Precalculus

Trigonometry

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Pythagorean Trigonometric Identities

A trigonometric identity in one variable is an equality that involves trigonometric functions and is true for all values of the variable for which both sides of the equality are defined.

Recall the Pythagorean theorem that relates the lengths of the sides of a right triangle:

\[{a^2} + {b^2} = {c^2},\]

where \(a,b\) are the lengths of the triangle’s legs and \(c\) is the length of its hypotenuse.

A right triangle illustrating the Pythagorean theorem
Figure 1.

The sine and cosine functions of an acute angle \(\alpha\) in the right triangle are expressed in terms of the sides \(a,b, \text{and } c:\)

\[{\sin \alpha = \frac{a}{c},\;\;}\kern0pt{\cos \alpha = \frac{b}{c}.}\]

By squaring and adding these equations, we get the famous relationship

\[{{\sin ^2}\alpha + {\cos ^2}\alpha }={ \frac{{{a^2}}}{{{c^2}}} + \frac{{{b^2}}}{{{c^2}}} }={ \frac{{{a^2} + {b^2}}}{{{c^2}}} }={ \frac{{{c^2}}}{{{c^2}}} = 1,}\]

which is called the Pythagorean trigonometric identity. Thus, for any angle \(\alpha\) in the range \(0 \le \alpha \le \large{\frac{\pi}{2}}\normalsize,\) we have

Pythagorean trigonometric identity

Later we can see that this identity is valid for any real value of \(\alpha.\)

Assuming \(\cos \alpha \ne 0\) (so that \(\alpha \ne \large{\frac{\pi }{2}}\normalsize\)), we can divide both sides of the equation by \({\cos^2}\alpha.\) This yields:

\[\require{cancel}{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{\cancel{{{\cos }^2}\alpha }}{\cancel{{{\cos }^2}\alpha }} = \frac{1}{{{{\cos }^2}\alpha }},}\]

or

Pythagorean trig identity involving tangent and secant.

This identity relates the tangent and secant functions of an angle.

Similarly, if \(\sin \alpha \ne 0,\) that is \(\alpha \ne 0,\) we can divide the initial Pythagorean trigonometric identity \({\sin^2}\alpha + {\cos^2}\alpha = 1\) by \({\sin^2}\alpha\) to obtain an equation relating the cotangent and cosecant functions:

\[\frac{\cancel{{{\sin }^2}\alpha }}{\cancel{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{1}{{{{\sin }^2}\alpha }},\]

or

Pythagorean trig identity involving cotangent and cosecant.

The last two formulas are also called Pythagorean trigonometric identities.

These identities allow us to convert between the trigonometric functions of an angle, without knowing the angle itself, and simplify trig expressions.


Solved Problems

Click or tap a problem to see the solution.

Example 1

Suppose \(\sin \alpha = \large{\frac{4}{5}}\normalsize,\) where \(0 \lt \alpha \lt \large{\frac{\pi }{2}}\normalsize.\) Find the \(5\) other trigonometric functions.

Example 2

Suppose \(\sec \alpha = \large{\frac{25}{7}}\normalsize,\) where \(0 \lt \alpha \lt \large{\frac{\pi }{2}}\normalsize.\) Find the \(5\) other trigonometric functions.

Example 3

Simplify the expression \[\tan \alpha + \frac{{\cos \alpha }}{{1 + \sin \alpha }}.\]

Example 4

Simplify the expression \[{\tan ^2}\alpha – {\sin ^2}\alpha – {\tan ^2}\alpha \,{\sin ^2}\alpha .\]

Example 5

Prove the identity \[\sec \alpha – \sin \alpha \tan \alpha = \cos \alpha .\]

Example 6

Prove the identity \[\frac{{{{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }} \cdot \frac{{1 + {{\cot }^2}\alpha }}{{{{\cot }^2}\alpha }} = {\tan ^2}\alpha .\]

Example 7

Suppose \(\tan \alpha = \frac{2}{{15}}.\) Calculate the value of the expression \[\frac{{5\sin \alpha + 6\cos \alpha }}{{4\cos \alpha – 3\sin \alpha }}.\]

Example 8

Let \(\tan \alpha + \cot \alpha = n.\) Find \({\tan ^2}\alpha + {\cot ^2}\alpha .\)

Example 9

The length of the hypotenuse of a triangle is \(1,\) and the length of one of the legs is \(a.\) Find the \(6\) trigonometric functions of an angle \(\alpha\) opposite to this leg.

Example 10

Let \(\sin \alpha + \cos \alpha = k.\) Calculate \({\sin ^3}\alpha + {\cos ^3}\alpha. \)

Example 11

Let \(\sin \alpha + \cos \alpha = m.\) Calculate \({\sin ^4}\alpha + {\cos ^4}\alpha. \)

Example 12

Let \(\tan \alpha = \sqrt{3}.\) Calculate \({\sin ^4}\alpha + {\cos ^4}\alpha .\)

Example 1.

Suppose \(\sin \alpha = \large{\frac{4}{5}}\normalsize,\) where \(0 \lt \alpha \lt \large{\frac{\pi }{2}}\normalsize.\) Find the \(5\) other trigonometric functions.

Solution.

Using the identity \({\sin^2}\alpha + {\cos^2}\alpha = 1,\) we calculate the cosine function:

\[{{\sin ^2}\alpha + {\cos ^2}\alpha = 1,}\;\; \Rightarrow {{\cos ^2}\alpha = 1 – {\sin ^2}\alpha ,}\;\;\Rightarrow {\cos \alpha = \sqrt {1 – {{\sin }^2}\alpha } }={ \sqrt {1 – {{\left( {\frac{4}{5}} \right)}^2}} }={ \sqrt {1 – \frac{{16}}{{25}}} }={ \sqrt {\frac{9}{{25}}} }={ \frac{3}{5}.}\]

The tangent function is the the ratio of the sine and cosine:

\[{\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} }={ \frac{{\frac{4}{5}}}{{\frac{3}{5}}} }={ \frac{4}{3}.}\]

The cotangent is the reciprocal of the tangent:

\[{\cot \alpha = \frac{1}{{\tan \alpha }} }={ \frac{1}{{\frac{4}{3}}} }={ \frac{3}{4}.}\]

The secant and cosecant are the reciprocals of the cosine and sine, respectively:

\[{\sec \alpha = \frac{1}{{\cos \alpha }} }={ \frac{1}{{\frac{3}{5}}} }={ \frac{5}{3},}\]

\[{\csc \alpha = \frac{1}{{\sin \alpha }} }={ \frac{1}{{\frac{4}{5}}} }={ \frac{5}{4}.}\]

Example 2.

Suppose \(\sec \alpha = \large{\frac{25}{7}}\normalsize,\) where \(0 \lt \alpha \lt \large{\frac{\pi }{2}}\normalsize.\) Find the \(5\) other trigonometric functions.

Solution.

By definition, \(\sec \alpha = \large{\frac{1}{{\cos \alpha }}}\normalsize.\) Therefore,

\[{\cos \alpha = \frac{1}{{\sec \alpha }} }={ \frac{1}{{\frac{{25}}{7}}} }={ \frac{7}{{25}}.}\]

Using the identity \({\sin ^2}\alpha + {\cos ^2}\alpha = 1,\) we find \(\sin \alpha:\)

\[{\sin \alpha }={ \sqrt {1 – {{\cos }^2}\alpha } }={ \sqrt {1 – {{\left( {\frac{7}{{25}}} \right)}^2}} }={ \sqrt {1 – \frac{{49}}{{625}}} }={ \sqrt {\frac{{576}}{{625}}} }={ \frac{{24}}{{25}}.}\]

The tangent function can be expressed in terms of the sine and cosine:

\[{\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} }={ \frac{{\frac{{24}}{{25}}}}{{\frac{7}{{25}}}} }={ \frac{{24}}{7}.}\]

The cotangent is the reciprocal of the tangent:

\[{\cot \alpha = \frac{1}{{\tan \alpha }} }={ \frac{1}{{\frac{{24}}{7}}} }={ \frac{7}{{24}}.}\]

Finally, we compute the value of the cosecant:

\[{\csc \alpha = \frac{1}{{\sin \alpha }} }={ \frac{1}{{\frac{{24}}{{25}}}} }={ \frac{{25}}{{24}}.}\]

Example 3.

Simplify the expression \[\tan \alpha + \frac{{\cos \alpha }}{{1 + \sin \alpha }}.\]

Solution.

\[{\tan \alpha + \frac{{\cos \alpha }}{{1 + \sin \alpha }} }={ \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{1 + \sin \alpha }} }={ \frac{{\sin \alpha \left( {1 + \sin \alpha } \right) + \cos \alpha \cos \alpha }}{{\cos \alpha \left( {1 + \sin \alpha } \right)}} }={ \frac{{\sin \alpha + \overbrace {{{\sin }^2}\alpha + {{\cos }^2}\alpha }^{ = 1}}}{{\cos \alpha \left( {1 + \sin \alpha } \right)}} }={ \frac{{\sin \alpha + 1}}{{\cos \alpha \left( {\sin \alpha + 1} \right)}} }={ \frac{1}{{\cos \alpha }} }={ \sec \alpha .}\]

Example 4.

Simplify the expression \[{\tan ^2}\alpha – {\sin ^2}\alpha – {\tan ^2}\alpha \,{\sin ^2}\alpha .\]

Solution.

We denote this expression by \(A.\)

\[{A = {\tan ^2}\alpha – {\sin ^2}\alpha – {\tan ^2}\alpha {\sin ^2}\alpha }={ {\tan ^2}\alpha – {\sin ^2}\alpha \left( {1 + {{\tan }^2}\alpha } \right).}\]

Using the identity \(1 + {\tan ^2}\alpha = {\sec ^2}\alpha ,\) we have

\[A = {\tan ^2}\alpha – {\sin ^2}\alpha \,{\sec ^2}\alpha .\]

By definition, \({\sec ^2}\alpha = \large{\frac{1}{{{{\cos }^2}\alpha }}}\normalsize.\) Therefore,

\[\require{cancel}{A = {\tan ^2}\alpha – \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} }={ \cancel{{\tan ^2}\alpha} – \cancel{{\tan ^2}\alpha} }={ 0.}\]

Example 5.

Prove the identity \[\sec \alpha – \sin \alpha \tan \alpha = \cos \alpha .\]

Solution.

By definition,

\[\sec \alpha = \frac{1}{{\cos \alpha }} \;\text{ and }\; \tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\]

Hence,

\[{\sec \alpha – \sin \alpha \tan \alpha = \cos \alpha ,}\;\; \Rightarrow {\frac{1}{{\cos \alpha }} – \frac{{{{\sin }^2}\alpha }}{{\cos \alpha }} = \cos \alpha ,}\;\; \Rightarrow {\frac{{1 – {{\sin }^2}\alpha }}{{\cos \alpha }} = \cos \alpha ,}\;\; \Rightarrow {\frac{{{{\cos }^{\cancel{2}}}\alpha }}{{\cancel{\cos \alpha} }} = \cos \alpha ,}\;\; \Rightarrow {\cos \alpha = \cos \alpha .}\]

Example 6.

Prove the identity \[\frac{{{{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }} \cdot \frac{{1 + {{\cot }^2}\alpha }}{{{{\cot }^2}\alpha }} = {\tan ^2}\alpha .\]

Solution.

We use the Pythagorean identities

\[{1 + {\tan ^2}\alpha = {\sec ^2}\alpha }={ \frac{1}{{{{\cos }^2}\alpha }} \;\text{ and }}\]

\[{1 + {\cot ^2}\alpha = {\csc ^2}\alpha }={ \frac{1}{{{{\sin }^2}\alpha }}.}\]

Then we write the initial identity in the form

\[\frac{{{{\tan }^2}\alpha }}{{\frac{1}{{{{\cos }^2}\alpha }}}} \cdot \frac{{\frac{1}{{{{\sin }^2}\alpha }}}}{{{{\cot }^2}\alpha }} = {\tan ^2}\alpha , \;\text{ or }\]

\[\frac{{{{\tan }^2}\alpha \,{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha \,{{\cot }^2}\alpha }} = {\tan ^2}\alpha .\]

Since \({\cot ^2}\alpha = \large{\frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}}\normalsize,\) we have

\[{\frac{{{{\tan }^2}\alpha \cancel{{{\cot }^2}\alpha} }}{{\cancel{{{\cot }^2}\alpha} }} = {\tan ^2}\alpha ,}\;\; \Rightarrow {{\tan ^2}\alpha = {\tan ^2}\alpha .}\]

Example 7.

Suppose \(\tan \alpha = \frac{2}{{15}}.\) Calculate the value of the expression \[\frac{{5\sin \alpha + 6\cos \alpha }}{{4\cos \alpha – 3\sin \alpha }}.\]

Solution.

Since \(\tan \alpha\) is a finite quantity, then \(\cos \alpha \ne 0.\) Therefore, we can divide the numerator and denominator of the expression by \(\cos \alpha:\)

\[{\frac{{5\sin \alpha + 6\cos \alpha }}{{4\cos \alpha – 3\sin \alpha }} }={ \frac{{\frac{{5\sin \alpha + 6\cos \alpha }}{{\cos \alpha }}}}{{\frac{{4\cos \alpha – 3\sin \alpha }}{{\cos \alpha }}}} }={ \frac{{\frac{{5\sin \alpha }}{{\cos \alpha }} + \frac{{6\cancel{\cos \alpha} }}{{\cancel{\cos \alpha} }}}}{{\frac{{4\cancel{\cos \alpha} }}{{\cancel{\cos \alpha} }} – \frac{{3\sin \alpha }}{{\cos \alpha }}}} }={ \frac{{5\tan \alpha + 6}}{{4 – 3\tan \alpha }} }={ \frac{{5 \times \frac{2}{{15}} + 6}}{{4 – 3 \times \frac{2}{{15}}}} }={ \frac{{\frac{2}{3} + 6}}{{4 – \frac{2}{5}}} }={ \frac{{\frac{{2 + 18}}{3}}}{{\frac{{20 – 2}}{5}}} }={ \frac{{\frac{{20}}{3}}}{{\frac{{18}}{5}}} }={ \frac{{100}}{{54}} }={ \frac{{50}}{{27}}.}\]

Example 8.

Let \(\tan \alpha + \cot \alpha = n.\) Find \({\tan ^2}\alpha + {\cot ^2}\alpha .\)

Solution.

By squaring both sides of the equation \(\tan \alpha + \cot \alpha = n,\) we obtain

\[{{\left( {\tan \alpha + \cot \alpha } \right)^2} = {n^2},}\;\; \Rightarrow {{\tan ^2}\alpha + 2\,\underbrace {\tan \alpha \cot \alpha }_{ = 1} + {\cot ^2}\alpha = {n^2},}\;\; \Rightarrow {{\tan ^2}\alpha + 2 + {\cot ^2}\alpha = {n^2},}\;\; \Rightarrow {{\tan ^2}\alpha + {\cot ^2}\alpha = {n^2} – 2.}\]

Example 9.

The length of the hypotenuse of a triangle is \(1,\) and the length of one of the legs is \(a.\) Find the \(6\) trigonometric functions of an angle \(\alpha\) opposite to this leg.

Solution.

A right triangle with one leg a and hypotenuse equal to 1.
Figure 2.

Using the Pythagorean theorem, we first find the length of the other leg \(b\), adjacent to \(\alpha:\)

\[{{a^2} + {b^2} = 1,}\;\; \Rightarrow {b = \sqrt {1 – {a^2}} .}\]

The sine and cosine of the angle \(\alpha\) are given by

\[{\sin \alpha = \frac{a}{c} = a,\;\;}\kern0pt{\cos \alpha = \frac{b}{c} = \sqrt {1 – {a^2}} .}\]

Determine the tangent and cotangent of \(\alpha:\)

\[{\tan \alpha = \frac{a}{b} = \frac{a}{{\sqrt {1 – {a^2}} }},\;\;}\kern0pt{\cot \alpha = \frac{b}{a} = \frac{{\sqrt {1 – {a^2}} }}{a}.}\]

Calculate the values of the secant and cosecant:

\[{\sec \alpha = \frac{c}{b} = \frac{1}{{\sqrt {1 – {a^2}} }},\;\;}\kern0pt{\csc \alpha = \frac{c}{a} = \frac{1}{a}.}\]

Example 10.

Let \(\sin \alpha + \cos \alpha = k.\) Calculate \({\sin ^3}\alpha + {\cos ^3}\alpha. \)

Solution.

We factor the sum of two cubes:

\[{{\sin ^3}\alpha + {\cos ^3}\alpha }={ \left( {\sin \alpha + \cos \alpha } \right)}\kern0pt{\left( {{{\sin }^2}\alpha – \sin \alpha \cos \alpha + {{\cos }^2}\alpha } \right).}\]

Since \({\sin ^2}\alpha + {\cos ^2}\alpha = 1,\) we represent the last expression in the form

\[{{\sin ^3}\alpha + {\cos ^3}\alpha }={ \left( {\sin \alpha + \cos \alpha } \right)\left( {1 – \sin \alpha \cos \alpha } \right).}\]

Now, to find the product \(\sin \alpha \cos \alpha,\) we square both sides of the equation \(\sin \alpha + \cos \alpha = k:\)

\[{{\left( {\sin \alpha + \cos \alpha } \right)^2} = {k^2},}\;\; \Rightarrow {{\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = {k^2},}\;\; \Rightarrow {1 + 2\sin \alpha \cos \alpha = {k^2},}\;\; \Rightarrow {\sin \alpha \cos \alpha = \frac{{{k^2} – 1}}{2}.}\]

Hence,

\[{{\sin ^3}\alpha + {\cos ^3}\alpha }={ \left( {\sin \alpha + \cos \alpha } \right)}\kern0pt{\left( {1 – \sin \alpha \cos \alpha } \right) }={ k\left( {1 – \frac{{{k^2} – 1}}{2}} \right) }={ \frac{{k\left( {3 – {k^2}} \right)}}{2}.}\]

Example 11.

Let \(\sin \alpha + \cos \alpha = m.\) Calculate \({\sin ^4}\alpha + {\cos ^4}\alpha. \)

Solution.

We square both sides of the Pythagorean trig identity:

\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1,\]

\[\Rightarrow {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = 1,\]

\[{\Rightarrow {\sin ^4}\alpha + 2{\sin ^2}\alpha {\cos ^2}\alpha + {\cos ^4}\alpha }={ 1.}\]

Then

\[{{\sin ^4}\alpha + {\cos ^4}\alpha }={ 1 – 2{\left( {\sin \alpha \cos \alpha } \right)^2}.}\]

Now we find \(\sin \alpha \cos \alpha.\) By condition, \(\sin \alpha + \cos \alpha = m.\) So,

\[{{\left( {\sin \alpha + \cos \alpha } \right)^2} = {m^2},}\;\; \Rightarrow {{\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = {m^2},}\;\; \Rightarrow {1 + 2\sin \alpha \cos \alpha = {m^2},}\;\; \Rightarrow {\sin \alpha \cos \alpha = \frac{{{m^2} – 1}}{2}.}\]

Therefore,

\[{{\sin ^4}\alpha + {\cos ^4}\alpha }={ 1 – 2{\left( {\frac{{{m^2} – 1}}{2}} \right)^2} }={ 1 – 2 \cdot \frac{{{m^4} – 2{m^2} + 1}}{4} }={ \frac{{1 + 2{m^2} – {m^4}}}{2}.}\]

Example 12.

Let \(\tan \alpha = \sqrt{3}.\) Calculate \({\sin ^4}\alpha + {\cos ^4}\alpha .\)

Solution.

Using the identity \({\sin ^2}\alpha + {\cos ^2}\alpha = 1,\) we have

\[{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = 1,}\;\; \Rightarrow {{\sin ^4}\alpha + 2\,{\sin ^2}\alpha\, {\cos ^2}\alpha + {\cos ^4}\alpha = 1,}\;\; \Rightarrow {{\sin ^4}\alpha + {\cos ^4}\alpha }={ 1 – 2\,{\sin ^2}\alpha \,{\cos ^2}\alpha .}\]

By condition,

\[{\tan \alpha = \sqrt 3 ,}\;\; \Rightarrow {{\tan ^2}\alpha = 3.}\]

Given the tangent function, find the cosine squared:

\[{{\tan ^2}\alpha + 1 = {\sec ^2}\alpha ,}\;\; \Rightarrow {{\sec ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} = 3 + 1 = 4,}\;\; \Rightarrow {{\cos ^2}\alpha = \frac{1}{4}.}\]

The sine squared is given by

\[{{\sin ^2}\alpha = 1 – {\cos ^2}\alpha }={ 1 – \frac{1}{4} }={ \frac{3}{4}.}\]

Then

\[{{\sin ^4}\alpha + {\cos ^4}\alpha }={ 1 – 2{\sin ^2}\alpha {\cos ^2}\alpha }={ 1 – 2 \times \frac{3}{4} \times \frac{1}{4} }={ 1 – \frac{3}{8} }={ \frac{5}{8}.}\]