Precalculus

Trigonometry

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Trigonometric Functions of a General Angle

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the value of the expression \[\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3}.\]

Example 2

Find the value of the expression \[\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6}.\]

Example 3

Calculate the values of the six trigonometric functions of α = π/6.

Example 4

Calculate the values of the six trigonometric functions of α = π/4.

Example 5

The terminal side of an angle \(\alpha\) in standard position contains the point \(P\left( { - 2,3} \right).\) Find the six trigonometric functions of the angle \(\alpha.\)

Example 6

The terminal side of an angle \(\beta\) in standard position contains the point \(Q\left( { - 8,-6} \right).\) Find the six trigonometric functions of the angle \(\beta.\)

Example 7

Calculate the sum of the series \[1 + \sin \frac{\pi }{6} + {\sin ^2}\frac{\pi }{6} + {\sin ^3}\frac{\pi }{6} + \ldots \]

Example 8

Calculate the sum of the series \[1 - \cos \frac{\pi }{4} + {\cos ^2}\frac{\pi }{4} - {\cos ^3}\frac{\pi }{4} + \ldots \]

Example 1.

Find the value of the expression \[\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3}.\]

Solution.

This expression contains trig functions of special angles. The values of these functions are given in the table above:

\[\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;\cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2},\;\;\tan \frac{\pi }{3} = \sqrt 3 .\]

Hence

\[\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3} = \frac{{\sqrt 2 }}{2} \times \frac{{\sqrt 3 }}{2} \times \sqrt 3 = \frac{{3\sqrt 2 }}{4}.\]

Example 2.

Find the value of the expression \[\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6}.\]

Solution.

Using the table above, we find that

\[\cot \frac{\pi }{6} = \sqrt 3 ,\;\;\sec \frac{\pi }{3} = 2,\;\;\csc \frac{\pi }{4} = \sqrt 2 ,\;\;\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}.\]

Substitute these values into our expression:

\[\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6} = \sqrt 3 \times 2 \times \sqrt 2 \times \frac{1}{{\sqrt 3 }} = 2\sqrt 2 .\]

Example 3.

Calculate the values of the six trigonometric functions of \(\alpha = \frac{\pi }{6}.\)

Solution.

Trigonometric functions of the angle of 30 degrees.
Figure 6.

The right triangle \(OAM\) is a special \(30\text{-}60\text{-}90\) triangle in which the hypotenuse is twice the length of the shorter leg. Therefore, we have

\[AM = \frac{{OM}}{2},\;\; \Rightarrow \sin \frac{\pi }{6} = \frac{r}{2} = \frac{1}{2}.\]

The cosine of \(\alpha = \frac{\pi }{6}\) can be found by the Pythagorean trig identity:

\[\cos \frac{\pi }{6} = \sqrt {1 - {{\sin }^2}\left( {\frac{\pi }{6}} \right)} = \sqrt {1 - \left( {\frac{1}{2}} \right)} = \sqrt {1 - \frac{1}{4}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}.\]

The other trigonometric functions of \(\alpha = \frac{\pi }{6}\) are given by

\[\tan \frac{\pi }{6} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{6}}} = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }};\]
\[\cot \frac{\pi }{6} = \frac{{\cos \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}}} = \sqrt 3 ;\]
\[\sec \frac{\pi }{6} = \frac{1}{{\cos \frac{\pi }{6}}} = \frac{1}{{\frac{{\sqrt 3 }}{2}}} = \frac{2}{{\sqrt 3 }};\]
\[\csc \frac{\pi }{6} = \frac{1}{{\sin \frac{\pi }{6}}} = \frac{1}{{\frac{1}{2}}} = 2.\]

The table above with the values of trig functions for special angles is composed on the basis of these calculations.

Example 4.

Calculate the values of the six trigonometric functions of \(\alpha = \frac{\pi }{4}.\)

Solution.

Trigonometric functions of the angle of 45 degrees.
Figure 7.

We deal here with a \(45\text{-}45\text{-}90\) triangle. This is a right isosceles triangle, so it has equal legs. Suppose the length of a leg be \(x.\) By the Pythagorean theorem,

\[{x^2} + {x^2} = {r^2},\;\; \Rightarrow {x^2} = \frac{{{r^2}}}{2},\;\; \Rightarrow x = \frac{r}{{\sqrt 2 }} = \frac{{r\sqrt 2 }}{2}.\]

Since \(r = 1,\) we have

\[x = \sin \frac{\pi }{4} = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}.\]

Compute the values of the other trig functions:

\[\require{cancel} \tan \frac{\pi }{4} = \frac{{\sin \frac{\pi }{4}}}{{\cos \frac{\pi }{4}}} = \frac{\cancel{\frac{{\sqrt 2 }}{2}}}{\cancel{\frac{{\sqrt 2 }}{2}}} = 1;\]
\[\cot \frac{\pi }{4} = \frac{{\cos \frac{\pi }{4}}}{{\sin \frac{\pi }{4}}} = \frac{\cancel{\frac{{\sqrt 2 }}{2}}}{\cancel{\frac{{\sqrt 2 }}{2}}} = 1;\]
\[\sec \frac{\pi }{4} = \frac{1}{{\cos \frac{\pi }{4}}} = \frac{1}{{\frac{{\sqrt 2 }}{2}}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 ;\]
\[\csc \frac{\pi }{4} = \frac{1}{{\sin \frac{\pi }{4}}} = \frac{1}{{\frac{{\sqrt 2 }}{2}}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 .\]

Example 5.

The terminal side of an angle \(\alpha\) in standard position contains the point \(P\left( { - 2,3} \right).\) Find the six trigonometric functions of the angle \(\alpha.\)

Solution.

Angle in standard position passing through the point P(-2,3).
Figure 8.

Here \(x = -2,\) \(y = 3.\) The distance of the point \(P\left( { - 2,3} \right)\) from the origin is equal to

\[r = \sqrt {{x^2} + {y^2}} = \sqrt {{{\left( { - 2} \right)}^2} + {3^2}} = \sqrt {4 + 9} = \sqrt {13}.\]

The trig functions of the angle \(\alpha\) are given by

\[\sin \alpha = \frac{y}{r} = \frac{{ - 2}}{{\sqrt {13} }} = - \frac{{2\sqrt {13} }}{{13}};\]
\[\cos \alpha = \frac{x}{r} = \frac{{3}}{{\sqrt {13} }} = \frac{{3\sqrt {13} }}{{13}};\]
\[\tan \alpha = \frac{y}{x} = \frac{3}{{ - 2}} = - \frac{3}{2};\]
\[\cot \alpha = \frac{x}{y} = \frac{-2}{{ 3}} = - \frac{2}{3};\]
\[\sec \alpha = \frac{r}{x} = \frac{{\sqrt {13} }}{{ - 2}} = - \frac{{\sqrt {13} }}{2};\]
\[\csc \alpha = \frac{r}{y} = \frac{{\sqrt {13} }}{{3}}.\]

Example 6.

The terminal side of an angle \(\beta\) in standard position contains the point \(Q\left( { - 8,-6} \right).\) Find the six trigonometric functions of the angle \(\beta.\)

Solution.

Angle in standard position passing through the point Q(-8,-6).
Figure 9.

Determine the distance \(r\) from the origin to the point \(Q\left( { - 8,-6} \right):\)

\[r = \sqrt {{x^2} + {y^2}} = \sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {64 + 36} = \sqrt {100} = 10.\]

Calculate the values of the trig functions:

\[\sin \beta = \frac{y}{r} = \frac{{ - 6}}{{10}} = - \frac{3}{5};\]
\[\cos \beta = \frac{x}{r} = \frac{{ - 8}}{{10}} = - \frac{4}{5};\]
\[\tan \beta = \frac{y}{x} = \frac{{ - 6}}{{-8}} = \frac{3}{4};\]
\[\cot \beta = \frac{x}{y} = \frac{{ - 8}}{{- 6}} = \frac{4}{3};\]
\[\sec \beta = \frac{r}{x} = \frac{{ 10}}{{-8}} = - \frac{5}{4};\]
\[\csc \beta = \frac{r}{y} = \frac{{ 10}}{{-6}} = - \frac{5}{3}.\]

Example 7.

Calculate the sum of the series \[1 + \sin \frac{\pi }{6} + {\sin ^2}\frac{\pi }{6} + {\sin ^3}\frac{\pi }{6} + \ldots \]

Solution.

Since \(\sin \frac{\pi }{6} = \frac{1}{2},\) we can write this expression in the form:

\[1 + \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^3} + \ldots \]

We have here an infinite geometric series with the initial term \(a_1 =1\) and the common ratio \(q = \frac{1}{2}.\) The sum of the geometric series is given by

\[S = \frac{{{a_1}}}{{1 - q}} = \frac{1}{{1 - \frac{1}{2}}} = \frac{1}{{\frac{1}{2}}} = 2.\]

Example 8.

Calculate the sum of the series \[1 - \cos \frac{\pi }{4} + {\cos ^2}\frac{\pi }{4} - {\cos ^3}\frac{\pi }{4} + \ldots \]

Solution.

We have here an infinite geometric series with the initial term \(a_1 = 1\) and the negative common ratio \(q = - \cos \frac{\pi }{4} = - \frac{{\sqrt 2 }}{2}.\) Determine the sum of the series:

\[S = \frac{{{a_1}}}{{1 - q}} = \frac{1}{{1 - \left( { - \frac{{\sqrt 2 }}{2}} \right)}} = \frac{1}{{1 + \frac{1}{{\sqrt 2 }}}} = \frac{1}{{\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}}} = \frac{{\sqrt 2 }}{{\sqrt 2 + 1}}.\]
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