Evaluating Trigonometric Expressions
Solved Problems
Example 9.
Evaluate the trigonometric expression:
\[\sin 18^\circ \sin 54^\circ .\]
Solution.
We divide and multiply this expression by \(2\cos 18^\circ.\) Then using the double-angle formula for sine, we get
\[A = \sin 18^\circ \sin 54^\circ = \frac{{\overbrace {2\cos 18^\circ \sin 18^\circ }^{\sin 36^\circ }\sin 54^\circ }}{{2\cos 18^\circ }} = \frac{{\sin 36^\circ \sin 54^\circ }}{{2\cos 18^\circ }}.\]
By the cofunction identity, \(\sin 54^\circ = \cos 36^\circ .\) So
\[A = \frac{{\sin 36^\circ \cos 36^\circ }}{{2\cos 18^\circ }} = \frac{{2\sin 36^\circ \cos 36^\circ }}{{4\cos 18^\circ }} = \frac{{\sin 72^\circ }}{{4\cos 18^\circ }}.\]
Again,\(\sin 72^\circ = \cos 18^\circ .\) Hence,
\[A = \frac{{\sin 72^\circ }}{{4\cos 18^\circ }} = \frac{{\cancel{{\cos 18^\circ }}}}{{4\cancel{{\cos 18^\circ }}}} = \frac{1}{4}.\]
Example 10.
Calculate:
\[\sin \frac{\pi }{{10}} + \sin \frac{{13\pi }}{{10}}.\]
Solution.
First we use the sum-to-product identity:
\[A = \sin \frac{\pi }{{10}} + \sin \frac{{13\pi }}{{10}} = 2\sin \frac{{\frac{\pi }{{10}} + \frac{{13\pi }}{{10}}}}{2}\cos \frac{{\frac{\pi }{{10}} - \frac{{13\pi }}{{10}}}}{2} = 2\sin \frac{{14\pi }}{{20}}\cos \left( { - \frac{{12\pi }}{{20}}} \right) = 2\sin \frac{{7\pi }}{{10}}\cos \left( { - \frac{{6\pi }}{{10}}} \right).\]
The cosine is a periodic function, so we can write:
\[A = 2\sin \frac{{7\pi }}{{10}}\cos \left( { - \frac{{6\pi }}{{10}} + 2\pi } \right) = 2\sin \frac{{7\pi }}{{10}}\cos \frac{{14\pi }}{{10}}.\]
Now we divide and multiply the latter expression by \(\cos \frac{{7\pi }}{{10}}\) and apply the double-angle formula:
\[A = \frac{{\overbrace {2\sin \frac{{7\pi }}{{10}}\cos \frac{{7\pi }}{{10}}}^{\sin \frac{{14\pi }}{{10}}}\cos \frac{{14\pi }}{{10}}}}{{\cos \frac{{7\pi }}{{10}}}} = \frac{{\sin \frac{{14\pi }}{{10}}\cos \frac{{14\pi }}{{10}}}}{{\cos \frac{{7\pi }}{{10}}}} = \frac{{2\sin \frac{{14\pi }}{{10}}\cos \frac{{14\pi }}{{10}}}}{{2\cos \frac{{7\pi }}{{10}}}} = \frac{{\sin \frac{{28\pi }}{{10}}}}{{2\cos \frac{{7\pi }}{{10}}}} = \frac{{\sin \left( {\frac{{8\pi }}{{10}} + 2\pi } \right)}}{{2\cos \frac{{7\pi }}{{10}}}} = \frac{{\sin \frac{{8\pi }}{{10}}}}{{2\cos \frac{{7\pi }}{{10}}}}.\]
Recall the cofunction identity \(\sin \left( {\frac{{3\pi }}{2} - \beta } \right) = - \cos \beta .\) Then
\[A = \frac{{\sin \frac{{8\pi }}{{10}}}}{{2\cos \frac{{7\pi }}{{10}}}} = \frac{{\sin \left( {\frac{{15\pi }}{{10}} - \frac{{7\pi }}{{10}}} \right)}}{{2\cos \frac{{7\pi }}{{10}}}} = \frac{{ - \cancel{{\cos \frac{{7\pi }}{{10}}}}}}{{2\cancel{{\cos \frac{{7\pi }}{{10}}}}}} = - \frac{1}{2}.\]
Example 11.
Calculate:
\[\cot \frac{{13\pi }}{{12}} - \cot \frac{{5\pi }}{{12}}.\]
Solution.
The cotangent has a period of \(\pi.\) Therefore
\[\cot \frac{{13\pi }}{{12}} = \cot \left( {\frac{\pi }{{12}} + \pi } \right) = \cot \frac{\pi }{{12}}.\]
Using the cofunction identity, we rewrite \(\cot \frac{{5\pi }}{{12}}\) as
\[\cot \frac{{5\pi }}{{12}} = \cot \left( {\frac{\pi }{2} - \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{{12}}.\]
Then the original expression takes the form
\[A = \cot \frac{{13\pi }}{{12}} - \cot \frac{{5\pi }}{{12}} = \cot \frac{\pi }{{12}} - \tan \frac{\pi }{{12}}.\]
Applying the definition of tangent and cotangent and double-angle formulas, we get
\[A = \frac{{\cos \frac{\pi }{{12}}}}{{\sin \frac{\pi }{{12}}}} - \frac{{\sin \frac{\pi }{{12}}}}{{\cos \frac{\pi }{{12}}}} = \frac{{{{\cos }^2}\frac{\pi }{{12}} - {{\sin }^2}\frac{\pi }{{12}}}}{{\sin \frac{\pi }{{12}}\cos \frac{\pi }{{12}}}} = \frac{{\cos \frac{\pi }{6}}}{{\frac{1}{2}\sin \frac{\pi }{6}}} = 2\cot \frac{\pi }{6} = 2\sqrt 3 .\]
Example 12.
Calculate \({\sin ^3}\alpha - {\cos ^3}\alpha \) if
\[\sin \alpha - \cos \alpha = p.\]
Solution.
We factor the difference of cubes:
\[{\sin ^3}\alpha - {\cos ^3}\alpha = \left( {\sin \alpha - \cos \alpha } \right)\left( {{{\sin }^2}\alpha + \sin \alpha \cos \alpha + {{\cos }^2}\alpha } \right) = \left( {\sin \alpha - \cos \alpha } \right)\left( {1 + \sin \alpha \cos \alpha } \right).\]
To find \(\sin \alpha \cos \alpha,\) we square the expression for \(p:\)
\[{p^2} = {\left( {\sin \alpha - \cos \alpha } \right)^2} = {\sin ^2}\alpha - 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = 1 - 2\sin \alpha \cos \alpha .\]
Consequently,
\[\sin \alpha \cos \alpha = \frac{{1 - {p^2}}}{2}.\]
As a result, we have
\[{\sin ^3}\alpha - {\cos ^3}\alpha = \left( {\sin \alpha - \cos \alpha } \right)\left( {1 + \sin \alpha \cos \alpha } \right) = p\left( {1 + \frac{{1 - {p^2}}}{2}} \right) = p \cdot \frac{{2 + 1 - {p^2}}}{2} = \frac{{p\left( {3 - {p^2}} \right)}}{2} = \frac{{3p - {p^3}}}{2}.\]
Example 13.
Calculate \({\sin ^6}\alpha + {\cos ^6}\alpha \) if
\[\sin \alpha + \cos \alpha = m.\]
Solution.
We apply the cube of the sum formula:
\[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}.\]
Hence
\[1 = {1^3} = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} = {\sin ^6}\alpha + 3\,{\sin ^4}\alpha \,{\cos ^2}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^4}\alpha + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha \underbrace {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}_1 + {\cos ^6}\alpha = {\sin ^6}\alpha + 3\,{\sin ^2}\alpha\, {\cos ^2}\alpha + {\cos ^6}\alpha .\]
It follows from here that
\[{\sin ^6}\alpha + {\cos ^6}\alpha = 1 - 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha .\]
Let's determine \({\sin ^2}\alpha\, {\cos ^2}\alpha :\)
\[\sin \alpha + \cos \alpha = m, \Rightarrow {m^2} = {\left( {\sin \alpha + \cos \alpha } \right)^2} = {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = 1 + 2\sin \alpha \cos \alpha .\]
Therefore
\[\sin \alpha \cos \alpha = \frac{{{m^2} - 1}}{2}, \Rightarrow {\sin ^2}\alpha \,{\cos ^2}\alpha = \frac{{{{\left( {{m^2} - 1} \right)}^2}}}{4}.\]
Thus,
\[{\sin ^6}\alpha + {\cos ^6}\alpha = 1 - 3\,{\sin ^2}\alpha \,{\cos ^2}\alpha = 1 - \frac{{3{{\left( {{m^2} - 1} \right)}^2}}}{4} = \frac{{4 - 3\left( {{m^4} - 2{m^2} + 1} \right)}}{4} = \frac{{4 - 3{m^4} + 6{m^2} - 3}}{4} = \frac{{1 + 6{m^2} - 3{m^4}}}{4}.\]
Example 14.
Evaluate the trigonometric expression:
\[\cos 260^\circ \sin 130^\circ \cos 160^\circ .\]
Solution.
We divide and multiply this expression by \(2\cos 130^\circ\) and apply the double-angle formula for sine:
\[A = \cos 260^\circ \sin 130^\circ \cos 160^\circ = \frac{{\overbrace {2\cos 130^\circ \sin 130^\circ }^{\sin 260^\circ }\cos 260^\circ \cos 160^\circ }}{{2\cos 130^\circ }} = \frac{{\sin 260^\circ \cos 260^\circ \cos 160^\circ }}{{2\cos 130^\circ }} = \frac{{\overbrace {2\sin 260^\circ \cos 260^\circ }^{\sin 520^\circ }\cos 160^\circ }}{{4\cos 130^\circ }} = \frac{{\sin 520^\circ \cos 160^\circ }}{{4\cos 130^\circ }}.\]
The sine function is periodic with period \(360^\circ.\) Therefore
\[\sin 520^\circ = \sin \left( {160^\circ + 360^\circ } \right) = \sin 160^\circ .\]
This yields:
\[A = \frac{{\sin 520^\circ \cos 160^\circ }}{{4\cos 130^\circ }} = \frac{{\sin 160^\circ \cos 160^\circ }}{{4\cos 130^\circ }} = \frac{{2\sin 160^\circ \cos 160^\circ }}{{8\cos 130^\circ }} = \frac{{\sin 320^\circ }}{{8\cos 130^\circ }}.\]
Again, using properties of the sine function and cofunction identity, we have
\[\sin 320^\circ = \sin \left( {360^\circ - 40^\circ } \right) = \sin \left( { - 40^\circ } \right) = - \sin 40^\circ ;\]
\[\cos 130^\circ = \cos \left( {90^\circ + 40^\circ } \right) = - \sin 40^\circ .\]
The answer is
\[A = \frac{{\sin 320^\circ }}{{8\cos 130^\circ }} = \frac{{\cancel{{ - \sin 40^\circ }}}}{{8\cancel{{\left( { - \sin 40^\circ } \right)}}}} = \frac{1}{8}.\]
