Proving Trigonometric Identities
Solved Problems
Example 9.
Prove the identity
\[\frac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha - 1}}{{{{\sin }^6}\alpha + {{\cos }^6}\alpha - 1}} = \frac{2}{3}.\]
Solution.
We take the Pythagorean identity and square it:
\[1 = {1^2} = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = {\sin ^4}\alpha + 2\,{\sin ^2}\alpha \,{\cos ^2}\alpha + {\cos ^4}\alpha .\]
Then the numerator in the \(LHS\) of original expression can be written as
\[{\sin ^4}\alpha + {\cos ^4}\alpha - 1 = - 2\,{\sin ^2}\alpha \,{\cos ^2}\alpha .\]
Similarly, we cube the Pythagorean identity to get:
\[1 = {1^3} = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} = {\sin ^6}\alpha + 3\,{\sin ^4}\alpha \,{\cos ^2}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^4}\alpha + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha \underbrace {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}_1 + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha + {\cos ^6}\alpha .\]
Hence, the denominator of the \(LHS\) is given by
\[{\sin ^6}\alpha + {\cos ^6}\alpha - 1 = - 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha .\]
Substitute this into the original equation:
\[LHS = \frac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha - 1}}{{{{\sin }^6}\alpha + {{\cos }^6}\alpha - 1}} = \frac{{ - 2\cancel{{{{\sin }^2}\alpha \,{{\cos }^2}\alpha }}}}{{ - 3\cancel{{{{\sin }^2}\alpha \,{{\cos }^2}\alpha }}}} = \frac{2}{3} = RHS.\]
Example 10.
Prove the trigonometric identity:
\[{\sin ^6}\alpha + {\cos ^6}\alpha = \frac{{5 + 3\cos 4\alpha }}{8}.\]
Solution.
We transform the left-hand side to express it in terms of the angle \(2\alpha.\) Note that
\[1 = {1^3} = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} = {\sin ^6}\alpha + 3\,{\sin ^4}\alpha \,{\cos ^2}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^4}\alpha + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha \underbrace {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}_1 + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha + {\cos ^6}\alpha .\]
Therefore
\[LHS = {\sin ^6}\alpha + {\cos ^6}\alpha = 1 - 3\,{\sin ^2}\alpha\, {\cos ^2}\alpha = 1 - \frac{3}{4} \cdot 4\,{\sin ^2}\alpha \,{\cos ^2}\alpha = 1 - \frac{3}{4}{\sin ^2}2\alpha .\]
Let's also express the right-hand side in terms of \(2\alpha\) using the double-angle identity:
\[RHS = \frac{{5 + 3\cos 4\alpha }}{8} = \frac{{5 + 3\left( {1 - 2\,{{\sin }^2}2\alpha } \right)}}{8} = \frac{{5 + 3 - 6\,{{\sin }^2}2\alpha }}{8} = 1 - \frac{3}{4}{\sin ^2}2\alpha .\]
As you can see, the left-hand side is equal to the right-hand side. Hence, the original identity is proved.
Example 11.
Prove the trigonometric identity:
\[{\cos ^2}\left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\frac{{2\pi }}{3} - \alpha } \right)\sin \left( {\alpha - \frac{\pi }{3}} \right) = \frac{3}{4}.\]
Solution.
By the cofunction identity,
\[\cos \left( {\frac{\pi }{2} - \alpha } \right) = \sin \alpha .\]
Transform the product of sines:
\[\sin \left( {\frac{{2\pi }}{3} - \alpha } \right)\sin \left( {\alpha - \frac{\pi }{3}} \right) = \frac{1}{2}\left[ {\cos \left( {\frac{{2\pi }}{3} - \alpha - \alpha + \frac{\pi }{3}} \right) - \cos \left( {\frac{{2\pi }}{3} - \cancel{\alpha } + \cancel{\alpha } - \frac{\pi }{3}} \right)} \right] = \frac{1}{2}\left[ {\cos \left( {\pi - 2\alpha } \right) - \frac{\pi }{3}} \right] = \frac{1}{2}\left( { - \cos 2\alpha - \frac{1}{2}} \right) = - \left( {\frac{1}{2}\cos 2\alpha + \frac{1}{4}} \right).\]
Then the left-hand side of the identity is written as
\[LHS = {\sin ^2}\alpha + \frac{1}{2}\cos 2\alpha + \frac{1}{4}.\]
Use the double-angle formula for cosine:
\[LHS = {\sin ^2}\alpha + \frac{1}{2}\left( {1 - 2{{\sin }^2}\alpha } \right) + \frac{1}{4} = \cancel{{{{\sin }^2}\alpha }} + \frac{1}{2} - \cancel{{{{\sin }^2}\alpha }} + \frac{1}{4} = \frac{3}{4} = RHS,\]
which proves the identity.
Example 12.
Prove the identity:
\[\frac{{\sqrt {\cot \alpha } + \sqrt {\tan \alpha } }}{{\sqrt {\cot \alpha } - \sqrt {\tan \alpha } }} = \cot \left( {\frac{\pi }{4} - \alpha } \right).\]
Solution.
Using some algebra and trig identities, simplify the left-hand side:
\[LHS = \frac{{\sqrt {\cot \alpha } + \sqrt {\tan \alpha } }}{{\sqrt {\cot \alpha } - \sqrt {\tan \alpha } }} = \frac{{{{\left( {\sqrt {\cot \alpha } + \sqrt {\tan \alpha } } \right)}^2}}}{{\left( {\sqrt {\cot \alpha } - \sqrt {\tan \alpha } } \right)\left( {\sqrt {\cot \alpha } + \sqrt {\tan \alpha } } \right)}} = \frac{{\cot \alpha + 2\sqrt {\cot \alpha \tan \alpha } + \tan \alpha }}{{\cot \alpha - \tan \alpha }} = \frac{{\cot \alpha + 2 + \tan \alpha }}{{\cot \alpha - \tan \alpha }} = \frac{{\frac{{\cos \alpha }}{{\sin \alpha }} + 2 + \frac{{\sin \alpha }}{{\cos \alpha }}}}{{\frac{{\cos \alpha }}{{\sin \alpha }} - \frac{{\sin \alpha }}{{\cos \alpha }}}} = \frac{{\frac{{{{\cos }^2}\alpha + 2\sin \alpha \cos \alpha + {{\sin }^2}\alpha }}{{\cancel{{\sin \alpha \cos \alpha }}}}}}{{\frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\cancel{{\sin \alpha \cos \alpha }}}}}} = \frac{{1 + \sin 2\alpha }}{{\cos 2\alpha }}.\]
Using the half-angle formula and cofunction identities, we represent the right-hand side as
\[RHS = \cot \left( {\frac{\pi }{4} - \alpha } \right) = \frac{{1 + \cos \left( {\frac{\pi }{2} - 2\alpha } \right)}}{{\sin \left( {\frac{\pi }{2} - 2\alpha } \right)}} = \frac{{1 + \sin 2\alpha }}{{\cos 2\alpha }} = LHS.\]
The original identity is proved as \(RHS = LHS.\)
Example 13 .
Prove the identity
\[4\cos \left( {\frac{\pi }{6} - \alpha } \right)\sin \left( {\frac{\pi }{3} - \alpha } \right) = \frac{{\sin 3\alpha }}{{\sin \alpha }}.\]
Solution.
Using the sine and cosine subtraction formulas we represent the functions in the left-hand side as follows:
\[\cos \left( {\frac{\pi }{6} - \alpha } \right) = \cos \frac{\pi }{6}\cos \alpha + \sin \frac{\pi }{6}\sin \alpha = \frac{{\sqrt 3 }}{2}\cos \alpha + \frac{1}{2}\sin \alpha ,\]
\[\sin \left( {\frac{\pi }{3} - \alpha } \right) = \sin \frac{\pi }{3}\cos \alpha - \cos \frac{\pi }{3}\sin \alpha = \frac{{\sqrt 3 }}{2}\cos \alpha - \frac{1}{2}\sin \alpha .\]
Then
\[LHS = 4\cos \left( {\frac{\pi }{6} - \alpha } \right)\sin \left( {\frac{\pi }{3} - \alpha } \right) = 4\left( {\frac{{\sqrt 3 }}{2}\cos \alpha + \frac{1}{2}\sin \alpha } \right)\left( {\frac{{\sqrt 3 }}{2}\cos \alpha - \frac{1}{2}\sin \alpha } \right) = 4\left[ {{{\left( {\frac{{\sqrt 3 }}{2}\cos \alpha } \right)}^2} - {{\left( {\frac{1}{2}\sin \alpha } \right)}^2}} \right] = 4\left( {\frac{3}{4}{{\cos }^2}\alpha - \frac{1}{4}{{\sin }^2}\alpha } \right) = 3\,{\cos ^2}\alpha - {\sin ^2}\alpha .\]
By the Pythagorean trigonometric identity,
\[LHS = 3\,{\cos ^2}\alpha - {\sin ^2}\alpha = 3\left( {1 - {{\sin }^2}\alpha } \right) - {\sin ^2}\alpha = 3 - 4\,{\sin ^2}\alpha .\]
Consider now the right-hand side \(\left( {RHS} \right)\) and apply the triple-angle formula for sine:
\[RHS = \frac{{\sin 3\alpha }}{{\sin \alpha }} = \frac{{3\sin \alpha - 4\,{{\sin }^3}\alpha }}{{\sin \alpha }} = 3 - 4\,{\sin ^2}\alpha .\]
Since \(LHS = RHS,\) the identity is proved.