Signs of Trigonometric Functions
Solved Problems
Example 1.
Calculate \(\sin \alpha\) and \(\cot \alpha\) if \(\cos \alpha = \frac{5}{{13}}\) and the angle \(\alpha\) lies in the \(4\text{th}\) quadrant.
Solution.
The sine function is negative in quadrant \(IV.\) Using the Pythagorean trigonometric identity, we get
\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1,\;\; \Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha ,\;\; \Rightarrow \sin \alpha = - \sqrt {1 - {{\cos }^2}\alpha } = - \sqrt {1 - {{\left( {\frac{5}{{13}}} \right)}^2}} = - \sqrt {1 - \frac{{25}}{{169}}} = - \sqrt {\frac{{144}}{{169}}} = - \frac{{12}}{{13}}.\]
The cotangent function is given by
\[\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{5}{{13}}}}{{ - \frac{{12}}{{13}}}} = - \frac{5}{{12}}.\]
Example 2.
Calculate \(\cos \theta\) and \(\tan \theta\) if \(\sin \theta = -\frac{24}{{25}}\) and the angle \(\theta\) lies in the \(3\text{rd}\) quadrant.
Solution.
In the \(3\text{rd}\) quadrant, the cosine function is negative. Therefore,
\[{\sin ^2}\theta + {\cos ^2}\theta = 1,\;\; \Rightarrow \cos \theta = - \sqrt {1 - {{\sin }^2}\theta } = - \sqrt {1 - {{\left( { - \frac{{24}}{{25}}} \right)}^2}} = - \sqrt {1 - \frac{{576}}{{625}}} = - \sqrt {\frac{{49}}{{169}}} = - \frac{7}{{25}}.\]
Using the definition of tangent,we have
\[\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{ - \frac{{24}}{{25}}}}{{ - \frac{7}{{25}}}} = \frac{{24}}{7}.\]
Example 3.
Find \(\cos \alpha\) if \(\tan \alpha = 5\) and the angle \(\alpha\) lies in the \(3\text{rd}\) quadrant.
Solution.
We use the identity
\[{{\tan ^2}\alpha + 1 = {\sec ^2}\alpha }.\]
Hence,
\[{\sec ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} = {5^2} + 1 = 26,\;\; \Rightarrow {\cos ^2}\alpha = \frac{1}{{26}}.\]
The cosine has a negative sign in the \(3\text{rd}\) quadrant. So,
\[\cos \alpha = - \sqrt {{{\cos }^2}\alpha } = - \frac{1}{{\sqrt {26} }}.\]
Example 4.
Find \(\sin \beta\) if \(\cot \beta = -3\) and the angle \(\beta\) lies in the \(2\text{nd}\) quadrant.
Solution.
Using the trigonometric identity
\[{\cot ^2}\beta + 1 = {\csc ^2}\beta ,\]
we express the sine in terms of cotangent:
\[{\csc ^2}\beta = \frac{1}{{{{\sin }^2}\beta }} = {\left( { - 3} \right)^2} + 1 = 10,\;\; \Rightarrow {\sin ^2}\beta = \frac{1}{{10}}.\]
In the \(2\text{nd}\) quadrant, the sine is positive. Therefore,
\[\sin \beta = \sqrt {{{\sin }^2}\beta } = \frac{1}{{\sqrt {10} }}.\]
Example 5.
Find the values of the six trigonometric functions of \(\alpha = \frac{{2\pi }}{3}.\)
Solution.
The angle \(\alpha = \frac{{2\pi }}{3}\) lies in \(2\text{nd}\) quadrant.
Figure 6.
Determine the reference angle for \(\alpha = \frac{{2\pi }}{3}:\)
\[\alpha^\prime = \pi - \alpha = \pi - \frac{{2\pi }}{3} = \frac{\pi }{3}.\]
The reference angle \(\alpha^\prime\) is a special angle. We can easily find the values of the trig functions for it:
\[\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2},\;\;\cos \frac{\pi }{3} = \frac{1}{2},\;\;\tan \frac{\pi }{3} = \sqrt 3 ,\;\;\cot \frac{\pi }{3} = \frac{1}{{\sqrt 3 }},\;\;\sec \frac{\pi }{3} = 2,\;\;\csc \frac{\pi }{3} = \frac{2}{{\sqrt 3 }}.\]
In the \(2\text{nd}\) quadrant, the sine and cosecant are positive while cosine, secant, tangent, and cotangent are negative. Therefore,
\[\sin \frac{{2\pi }}{3} = \frac{{\sqrt 3 }}{2},\;\;\cos \frac{{2\pi }}{3} = - \frac{1}{2},\;\;\tan \frac{{2\pi }}{3} = - \sqrt 3 ,\;\;\cot \frac{{2\pi }}{3} = - \frac{1}{{\sqrt 3 }},\;\;\sec \frac{{2\pi }}{3} = - 2,\;\;\csc \frac{{2\pi }}{3} = \frac{2}{{\sqrt 3 }}.\]
Example 6.
Find the values of the six trigonometric functions of \(\beta = \frac{{5\pi }}{4}.\)
Solution.
The terminal side of the angle \(\beta = \frac{{5\pi }}{4}\) is in the \(3\text{rd}\) quadrant.
Figure 7.
The reference angle of \(\beta\) is equal to
\[\beta^\prime = \beta - \pi = \frac{{5\pi }}{4} - \pi = \frac{\pi }{4}.\]
Find the values of trig functions of \(\beta^\prime:\)
\[\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;\cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;\tan \frac{\pi }{4} = 1,\;\;\cot \frac{\pi }{4} = 1,\;\;\sec \frac{\pi }{4} = \sqrt 2 ,\;\;\csc \frac{\pi }{4} = \sqrt 2 .\]
In the \(3\text{rd}\) quadrant, the tangent and cotangent are positive while all other trigonometric functions are negative. Hence,
\[\sin \frac{{5\pi }}{4} = - \frac{{\sqrt 2 }}{2},\;\;\cos \frac{{5\pi }}{4} = - \frac{{\sqrt 2 }}{2},\;\;\tan \frac{{5\pi }}{4} = 1,\;\;\cot \frac{{5\pi }}{4} = 1,\;\;\sec \frac{{5\pi }}{4} = - \sqrt 2 ,\;\;\csc \frac{{5\pi }}{4} = - \sqrt 2 .\]
Example 7.
Determine the sign of the expression
\(\sin \frac{{13\pi }}{6}\cos \frac{{11\pi }}{7}\tan \frac{{9\pi }}{8}.\)
Solution.
The angle \(\frac{{13\pi }}{6}\) lies in the \(1\text{st}\) quadrant:
\[\frac{{12\pi }}{6} \lt \frac{{13\pi }}{6} \lt \frac{{15\pi }}{6},\;\; \Rightarrow 2\pi \lt \frac{{13\pi }}{6} \lt 2\pi + \frac{\pi }{2}.\]
Therefore \(\sin \frac{{13\pi }}{6}\) has a positive sign.
The angle \(\frac{{11\pi }}{7}\) is in the \(4\text{th}\) quadrant:
\[\frac{{21\pi }}{{14}} \lt \frac{{22\pi }}{{14}} \lt \frac{{28\pi }}{{14}},\;\; \Rightarrow \frac{{3\pi }}{2} \lt \frac{{22\pi }}{{14}} \lt 2\pi ,\;\; \Rightarrow \frac{{3\pi }}{2} \lt \frac{{11\pi }}{7} \lt 2\pi .\]
Hence, \(\cos \frac{{11\pi }}{7}\) has a positive sign.
The angle \(\frac{{9\pi }}{8}\) is obviously in the \(3\text{rd}\) quadrant, so \(\tan \frac{{9\pi }}{8}\) has a positive sign.
Since all three components are positive, their product is also positive:
\[\sin \frac{{13\pi }}{6}\cos \frac{{11\pi }}{7}\tan \frac{{9\pi }}{8} \gt 0.\]
Example 8.
Determine the sign of the expression
\(\tan \left( { - \frac{{2\pi }}{5}} \right)\cot \left( { - \frac{{5\pi }}{7}} \right)\) \(\sec \left( { - \frac{{8\pi }}{9}} \right).\)
Solution.
The angle \({ - \frac{{2\pi }}{5}}\) lies in the \(4\text{th}\) quadrant where tangent is negative. Indeed,
\[ - \frac{{5\pi }}{{10}} \lt - \frac{{4\pi }}{{10}} \lt 0,\;\; \Rightarrow - \frac{\pi }{2} \lt - \frac{{4\pi }}{{10}} \lt 0,\;\; \Rightarrow - \frac{\pi }{2} \lt - \frac{{2\pi }}{5} \lt 0.\]
The angle \({ - \frac{{5\pi }}{7}}\) is in the \(3\text{rd}\) quadrant where cotangent is positive:
\[- \frac{{14\pi }}{{14}} \lt - \frac{{10\pi }}{{14}} \lt - \frac{{7\pi }}{{14}},\;\; \Rightarrow - \pi \lt - \frac{{5\pi }}{7} \lt - \frac{\pi }{2}.\]
The angle \({ - \frac{{8\pi }}{9}}\) also belongs to the \(3\text{rd}\) quadrant where secant is negative:
\[ - \frac{{18\pi }}{{18}} \lt - \frac{{16\pi }}{{18}} \lt - \frac{{9\pi }}{{18}},\;\; \Rightarrow - \pi \lt - \frac{{8\pi }}{9} \lt - \frac{\pi }{2}.\]
So, \(2\) of the \(3\) factors are negative and one is positive. It is clear that their product is positive:
\[\tan \left( { - \frac{{2\pi }}{5}} \right)\cot \left( { - \frac{{5\pi }}{7}} \right)\sec \left( { - \frac{{8\pi }}{9}} \right) \gt 0.\]
