{1 - \csc (t)}\)
\(\frac {1 - \cot (x)}{1+ \cot (x)} = \frac {\tan (x) - 1}{\tan (x) + 1}\)
\(\frac {1 - \tan (t)}{1+ \tan (t)} = \frac {\cos (t) - \sin (t)}{\cos (t) + \sin (t)}\)
\(\tan (\theta ) + \cot (\theta ) = \sec (\theta )\csc (\theta )\)
\(\csc (t) - \sin (t) = \cot (t)\cos (t)\)
\(\cos (x) - \sec (x) = -\tan (x)\sin (x)\)
\(\cos (x)(\tan (x) + \cot (x)) = \csc (x)\)
\(\sin (t)(\tan (t) + \cot (t)) = \sec (t)\)
\(\frac {1}{1-\cos (\theta )} + \frac {1}{1+\cos (\theta )} = 2\csc ^{2}(\theta )\)
\(\frac {1}{\sec (t) + 1} + \frac {1}{\sec (t)-1} = 2 \csc (t) \cot (t)\)
\(\frac {1}{\csc (x) + 1} + \frac {1}{\csc (x)-1} = 2 \sec (x) \tan (x)\)
\(\frac {1}{\csc (t)-\cot (t)} - \frac {1}{\csc (t) + \cot (t)} = 2 \cot (t)\)
\(\frac {\cos (\theta )}{1 - \tan (\theta )} + \frac {\sin (\theta )}{1 - \cot (\theta )} = \sin (\theta ) + \cos (\theta )\)
\(\frac {1}{\sec (t) + \tan (t)} = \sec (t) - \tan (t)\)
\(\frac {1}{\sec (x) - \tan (x)} = \sec (x) + \tan (x)\)
\(\frac {1}{\csc (t) - \cot (t)} = \csc (t) + \cot (t)\)
\(\frac {1}{\csc (\theta ) + \cot (\theta )} = \csc (\theta ) - \cot (\theta )\)
\(\frac {1}{1-\sin (x)} = \sec ^{2}(x) + \sec (x) \tan (x)\)
\(\frac {1}{1+\sin (t)} = \sec ^{2}(t) - \sec (t) \tan (t)\)
\(\frac {1}{1-\cos (\theta )} = \csc ^{2}(\theta ) + \csc (\theta ) \cot (\theta )\)
\(\frac {1}{1+\cos (x)} = \csc ^{2}(x) - \csc (x) \cot (x)\)
\(\frac {\cos (t)}{1 + \sin (t)} = \frac {1-\sin (t)}{\cos (t)}\)
\(\csc (\theta ) - \cot (\theta ) = \frac {\sin (\theta )}{1 + \cos (\theta )}\)
\(\frac {1 - \sin (x)}{1 + \sin (x)} = (\sec (x) - \tan (x))^{2}\)
In Exercises logcircidenfirst - logcircidenlast, verify the identity. You may need to consult Sections AbsoluteValueFunctions and PropertiesofLogarithms for a review of the properties of absolute value and logarithms before proceeding.
\(\quad \ln |\sec (x)| = -\ln |\cos (x)|\)
\(-\ln |\csc (x)| = \ln |\sin (x)|\)
\(-\ln |\sec (x) - \tan (x)| = \ln |\sec (x)+\tan (x)|\)
\(-\ln |\csc (x) + \cot (x)|= \ln |\csc (x) - \cot (x)|\)

  1. What indeterminate form is present in the limit

    \[\lim _{\theta \rightarrow 0} \frac {1 - \cos (\theta )}{\theta }?\]

    As \(\theta \rightarrow 0\), \(\frac {1 - \cos (\theta )}{\theta } \rightarrow \frac {0}{0}\).

  2. Graph \(f(\theta ) = \frac {1 - \cos (\theta )}{\theta }\) near \(\theta = 0\). What appears to be

    \[\lim _{\theta \rightarrow 0} \frac {1 - \cos (\theta )}{\theta }?\]

    The graph of \(f(\theta ) = \frac {1 - \cos (\theta )}{\theta }\) approaches \((0,0)\), so \(\ds {\lim _{\theta \rightarrow 0}}\) \(\frac {1 - \cos (\theta )}{\theta }\) appears to be \(0\).
  3. Verify the identity: \(\frac {1 - \cos (\theta )}{\theta } = \frac {\sin (\theta )}{1 + \cos (\theta )} \, \frac {\sin (\theta )}{\theta }\).

  4. Use the fact that

    \[\lim _{\theta \rightarrow 0} \frac {\sin (\theta )}{\theta } = 1\]
    along with part oneminuscosinerewrite to help you find \(\lim _{\theta \rightarrow 0} \frac {1 - \cos (\theta )}{\theta }\).

    \(\ds {\lim _{\theta \rightarrow 0}}\) \(\frac {1 - \cos (\theta )}{\theta }\) \(= \ds {\lim _{\theta \rightarrow 0}}\) \(\frac {\sin (\theta )}{1 + \cos (\theta )} \, \frac {\sin (\theta )}{\theta } = \left (\frac {0}{1+\cos (0)}\right )(1) = \left (\frac {0}{2}\right )(1) = 0\).