Simplifying Trigonometric Expressions
Solved Problems
Example 9.
Simplify:
\[{\sin ^2}\left( {\frac{\alpha }{2} + 2\beta } \right) - {\sin ^2}\left( {\frac{\alpha }{2} - 2\beta } \right).\]
Solution.
Factoring the expression, we can write it in the form
\[A = {\sin ^2}\left( {\frac{\alpha }{2} + 2\beta } \right) - {\sin ^2}\left( {\frac{\alpha }{2} - 2\beta } \right) = \left[ {\sin \left( {\frac{\alpha }{2} + 2\beta } \right) - \sin \left( {\frac{\alpha }{2} - 2\beta } \right)} \right]\left[ {\sin \left( {\frac{\alpha }{2} + 2\beta } \right) + \sin \left( {\frac{\alpha }{2} - 2\beta } \right)} \right].\]
We calculate the expressions in square brackets separately using sum-to-product identities:
\[\sin \left( {\frac{\alpha }{2} + 2\beta } \right) - \sin \left( {\frac{\alpha }{2} - 2\beta } \right) = 2\cos \frac{{\frac{\alpha }{2} + \cancel{{2\beta }} + \frac{\alpha }{2} - \cancel{{2\beta }}}}{2}\sin \frac{{\cancel{{\frac{\alpha }{2}}} + 2\beta - \cancel{{\frac{\alpha }{2}}} + 2\beta }}{2} = 2\cos \frac{\alpha }{2}\sin 2\beta ,\]
\[\sin \left( {\frac{\alpha }{2} + 2\beta } \right) + \sin \left( {\frac{\alpha }{2} - 2\beta } \right) = 2\sin \frac{{\frac{\alpha }{2} + \cancel{{2\beta }} + \frac{\alpha }{2} - \cancel{{2\beta }}}}{2}\cos \frac{{\cancel{{\frac{\alpha }{2}}} + 2\beta - \cancel{{\frac{\alpha }{2}}} + 2\beta }}{2} = 2\sin \frac{\alpha }{2}\cos 2\beta .\]
Then by the double-angle formula for sine,
\[A = 2\cos \frac{\alpha }{2}\sin 2\beta \cdot 2\sin \frac{\alpha }{2}\cos 2\beta = 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} \cdot 2\sin 2\beta \cos 2\beta = \sin \alpha \sin 4\beta .\]
Example 10.
Simplify:
\[{\cos ^2}\left( {\alpha + 2\beta } \right) + {\sin ^2}\left( {\alpha - 2\beta } \right) - 1.\]
Solution.
By the Pythagorean identity,
\[\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right) = 2\cos \frac{{\cancel{\alpha } - \cancel{\beta } + \cancel{\beta } - \cancel{\alpha }}}{2}\sin \frac{{\alpha - \beta - \left( {\beta - \alpha } \right)}}{2} = 2\cos 0\sin \left( {\alpha - \beta } \right) = 2\sin \left( {\alpha - \beta } \right).\]
Using the sum-to-product identity, the numerator can be written as
\[\sin \left( {\alpha - \beta } \right) - \sin \left( {\beta - \alpha } \right) = 2\cos \frac{{\cancel{\alpha } - \cancel{\beta } + \cancel{\beta } - \cancel{\alpha }}}{2}\sin \frac{{\alpha - \beta - \left( {\beta - \alpha } \right)}}{2} = 2\cos 0\sin \left( {\alpha - \beta } \right) = 2\sin \left( {\alpha - \beta } \right).\]
Example 11.
Simplify the trigonometric expression:
\[\frac{{\tan 2\alpha }}{{\tan 4\alpha - \tan 2\alpha }}.\]
Solution.
We write \({\tan 4\alpha }\) in terms of \({\tan 2\alpha }.\) This yields:
\[A = \frac{{\tan 2\alpha }}{{\tan 4\alpha - \tan 2\alpha }} = \frac{{\tan 2\alpha }}{{\frac{{2\tan 2\alpha }}{{1 - {{\tan }^2}2\alpha }} - \tan 2\alpha }} = \frac{{\tan 2\alpha }}{{\frac{{2\tan 2\alpha }}{{1 - {{\tan }^2}2\alpha }} - \frac{{\tan 2\alpha \left( {1 - {{\tan }^2}2\alpha } \right)}}{{1 - {{\tan }^2}2\alpha }}}} = \frac{{\tan 2\alpha }}{{\frac{{2\tan 2\alpha - \tan 2\alpha + {{\tan }^3}2\alpha }}{{1 - {{\tan }^2}2\alpha }}}} = \frac{{\tan 2\alpha \left( {1 - {{\tan }^2}2\alpha } \right)}}{{\tan 2\alpha + {{\tan }^3}2\alpha }} = \frac{{\cancel{{\tan 2\alpha }}\left( {1 - {{\tan }^2}2\alpha } \right)}}{{\cancel{{\tan 2\alpha }}\left( {1 + {{\tan }^2}2\alpha } \right)}} = \frac{{1 - {{\tan }^2}2\alpha }}{{1 + {{\tan }^2}2\alpha }}.\]
Recall that
\[\cos 2\alpha = \frac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}.\]
Hence,
\[A = \frac{{1 - {{\tan }^2}2\alpha }}{{1 + {{\tan }^2}2\alpha }} = \cos 4\alpha .\]
Example 12.
Simplify the trigonometric expression:
\[{\sin ^3}\alpha \cos 3\alpha + {\cos ^3}\alpha \sin 3\alpha .\]
Solution.
Using the sine addition formula, we have
\[\sin 4\alpha = \sin \left( {\alpha + 3\alpha } \right) = \sin \alpha \cos 3\alpha + \cos \alpha \sin 3\alpha .\]
Let's convert this expression as follows:
\[\sin 4\alpha = \sin 4\alpha \cdot 1 = \sin 4\alpha \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = \left( {\sin \alpha \cos 3\alpha + \cos \alpha \sin 3\alpha } \right)\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = {\sin ^3}\alpha \cos 3\alpha + {\cos ^3}\alpha \sin 3\alpha + {\sin ^2}\alpha \cos \alpha \sin 3\alpha + {\cos ^2}\alpha \sin \alpha \cos 3\alpha .\]
Applying the double-angle formula for sine and cosine subtraction identity, we rewrite the last two terms in the form:
\[{\sin ^2}\alpha \cos \alpha \sin 3\alpha + {\cos ^2}\alpha \sin \alpha \cos 3\alpha = \sin \alpha \cos \alpha \left( {\sin \alpha \sin 3\alpha + \cos \alpha \cos 3\alpha } \right) = \frac{1}{2}\sin 2\alpha \cos \left( {3\alpha - \alpha } \right) = \frac{1}{2}\sin 2\alpha \cos 2\alpha = \frac{1}{4}\sin 4\alpha .\]
So
\[\sin 4\alpha = {\sin ^3}\alpha \cos 3\alpha + {\cos ^3}\alpha \sin 3\alpha + \frac{1}{4}\sin 4\alpha .\]
It follows from here that
\[{\sin ^3}\alpha \cos 3\alpha + {\cos ^3}\alpha \sin 3\alpha = \sin 4\alpha - \frac{1}{4}\sin 4\alpha = \frac{3}{4}\sin 4\alpha .\]
Example 13.
Simplify the trigonometric expression:
\[\frac{{\cot \alpha + \tan \alpha }}{{1 + \tan 2\alpha \tan \alpha }}.\]
Solution.
We express \(\cot \alpha\) and \(\tan 2\alpha\) in terms of \(\tan \alpha\) by the formulas
\[\cot \alpha = \frac{1}{{\tan \alpha }},\;\;\tan 2\alpha = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}.\]
Then we get
\[A = \frac{{\cot \alpha + \tan \alpha }}{{1 + \tan 2\alpha \tan \alpha }} = \frac{{\frac{1}{{\tan \alpha }} + \tan \alpha }}{{1 + \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \cdot \tan \alpha }} = \frac{{\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}}}{{\frac{{1 - {{\tan }^2}\alpha + 2{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} = \frac{{\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}}}{{\frac{{1 + {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}}.\]
Using the double-angle formula, we represent the numerator in the form:
\[\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }} = \frac{1}{{\frac{{\tan \alpha }}{{1 + {{\tan }^2}\alpha }}}} = \frac{2}{{\frac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }}}} = \frac{2}{{\sin 2\alpha }}.\]
Similarly, the denominator can be written as
\[\frac{{1 + {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{1}{{\frac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}}} = \frac{1}{{\cos 2\alpha }}.\]
Therefore,
\[A = \frac{{\frac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}}}{{\frac{{1 + {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} = \frac{{\frac{2}{{\sin 2\alpha }}}}{{\frac{1}{{\cos 2\alpha }}}} = \frac{{2\cos 2\alpha }}{{\sin 2\alpha }} = 2\cot 2\alpha .\]
Example 14.
Simplify the trigonometric expression:
\[\frac{{{{\sin }^2}\alpha }}{{\sin \left( {\alpha - \beta } \right)}} + \frac{{{{\sin }^2}\beta }}{{\sin \left( {\beta - \alpha } \right)}}.\]
Solution.
The sine function is odd, that is, \(\sin \left( {\beta - \alpha } \right) = - \sin \left( {\alpha - \beta } \right).\) Hence,
\[A = \frac{{{{\sin }^2}\alpha }}{{\sin \left( {\alpha - \beta } \right)}} + \frac{{{{\sin }^2}\beta }}{{\sin \left( {\beta - \alpha } \right)}} = \frac{{{{\sin }^2}\alpha }}{{\sin \left( {\alpha - \beta } \right)}} - \frac{{{{\sin }^2}\beta }}{{\sin \left( {\alpha - \beta } \right)}} = \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{\sin \left( {\alpha - \beta } \right)}}.\]
Applying the product-to-sum identities in the numerator and double-angle formula in the denominator, we obtain:
\[A = \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{\sin \left( {\alpha - \beta } \right)}} = \frac{{\left( {\sin \alpha - \sin \beta } \right)\left( {\sin \alpha + \sin \beta } \right)}}{{\sin \left( {\alpha - \beta } \right)}} = \frac{{2\cos \frac{{\alpha + \beta }}{2}\cancel{{\sin \frac{{\alpha - \beta }}{2}}} \cdot \cancel{2}\sin \frac{{\alpha + \beta }}{2}\cancel{{\cos \frac{{\alpha - \beta }}{2}}}}}{{\cancel{2}\cancel{{\sin \frac{{\alpha - \beta }}{2}}}\cancel{{\cos \frac{{\alpha - \beta }}{2}}}}} = 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha + \beta }}{2} = \sin \left( {\alpha + \beta } \right).\]