# 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.

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.

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.

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.

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:

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