Precalculus

Trigonometry

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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.

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