# Pythagorean Trigonometric Identities

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Suppose sin α = 4/5, where 0 < α < π/2. Find the 5 other trigonometric functions.

### Example 2

Suppose sec α = 25/7 where 0 < α < π/2. 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 $$\sin \alpha = \frac{4}{5},$$ 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.

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

${\sin ^4}\alpha + {\cos ^4}\alpha = 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha = 1 - 2 \cdot \frac{3}{4} \cdot \frac{1}{4} = 1 - \frac{3}{8} = \frac{5}{8}.$