Pythagorean Trigonometric Identities
Solved Problems
Example 1.
Suppose \(\sin \alpha = \frac{4}{5},\) where \(0 \lt \alpha \lt \frac{\pi }{2}.\) 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 \(\sec \alpha = \frac{25}{7},\) where \(0 \lt \alpha \lt \frac{\pi }{2}.\) Find the \(5\) other trigonometric functions.
Solution.
By definition, \(\sec \alpha = \frac{1}{{\cos \alpha }}.\) 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, \({\sec ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}.\) 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 \({\cot ^2}\alpha = \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }},\) 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.
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
\[\sin \alpha = \frac{a}{c} = a,\;\;\cos \alpha = \frac{b}{c} = \sqrt {1 - {a^2}}.\]
Determine the tangent and cotangent of \(\alpha:\)
\[\tan \alpha = \frac{a}{b} = \frac{a}{{\sqrt {1 - {a^2}} }},\;\;\cot \alpha = \frac{b}{a} = \frac{{\sqrt {1 - {a^2}} }}{a}.\]
Calculate the values of the secant and cosecant:
\[\sec \alpha = \frac{c}{b} = \frac{1}{{\sqrt {1 - {a^2}} }},\;\;\csc \alpha = \frac{c}{a} = \frac{1}{a}.\]
Example 10.
Let \(\sin \alpha + \cos \alpha = k.\) Calculate \({\sin ^3}\alpha + {\cos ^3}\alpha. \)
Solution.
We factor the sum of two cubes:
\[{\sin ^3}\alpha + {\cos ^3}\alpha = \left( {\sin \alpha + \cos \alpha } \right) \left( {{{\sin }^2}\alpha - \sin \alpha \cos \alpha + {{\cos }^2}\alpha } \right).\]
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,
\[{\sin ^3}\alpha + {\cos ^3}\alpha = \left( {\sin \alpha + \cos \alpha } \right)\left( {1 - \sin \alpha \cos \alpha } \right) = k\left( {1 - \frac{{{k^2} - 1}}{2}} \right) = \frac{{k\left( {3 - {k^2}} \right)}}{2}.\]
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
\[{\sin ^4}\alpha + {\cos ^4}\alpha = 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha = 1 - 2 \cdot \frac{3}{4} \cdot \frac{1}{4} = 1 - \frac{3}{8} = \frac{5}{8}.\]
