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