Proving Trigonometric Identities
Trigonometric identity is a mathematical expression involving trigonometric functions that is true for all values of the occurring variables from the common domain.
For example, the Pythagorean trigonometric identity
\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1\]
holds true for any angles \(\alpha \in \mathbb{R}.\)
However, the identity
\[{\tan ^2}\alpha + 1 = {\sec ^2}\alpha \]
is only true when \(\alpha \ne \frac{\pi }{2} + \pi n,\;n \in \mathbb{Z}.\)
To prove a trigonometric identity, you need to show that its left and right sides are equal. This can be done by transforming either the left side or the right side, or both at the same time (it is usually better to start with the more complex side). For the transformation you can use any valid trigonometric and algebraic formulas, which should be well remembered. Some useful tips you can apply to transform trigonometric expressions are given on the page Simplifying Trigonometric Expressions.
Solved Problems
Example 1.
Prove the identity
\[\cos 4\alpha - \sin 4\alpha \cot 2\alpha = - 1.\]
Solution.
We use the double-angle formulas to express all functions in terms of one angle \(2\alpha.\) This yields:
\[LHS = \cos 4\alpha - \sin 4\alpha \cot 2\alpha = {\cos ^2}2\alpha - {\sin ^2}2\alpha - 2\sin 2\alpha \cos 2\alpha \cdot \frac{{\cos 2\alpha }}{{\sin 2\alpha }} = {\cos ^2}2\alpha - {\sin ^2}2\alpha - 2\,{\cos ^2}2\alpha = - {\sin ^2}2\alpha - {\cos ^2}2\alpha .\]
By the Pythagorean trig identity,
\[LHS = - {\sin ^2}2\alpha - {\cos ^2}2\alpha = - \underbrace {\left( {{{\sin }^2}2\alpha + {{\cos }^2}2\alpha } \right)}_1 = - 1 = RHS.\]
Example 2.
Prove the identity:
\[\cos 4\alpha = 8\,{\cos ^4}\alpha - 8\,{\cos ^2}\alpha + 1.\]
Solution.
We transform the left-hand side applying twice the double-angle formula for cosine:
\[LHS = \cos 4\alpha = 2\,{\cos ^2}2\alpha - 1 = 2{\left( {2\,{{\cos }^2}\alpha - 1} \right)^2} - 1 = 2\left( {4\,{{\cos }^4}\alpha - 4\,{{\cos }^2}\alpha + 1} \right) - 1 = 8\,{\cos ^4}\alpha - 8\,{\cos ^2}\alpha + 2 - 1 = 8\,{\cos ^4}\alpha - 8\,{\cos ^2}\alpha + 1 = RHS.\]
We see that \(LHS = RHS.\) Hence, the identity is proved.
Example 3.
Prove the identity:
\[\frac{{3 - 4\cos 2\alpha + \cos 4\alpha }}{{3 + 4\cos 2\alpha + \cos 4\alpha }} = {\tan ^4}\alpha .\]
Solution.
We transform the left-hand side to express all functions in terms of \(2\alpha:\)
\[LHS = \frac{{3 - 4\cos 2\alpha + 2\,{{\cos }^2}2\alpha - 1}}{{3 + 4\cos 2\alpha + 2\,{{\cos }^2}2\alpha - 1}} = \frac{{2 - 4\cos 2\alpha + 2\,{{\cos }^2}2\alpha }}{{2 + 4\cos 2\alpha + 2\,{{\cos }^2}2\alpha }} = \frac{{\cancel{2}\left( {1 - 2\cos 2\alpha + {{\cos }^2}2\alpha } \right)}}{{\cancel{2}\left( {1 + 2\cos 2\alpha + {{\cos }^2}2\alpha } \right)}} = \frac{{1 - 2\cos 2\alpha + {{\cos }^2}2\alpha }}{{1 + 2\cos 2\alpha + {{\cos }^2}2\alpha }} = \frac{{{{\left( {1 - \cos 2\alpha } \right)}^2}}}{{{{\left( {1 + \cos 2\alpha } \right)}^2}}} = {\left( {\frac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}} \right)^2}.\]
Let's apply the double-angle formula for cosine again:
\[LHS = {\left( {\frac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}} \right)^2} = {\left( {\frac{{\cancel{1} - \cancel{1} + 2\,{{\sin }^2}\alpha }}{{\cancel{1} + 2\,{{\cos }^2}\alpha - \cancel{1}}}} \right)^2} = {\left( {\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}} \right)^2} = {\left( {{{\tan }^2}\alpha } \right)^2} = {\tan ^4}\alpha = RHS.\]
The identity is proved.
Example 4.
Prove that
\[\sin \left( {\alpha + 2\beta } \right) = \sin \alpha \]
if \(\cot \left( {\alpha + \beta } \right) = 0.\)
Solution.
If \(\cot \left( {\alpha + \beta } \right) = 0,\) then \(\cos \left( {\alpha + \beta } \right) = 0.\)
We rewrite the identity in the form
\[\sin \left( {\alpha + 2\beta } \right) - \sin \alpha = 0\]
and apply the difference formula for sine. Then the left-hand side is given by
\[LHS = \sin \left( {\alpha + 2\beta } \right) - \sin \alpha = \sin \left( {\alpha + 2\beta } \right) - \sin \alpha = 2\cos \frac{{\alpha + 2\beta + \alpha }}{2}\sin \frac{{\alpha + 2\beta - \alpha }}{2} = 2\cos \left( {\alpha + \beta } \right)\sin \beta .\]
As \(\cos \left( {\alpha + \beta } \right) = 0,\) it is obvious that
\[LHS = 2\cos \left( {\alpha + \beta } \right)\sin \beta = 0 \cdot \sin \beta \equiv 0 = RHS.\]
Example 5.
Prove the identity
\[\frac{{\cos 4\alpha + 1}}{{\cot \alpha - \tan \alpha }} = \frac{1}{2}\sin 4\alpha .\]
Solution.
Let's simplify the left-hand side \(\left( {LHS} \right)\) of the expression. Using the double-angle formulas and the Pythagorean trig identity, we get
\[LHS = \frac{{\cos 4\alpha + 1}}{{\cot \alpha - \tan \alpha }} = \frac{{{{\cos }^2}2\alpha - \cancel{{{{\sin }^2}2\alpha }} + \cancel{{{{\sin }^2}2\alpha }} + {{\cos }^2}2\alpha }}{{\frac{1}{{\tan \alpha }} - \tan \alpha }} = \frac{{2\,{{\cos }^2}2\alpha }}{{\frac{{1 - {{\tan }^2}\alpha }}{{\tan \alpha }}}} = {\cos ^2}2\alpha \cdot \underbrace {\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}}_{\tan 2\alpha } = {\cos ^2}2\alpha \tan 2\alpha = \frac{{{{\cos }^{\cancel{2}}}2\alpha \sin 2\alpha }}{{\cancel{{\cos 2\alpha }}}} = \cos 2\alpha \sin 2\alpha = \frac{1}{2}\sin 4\alpha .\]
We have proved that the left-hand side \(\left( {LHS} \right)\) is equal to the right-hand side \(\left( {RHS} \right).\) Therefore, the identity is true.
Example 6.
Prove the trigonometric identity:
\[\cot \alpha - \tan \alpha - 2\tan 2\alpha = 4\cot 4\alpha .\]
Solution.
The left-hand side looks more complex so we try to simplify it. Using the double-angle formula for cotangent, we can write:
\[LHS = \cot \alpha - \tan \alpha - 2\tan 2\alpha = \frac{1}{{\tan \alpha }} - \tan \alpha - 2\tan 2\alpha = \frac{{1 - {{\tan }^2}\alpha }}{{\tan \alpha }} - 2\tan 2\alpha = 2\cot 2\alpha - 2\tan 2\alpha .\]
Сontinue in the same way:
\[LHS = 2\cot 2\alpha - 2\tan 2\alpha = 2\left( {\frac{1}{{\tan 2\alpha }} - \tan 2\alpha } \right) = 2 \cdot \frac{{1 - {{\tan }^2}2\alpha }}{{\tan 2\alpha }} = 4 \cdot \frac{{1 - {{\tan }^2}2\alpha }}{{2\tan 2\alpha }} = 4\cot 4\alpha = RHS .\]
The identity is proved since \(LHS = RHS.\)
Example 7.
Prove the trigonometric identity:
\[\tan \left( {\alpha - \frac{{3\pi }}{4}} \right)\left( {1 - \sin 2\alpha } \right) = \cos 2\alpha .\]
Solution.
We express the tangent function in terms of \(2\alpha\) by the half-angle formula:
\[\tan \left( {\alpha - \frac{{3\pi }}{4}} \right) = \frac{{\sin \left( {2\alpha - \frac{{3\pi }}{2}} \right)}}{{1 + \cos \left( {2\alpha - \frac{{3\pi }}{2}} \right)}}.\]
Using the cofunction identities, we can represent the tangent as
\[\tan \left( {\alpha - \frac{{3\pi }}{4}} \right) = \frac{{ - \sin \left( {\frac{{3\pi }}{2} - 2\alpha } \right)}}{{1 + \cos \left( {\frac{{3\pi }}{2} - 2\alpha } \right)}} = \frac{{\cos 2\alpha }}{{1 + \sin 2\alpha }}.\]
As a result, we have
\[LHS = \tan \left( {\alpha - \frac{{3\pi }}{4}} \right)\left( {1 - \sin 2\alpha } \right) = \frac{{\cos 2\alpha \cancel{{\left( {1 + \sin 2\alpha } \right)}}}}{{\cancel{{1 + \sin 2\alpha }}}} = \cos 2\alpha = RHS.\]
Example 8.
Prove the trigonometric identity:
\[\frac{{1 - 2\,{{\sin }^2}\alpha }}{{1 + \sin 2\alpha }} = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }}.\]
Solution.
Using the Pythagorean identity and double-angle formula, we rewrite the right-hand side as follows:
\[RHS = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} = \frac{{1 - \frac{{\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{\sin \alpha }}{{\cos \alpha }}}} = \frac{{\frac{{\cos \alpha - \sin \alpha }}{{\cancel{{\cos \alpha }}}}}}{{\frac{{\cos \alpha + \sin \alpha }}{{\cancel{{\cos \alpha }}}}}} = \frac{{\cos \alpha - \sin \alpha }}{{\cos \alpha + \sin \alpha }} = \frac{{\left( {\cos \alpha - \sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right)}}{{{{\left( {\cos \alpha + \sin \alpha } \right)}^2}}} = \frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha + 2\sin \alpha \cos \alpha + {{\sin }^2}\alpha }} = \frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{1 + 2\sin \alpha \cos \alpha }} = \frac{{1 - 2\,{{\sin }^2}\alpha }}{{1 + \sin 2\alpha }} = LHS.\]
As \(RHS = LHS,\) the identity is proved.
See more problems on Page 2.
