So far, we dealt with right triangles and considered the trigonometric functions of an acute angle. In this section, we extend the definition of the trigonometric functions to any angles.

Unit Circle and Angles

To find trig functions of an arbitrary angle, it is convenient to use a unit circle. A unit circle is a circle with a radius of one, centered at the origin of the Cartesian plane.

An angle is said to be in standard position if its vertex is located at the origin and the initial side lies on the positive \(x-\)axis. If the rotation is counterclockwise, the angle has a positive measure. Respectively, if the rotation is clockwise, the angle is negative.

The \(x-\) and \(y-\) axes divide the coordinate plane into \(4\) quarters called quadrants. The unit circle has \(360^\circ.\) Hence, each quadrant is equal to \(90^\circ.\) Angles in the \(1\text{st}\) quadrant range from \(0^\circ\) to \(90^\circ.\) The \(2\text{nd}\) quadrant angles range from \(90^\circ\) to \(180^\circ,\) and so on.

Angles that have a measure multiple of \(90^\circ\) do not belong to a quadrant. Their terminal sides lie on the \(x-\) or \(y-\)axis. Such angles are called quadrantal.

The reference angle of an angle \(\alpha\) is a positive acute angle formed by the terminal side of \(\alpha\) and the \(x-\)axis. For example, if an angle \(\alpha\) lies in the \(1\text{st}\) quadrant, its reference angle is equal to \(\alpha\) itself. For an angle \(\alpha\) in the \(2\text{nd}\) quadrant, the reference angle is equal to \(180^\circ – \alpha.\)

Unit Circle Definitions of Sine and Cosine

Let \(\alpha\) be any angle in standard position in the unit circle. Suppose that the terminal side of the angle intersects the circle at a point \(M\left( {x,y} \right).\)

The sine of the angle \(\alpha\) is defined by

Similarly, the cosine function is defined on the coordinate plane by the formula

These expressions conform the definition of sine and cosine in a right triangle. Indeed, the triangle \(OAM\) in Figure \(1\) is a right triangle. In this triangle, the radius \(r = OM\) is the hypotenuse, and the \(y-\)and \(x-\)coordinates form, respectively, the opposite and adjacent legs.

Since we consider a unit circle, the distance \(r\) is equal to \(1\) for any point \(\left( {x,y} \right)\) of the circle:

\[r = \sqrt {{x^2} + {y^2}} = 1.\]

Therefore, the sine and cosine of an angle \(\alpha\) can be simply defined as the coordinates \(y\) and \(x\) of the point \(M\left( {x,y} \right)\) determined by the angle \(\alpha:\)

\[{\sin \alpha = y ,\;\;}\kern0pt{\cos \alpha = x.}\]

When the angle \(\alpha\) is measured in radians, we can consider any real values of \(\alpha.\) So, the unit circle definition of sine and cosine is more general than in a right triangle.

Unit Circle Definitions of Tangent and Cotangent

Consider again the unit circle and an angle \(\alpha\) in standard position. The angle \(\alpha\) determines the location of a point \(M\left( {x,y} \right)\) on the circle.

The tangent of \(\alpha\) on the coordinate plane is defined as

The tangent function is expressed in terms of sine and cosine in the form

\[{\tan \alpha = \frac{y}{x} }={ \frac{{\frac{y}{r}}}{{\frac{x}{r}}} }={ \frac{{\sin \alpha }}{{\cos \alpha }}.}\]

As you can see, the tangent function is defined at the points where \(cos \alpha = x = 0.\)

On the unit circle in Figure \(3,\) the value of the tangent function is displayed by the vertical segment line \(BK\) passing through the point \(B\left( {1,0} \right).\) Indeed, the triangles \(OAM\) and \(OBK\) are similar. Therefore,

\[\frac{{AM}}{{OA}} = \frac{{BK}}{{OB}}.\]

Here \(AM= OS = y = \sin \alpha,\) \(OA = x = \cos \alpha,\) \(OB = 1.\) Hence,

\[{BK = \frac{{AM}}{{OA}} \times OB }={ \frac{{\sin \alpha }}{{\cos \alpha }} \times 1 }={ \tan \alpha .}\]

The cotangent of angle \(\alpha\) is given by

We can also express the cotangent function as the ratio of cosine and sine:

\[{\cot \alpha = \frac{x}{y} }={ \frac{{\frac{x}{r}}}{{\frac{y}{r}}} }={ \frac{{\cos \alpha }}{{\sin \alpha }}.}\]

The cotangent function is not defined at the points where \(\sin \alpha = y = 0.\)

In Figure \(3\) above, the cotangent of \(\alpha\) is numerically equal to the length of the horizontal segment line \(PL.\) Since the triangles \(OSM\) and \(OPL\) are similar, we have

\[\frac{{SM}}{{OS}} = \frac{{PL}}{{OP}}.\]

Then

\[{PL = \frac{{SM}}{{OS}} \times OP }={ \frac{{\cos \alpha }}{{\sin \alpha }} \times 1 }={ \cot \alpha .}\]

Unit Circle Definitions of Secant and Cosecant

The secant is the reciprocal of cosine. Therefore it is defined by the formula

In the unit circle below, the secant function of \(\alpha\) is represented by the segment line \(OK.\) This follows from the similarity of triangles \(OAM\) and \(OBK.\) For these triangles,

\[\frac{{OM}}{{OA}} = \frac{{OK}}{{OB}}.\]

Hence,

\[{OK = \frac{{OM}}{{OA}} \times OB }={ \frac{1}{{\cos \alpha }} \times 1 }={ \sec \alpha .}\]

The cosecant is the reciprocal of sine. So we have

The value of the cosecant of \(\alpha\) in Figure \(4\) is equal to the length of the segment line \(OL.\) Indeed, since \(\triangle OSM \sim \triangle OPL,\) we can write

\[\frac{{OM}}{{OS}} = \frac{{OL}}{{OP}}.\]

Therefore,

\[{OL = \frac{{OM}}{{OS}} \times OP }={ \frac{1}{{\sin \alpha }} \times 1 }={ \csc \alpha .}\]

Trigonometric Functions of Special Angles

There are certain common angles that are more frequently used in trigonometry. The following table contains the values of trigonometric functions for such angles.

Solved Problems

Click or tap a problem to see the solution.

Solution.

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

\[{\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2},\;\;}\kern0pt{\tan \frac{\pi }{3} = \sqrt 3 .}\]

Hence

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

Solution.

Using the table above, we find that

\[{\cot \frac{\pi }{6} = \sqrt 3 ,\;\;}\kern0pt{\sec \frac{\pi }{3} = 2,\;\;}\kern0pt{\csc \frac{\pi }{4} = \sqrt 2 ,\;\;}\kern0pt{\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}.}\]

Substitute these values into our expression:

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

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 \(\alpha = \large{\frac{\pi }{6}}\normalsize\) 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 \(\alpha = \large{\frac{\pi }{6}}\normalsize\) 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.

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:

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

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

\[{\sec \frac{\pi }{4} = \frac{1}{{\cos \frac{\pi }{4}}} }={ \frac{1}{{\frac{{\sqrt 2 }}{2}}} }={ \frac{2}{{\sqrt 2 }} }={ \sqrt 2 ;}\]

\[{\csc \frac{\pi }{4} = \frac{1}{{\sin \frac{\pi }{4}}} }={ \frac{1}{{\frac{{\sqrt 2 }}{2}}} }={ \frac{2}{{\sqrt 2 }} }={ \sqrt 2 .}\]

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

\[{\sin \alpha = \frac{y}{r} }={ \frac{{ – 2}}{{\sqrt {13} }} }={ – \frac{{2\sqrt {13} }}{{13}};}\]

\[{\cos \alpha = \frac{x}{r} }={ \frac{{3}}{{\sqrt {13} }} }={\frac{{3\sqrt {13} }}{{13}};}\]

\[{\tan \alpha = \frac{y}{x} }={ \frac{3}{{ – 2}} }={ – \frac{3}{2};}\]

\[{\cot \alpha = \frac{x}{y} }={ \frac{-2}{{ 3}} }={ – \frac{2}{3};}\]

\[{\sec \alpha = \frac{r}{x} }={ \frac{{\sqrt {13} }}{{ – 2}} }={ – \frac{{\sqrt {13} }}{2};}\]

\[{\csc \alpha = \frac{r}{y} }={ \frac{{\sqrt {13} }}{{3}}. }\]

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:

\[{\sin \beta = \frac{y}{r} }={ \frac{{ – 6}}{{10}} }={ – \frac{3}{5};}\]

\[{\cos \beta = \frac{x}{r} }={ \frac{{ – 8}}{{10}} }={ – \frac{4}{5};}\]

\[{\tan \beta = \frac{y}{x} }={ \frac{{ – 6}}{{-8}} }={ \frac{3}{4};}\]

\[{\cot \beta = \frac{x}{y} }={ \frac{{ – 8}}{{- 6}} }={ \frac{4}{3};}\]

\[{\sec \beta = \frac{r}{x} }={ \frac{{ 10}}{{-8}} }={ – \frac{5}{4};}\]

\[{\csc \beta = \frac{r}{y} }={ \frac{{ 10}}{{-6}} }={ – \frac{5}{3}.}\]

Solution.

Since \(\sin \large{\frac{\pi }{6}}\normalsize = \large{\frac{1}{2}}\normalsize,\) 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 \(q = \large{\frac{1}{2}}\normalsize.\) The sum of the geometric series is given by

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

Solution.

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

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