Precalculus

Trigonometry

Trigonometry Logo

Simplifying Trigonometric Expressions

To simplify trigonometric expressions it is necessary to know trigonometric identities and algebraic rules. It's helpful to follow some general rules:

  1. If trigonometric functions contain different angles, we try to reduce them to functions of only one angle using, for example, cofunction and reduction identities or double-angle formulas.
  2. If a trigonometric expression contains a large number of functions, it is necessary to reduce the number of functions to a minimum. For this purpose, we use reduction identities, Pythagorean trigonometric identities and other formulas.
  3. If we need to reduce the power of a component in a trig expression, we apply the half-angle identities or power-reduction formulas. It is only necessary to remember that when the power is reduced by \(2\) times, the argument doubles.

Using these identities and formulas, we can convert any trigonometric expression into a rational expression containing only one function with one argument.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Simplify the trigonometric expression:

\[\frac{{\sin 6\alpha }}{{\sin 2\alpha }} + \frac{{\cos \left( {6\alpha - \pi } \right)}}{{\cos 2\alpha }}.\]

Example 2

Simplify the trigonometric expression:

\[\frac{{1 - \cos \left( {8\alpha - 3\pi } \right)}}{{\tan 2\alpha - \cot 2\alpha }}.\]

Example 3

Simplify the trigonometric expression:

\[\frac{{1 + \cot 2\alpha \cot \alpha }}{{\tan \alpha + \cot \alpha }}.\]

Example 4

Simplify the trigonometric expression:

\[\frac{{1 - \tan \left( {\pi - 2\alpha } \right)\tan \alpha }}{{\tan \left( {\frac{{3\pi }}{2} - \alpha } \right) + \tan \alpha }}.\]

Example 5

Simplify:

\[{\cos ^4}2\alpha - 6\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha + {\sin ^4}2\alpha .\]

Example 6

Simplify:

\[\frac{{\sin \left( {4\alpha + \frac{{5\pi }}{2}} \right)}}{{1 + \cos \left( {4\alpha - \frac{{3\pi }}{2}} \right)}}.\]

Example 7

Simplify:

\[\frac{{\tan \alpha + \tan \beta }}{{\tan \left( {\alpha + \beta } \right)}} + \frac{{\tan \alpha - \tan \beta }}{{\tan \left( {\alpha - \beta } \right)}}.\]

Example 8

Simplify:

\[\frac{{\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right)}}{{\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \alpha } \right)}}.\]

Example 1.

Simplify the trigonometric expression:

\[\frac{{\sin 6\alpha }}{{\sin 2\alpha }} + \frac{{\cos \left( {6\alpha - \pi } \right)}}{{\cos 2\alpha }}.\]

Solution.

The cosine function is even, so \(\cos \left( {6\alpha - \pi } \right) = \cos \left( {\pi - 6\alpha } \right).\) Using the reduction identity, we have

\[A = \frac{{\sin 6\alpha }}{{\sin 2\alpha }} + \frac{{\cos \left( {6\alpha - \pi } \right)}}{{\cos 2\alpha }} = \frac{{\sin 6\alpha }}{{\sin 2\alpha }} + \frac{{\cos \left( {\pi - 6\alpha } \right)}}{{\cos 2\alpha }} = \frac{{\sin 6\alpha }}{{\sin 2\alpha }} - \frac{{\cos 6\alpha }}{{\cos 2\alpha }}.\]

Convert to a common denominator and apply the sine addition formula:

\[A = \frac{{\sin 6\alpha }}{{\sin 2\alpha }} - \frac{{\cos 6\alpha }}{{\cos 2\alpha }} = \frac{{\sin 6\alpha \cos 2\alpha - \cos 6\alpha \sin 2\alpha }}{{\sin 2\alpha \cos 2\alpha }} = \frac{{\sin \left( {6\alpha - 2\alpha } \right)}}{{\sin 2\alpha \cos 2\alpha }} = \frac{{\sin 4\alpha }}{{\sin 2\alpha \cos 2\alpha }}.\]

By the double-angle identity, we get

\[A = \frac{{\sin 4\alpha }}{{\sin 2\alpha \cos 2\alpha }} = \frac{{2\cancel{{\sin 2\alpha }}\cancel{{\cos 2\alpha }}}}{{\cancel{{\sin 2\alpha }}\cancel{{\cos 2\alpha }}}} = 2.\]

Example 2.

Simplify the trigonometric expression:

\[\frac{{1 - \cos \left( {8\alpha - 3\pi } \right)}}{{\tan 2\alpha - \cot 2\alpha }}.\]

Solution.

Recall that cosine is an even and periodic function with a period of \(2\pi.\) Hence,

\[\cos \left( {8\alpha - 3\pi } \right) = \cos \left( {8\alpha - \pi } \right) = \cos \left( {\pi - 8\alpha } \right).\]

Using the reduction formula for cosine and expressing the tangent and cotangent in terms of sine and cosine, we have

\[A = \frac{{1 - \cos \left( {\pi - 8\alpha } \right)}}{{\tan 2\alpha - \cot 2\alpha }} = \frac{{1 + \cos 8\alpha }}{{\frac{{\sin 2\alpha }}{{\cos 2\alpha }} - \frac{{\cos 2\alpha }}{{\sin 2\alpha }}}} = \frac{{1 + \cos 8\alpha }}{{\frac{{{{\sin }^2}2\alpha - {{\cos }^2}2\alpha }}{{\sin 2\alpha \cos 2\alpha }}}} = - \frac{{\sin 2\alpha \cos 2\alpha \left( {1 + \cos 8\alpha } \right)}}{{{{\cos }^2}2\alpha - {{\sin }^2}2\alpha }}.\]

By the double-angle formulas,

\[\sin 2\alpha \cos 2\alpha = \frac{1}{2}\sin 4\alpha ,\]
\[1 + \cos 8\alpha = 2{\cos ^2}4\alpha ,\]
\[{\cos ^2}2\alpha - {\sin ^2}2\alpha = \cos 4\alpha .\]

Substitute this in our expression:

\[A = - \frac{{\sin 2\alpha \cos 2\alpha \left( {1 + \cos 8\alpha } \right)}}{{{{\cos }^2}2\alpha - {{\sin }^2}2\alpha }} = - \frac{{\frac{1}{2}\sin 4\alpha \cdot 2{{\cos }^2}4\alpha }}{{\cos 4\alpha }} = - \frac{{\sin 4\alpha {{\cos }^2}4\alpha }}{{\cos 4\alpha }} = - \sin 4\alpha \cos 4\alpha = - \frac{1}{2}\sin 8\alpha .\]

Example 3.

Simplify the trigonometric expression:

\[\frac{{1 + \cot 2\alpha \cot \alpha }}{{\tan \alpha + \cot \alpha }}.\]

Solution.

We denote this expression by \(A.\) Using the double-angle identity

\[\cot 2\alpha = \frac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }},\]

we can write:

\[A = \frac{{1 + \cot 2\alpha \cot \alpha }}{{\tan \alpha + \cot \alpha }} = \frac{{1 + \frac{{\left( {{{\cot }^2}\alpha - 1} \right)\cancel{{\cot \alpha }}}}{{2\cancel{{\cot \alpha }}}}}}{{\tan \alpha + \cot \alpha }} = \frac{{2 + {{\cot }^2}\alpha - 1}}{{2\left( {\tan \alpha + \cot \alpha } \right)}} = \frac{{1 + {{\cot }^2}\alpha }}{{2\left( {\tan \alpha + \cot \alpha } \right)}}.\]

Substitute \(\tan \alpha = \frac{1}{{\cot \alpha }}:\)

\[A = \frac{{1 + {{\cot }^2}\alpha }}{{2\left( {\frac{1}{{\cot \alpha }} + \cot \alpha } \right)}} = \frac{{\cancel{{1 + {{\cot }^2}\alpha }}}}{{\frac{{2\cancel{{\left( {1 + {{\cot }^2}\alpha } \right)}}}}{{\cot \alpha }}}} = \frac{{\cot \alpha }}{2}.\]

Example 4.

Simplify the trigonometric expression:

\[\frac{{1 - \tan \left( {\pi - 2\alpha } \right)\tan \alpha }}{{\tan \left( {\frac{{3\pi }}{2} - \alpha } \right) + \tan \alpha }}.\]

Solution.

Using the reduction and cofunction identities, we have

\[\tan \left( {\pi - 2\alpha } \right) = - \tan 2\alpha ,\]
\[\tan \left( {\frac{{3\pi }}{2} - \alpha } \right) = \cot \alpha = \frac{1}{{\tan \alpha }}.\]

Hence, the original expression is written as

\[A = \frac{{1 - \tan \left( {\pi - 2\alpha } \right)\tan \alpha }}{{\tan \left( {\frac{{3\pi }}{2} - \alpha } \right) + \tan \alpha }} = \frac{{1 + \tan 2\alpha \tan \alpha }}{{\frac{1}{{\tan \alpha }} + \tan \alpha }}.\]

Use the double-angle formula for tangent:

\[\tan 2\alpha = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}.\]

This yields:

\[A = \frac{{1 + \tan 2\alpha \tan \alpha }}{{\frac{1}{{\tan \alpha }} + \tan \alpha }} = \frac{{1 + \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \cdot \tan \alpha }}{{\frac{1}{{\tan \alpha }} + \tan \alpha }} = \frac{{\frac{{1 - {{\tan }^2}\alpha + 2{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}}} = \frac{{\frac{{1 + {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}}} = \frac{{\cancel{{\left( {1 + {{\tan }^2}\alpha } \right)}}\tan \alpha }}{{\left( {1 - {{\tan }^2}\alpha } \right)\cancel{{\left( {1 + {{\tan }^2}\alpha } \right)}}}} = \frac{{\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{1}{2} \cdot \underbrace {\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}}_{\tan 2\alpha } = \frac{{\tan 2\alpha }}{2}.\]

Example 5.

Simplify:

\[{\cos ^4}2\alpha - 6\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha + {\sin ^4}2\alpha .\]

Solution.

Notice that

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

Hence

\[A = {\cos ^4}2\alpha - 6\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha + {\sin ^4}2\alpha = \underbrace {{{\cos }^4}2\alpha + 2\,{{\cos }^2}2\alpha \,{{\sin }^2}2\alpha + {{\sin }^4}2\alpha }_{{{\left( {{{\cos }^2}2\alpha + {{\sin }^2}2\alpha } \right)}^2} = 1} - 8\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha = 1 - 8\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha .\]

Applying the double-angle formulas for sine and cosine and Pythagorean trigonometric identity, we get:

\[A = 1 - 8\,{\cos ^2}2\alpha \,{\sin ^2}2\alpha = 1 - 2{\left( {2\sin 2\alpha \cos 2\alpha } \right)^2} = 1 - 2\,{\sin ^2}4\alpha = \left( {{{\cos }^2}4\alpha + {{\sin }^2}4\alpha } \right) - 2\,{\sin ^2}4\alpha = {\cos ^2}4\alpha - {\sin ^2}4\alpha = \cos 8\alpha .\]

Example 6.

Simplify:

\[\frac{{\sin \left( {4\alpha + \frac{{5\pi }}{2}} \right)}}{{1 + \cos \left( {4\alpha - \frac{{3\pi }}{2}} \right)}}.\]

Solution.

Using periodicity of trig functions, we express them as functions of the same angle:

\[\sin \left( {4\alpha + \frac{{5\pi }}{2}} \right) = \sin \left( {4\alpha + \underbrace {2\pi + \frac{\pi }{2}}_{\frac{{5\pi }}{2}}} \right) = \sin \left( {4\alpha + \frac{\pi }{2}} \right),\]
\[\cos \left( {4\alpha - \frac{{3\pi }}{2}} \right) = \cos \left( {4\alpha - \frac{{3\pi }}{2} + 2\pi } \right) = \cos \left( {4\alpha + \frac{\pi }{2}} \right).\]

Then we can write:

\[A = \frac{{\sin \left( {4\alpha + \frac{{5\pi }}{2}} \right)}}{{1 + \cos \left( {4\alpha - \frac{{3\pi }}{2}} \right)}} = \frac{{\sin \left( {4\alpha + \frac{\pi }{2}} \right)}}{{1 + \cos \left( {4\alpha + \frac{\pi }{2}} \right)}}.\]

Apply now the tangent half-angle substitution:

\[A = \frac{{\sin \left( {4\alpha + \frac{\pi }{2}} \right)}}{{1 + \cos \left( {4\alpha + \frac{\pi }{2}} \right)}} = \frac{{\frac{{2\tan \left( {2\alpha + \frac{\pi }{4}} \right)}}{{1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}}}{{1 + \frac{{1 - {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}{{1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}}} = \frac{{\frac{{2\tan \left( {2\alpha + \frac{\pi }{4}} \right)}}{{1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}}}{{\frac{{1 + \cancel{{{{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}} + 1 - \cancel{{{{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}}}{{1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)}}}} = \frac{{\cancel{2}\tan \left( {2\alpha + \frac{\pi }{4}} \right)\cancel{{\left( {1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)} \right)}}}}{{\cancel{2}\cancel{{\left( {1 + {{\tan }^2}\left( {2\alpha + \frac{\pi }{4}} \right)} \right)}}}} = \tan \left( {2\alpha + \frac{\pi }{4}} \right).\]

Example 7.

Simplify:

\[\frac{{\tan \alpha + \tan \beta }}{{\tan \left( {\alpha + \beta } \right)}} + \frac{{\tan \alpha - \tan \beta }}{{\tan \left( {\alpha - \beta } \right)}}.\]

Solution.

Recall the addition and subtraction formulas for tangent:

\[\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }},\]
\[\tan \left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}.\]

Then we have:

\[\frac{{\tan \alpha + \tan \beta }}{{\tan \left( {\alpha + \beta } \right)}} + \frac{{\tan \alpha - \tan \beta }}{{\tan \left( {\alpha - \beta } \right)}} = \frac{{\cancel{{\tan \alpha + \tan \beta }}}}{{\frac{{\cancel{{\tan \alpha + \tan \beta }}}}{{1 - \tan \alpha \tan \beta }}}} + \frac{{\cancel{{\tan \alpha - \tan \beta }}}}{{\frac{{\cancel{{\tan \alpha - \tan \beta }}}}{{1 + \tan \alpha \tan \beta }}}} = 1 - \cancel{{\tan \alpha \tan \beta }} + 1 + \cancel{{\tan \alpha \tan \beta }} = 2.\]

Example 8.

Simplify:

\[\frac{{\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right)}}{{\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \alpha } \right)}}.\]

Solution.

Using the sum-to-product identity, the numerator can be written as

\[\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right) = 2\cos \frac{{\cancel{\alpha } - \cancel{\beta } + \cancel{\beta } - \cancel{\alpha }}}{2}\sin \frac{{\alpha - \beta - \left( {\beta - \alpha } \right)}}{2} = 2\cos 0\sin \left( {\alpha - \beta } \right) = 2\sin \left( {\alpha - \beta } \right).\]

Similarly, the denominator is represented in the form

\[\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \alpha } \right) = 2\cos \frac{{\cancel{\alpha } - \cancel{\beta } + \cancel{\beta } - \cancel{\alpha }}}{2}\cos \frac{{\alpha - \beta - \left( {\beta - \alpha } \right)}}{2} = 2\cos 0\cos \left( {\alpha - \beta } \right) = 2\cos \left( {\alpha - \beta } \right).\]

Hence,

\[\frac{{\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right)}}{{\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \alpha } \right)}} = \frac{{\cancel{2}\sin \left( {\alpha - \beta } \right)}}{{\cancel{2}\cos \left( {\alpha - \beta } \right)}} = \tan \left( {\alpha - \beta } \right).\]

See more problems on Page 2.

Page 1 Page 2