Cofunction and Reduction Identities
Solved Problems
Example 1.
Calculate \(\cos 840^\circ.\)
Solution.
As \({840^\circ} = {120^\circ} + {720^\circ},\) and cosine has a period of \(360^\circ,\) we have
\[\cos {840^\circ} = \cos \left( {{{120}^\circ} + {{720}^\circ}} \right) = \cos \left( {{{120}^\circ} + {{360}^\circ} \times 2} \right) = \cos {120^\circ} = \cos \left( {{{90}^\circ} + {{30}^\circ}} \right).\]
Now we use the identity \(\cos \left( {{{90}^\circ} + \beta } \right) = - \sin \beta :\)
\[\cos {840^\circ} = \cos \left( {{{90}^\circ} + {{30}^\circ}} \right) = - \sin {30^\circ} = - \frac{1}{2}.\]
Example 2.
Calculate \(\cot \left( { - {{765}^\circ}} \right).\)
Solution.
The angle \({ - {{765}^\circ}}\) can be expressed as \( - {765^\circ} = 135^\circ - 900^\circ.\) Therefore,
\[\cot \left( { - {{765}^\circ}} \right) = \cot \left( {{{135}^\circ} - {{900}^\circ}} \right) = \cot \left( {{{135}^\circ} - {{180}^\circ} \times 5} \right) = \cot {135^\circ} = \cot \left( {{{90}^\circ} + {{45}^\circ}} \right).\]
Using the identity \(\cot \left( {{{90}^\circ} + \beta } \right) = - \tan \beta ,\) we get
\[\cot \left( { - {{765}^\circ}} \right) = \cot \left( {{{90}^\circ} + {{45}^\circ}} \right) = - \tan {45^\circ} = - 1.\]
Example 3.
Find the value of the function \(\sin \frac{{31\pi }}{6}.\)
Solution.
The angle \(\frac{{31\pi }}{6}\) can be written as
\[\frac{{31\pi }}{6} = \frac{{7\pi + 24\pi }}{6} = \frac{{7\pi }}{6} + \frac{{24\pi }}{6} = \frac{{7\pi }}{6} + 4\pi .\]
Since the sine is a periodic function, with period \(2\pi,\) we have
\[\sin \frac{{31\pi }}{6} = \sin \left( {\frac{{7\pi }}{6} + 4\pi } \right) = \sin \left( {\frac{{7\pi }}{6} + 2\pi \cdot 2} \right) = \sin \frac{{7\pi }}{6} = \sin \left( {\pi + \frac{\pi }{6}} \right).\]
Using the reduction formula \(\sin \left( {\pi + \beta } \right) = - \sin \beta ,\) we obtain
\[\sin \frac{{31\pi }}{6} = \sin \left( {\pi + \frac{\pi }{6}} \right) = - \sin \frac{\pi }{6} = - \frac{1}{2}.\]
Example 4.
Determine the value of the function \(\tan \left( { - \frac{{21\pi }}{4}} \right).\)
Solution.
We represent the angle \({ - \frac{{23\pi }}{4}}\) in the form
\[ - \frac{{21\pi }}{4} = - \frac{{24\pi - 3\pi }}{4} = \frac{{3\pi }}{4} - \frac{{24\pi }}{4} = \frac{{3\pi }}{4} - 6\pi .\]
Then
\[\tan \left( { - \frac{{21\pi }}{4}} \right) = \tan \left( {\frac{{3\pi }}{4} - 6\pi } \right) = \tan \frac{{3\pi }}{4} = \tan \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right).\]
Finally, use the cofunction identity \(\tan \left( {\frac{\pi }{2} + \beta } \right) = - \cot \beta :\)
\[\tan \left( { - \frac{{21\pi }}{4}} \right) = \tan \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right) = - \cot \frac{\pi }{4} = - 1.\]
Example 5.
Simplify the expression \[\cos \left( {\pi - \alpha } \right)\tan \left( {\pi - \alpha } \right) - \sin \alpha .\]
Solution.
Applying the reduction formulas yields:
\[\cos \left( {\pi - \alpha } \right) = - \cos \alpha ,\;\;\tan \left( {\pi - \alpha } \right) = - \tan \alpha .\]
Then
\[\require{cancel} \cos \left( {\pi - \alpha } \right)\tan \left( {\pi - \alpha } \right) - \sin \alpha = - \cos \alpha \cdot \left( { - \tan \alpha } \right) - \sin \alpha = \frac{{\cancel{\cos \alpha} \sin \alpha }}{\cancel{\cos \alpha }} - \sin \alpha = \cancel{\sin \alpha} - \cancel{\sin \alpha} = 0.\]
Example 6.
Simplify the expression \[\sin \frac{{3\pi }}{4}\cos \frac{\pi }{4} - \tan \frac{{2\pi }}{3}\cos \frac{{5\pi }}{6}.\]
Solution.
Using the cofunction and reduction identities, we get
\[\sin \frac{{3\pi }}{4} = \sin \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\]
\[\tan \frac{{2\pi }}{3} = \tan \frac{{4\pi }}{6} = \tan \frac{{3\pi + \pi }}{6} = \tan \left( {\frac{\pi }{2} + \frac{\pi }{6}} \right) = - \cot \frac{\pi }{6} = - \sqrt 3 ,\]
\[\cos \frac{{5\pi }}{6} = \cos \left( {\pi - \frac{\pi }{6}} \right) = - \cos \frac{\pi }{6} = - \frac{{\sqrt 3 }}{2}.\]
Substituting these values into the original expression yields:
\[\sin \frac{{3\pi }}{4}\cos \frac{\pi }{4} - \tan \frac{{2\pi }}{3}\cos \frac{{5\pi }}{6} = \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - \left( { - \sqrt 3 } \right) \cdot \left( { - \frac{{\sqrt 3 }}{2}} \right) = \frac{1}{2} - \frac{3}{2} = - \frac{2}{2} = - 1.\]
Example 7.
Simplify the expression \(\sin \left( {\frac{\pi }{2} - \alpha } \right)\cos \left( {\pi + \alpha } \right)\) \(+ \sin \left( {\pi + \alpha } \right)\sin \alpha .\)
Solution.
Since
\[\sin \left( {\frac{\pi }{2} - \alpha } \right) = \cos \alpha ,\]
\[\cos \left( {\pi + \alpha } \right) = - \cos \alpha ,\]
\[\sin \left( {\pi + \alpha } \right) = - \sin \alpha ,\]
we have
\[\sin \left( {\frac{\pi }{2} - \alpha } \right)\cos \left( {\pi + \alpha } \right) + \sin \left( {\pi + \alpha } \right)\sin \alpha = \cos \alpha \cdot \left( { - \cos \alpha } \right) + \left( { - \sin \alpha } \right) \cdot \sin \alpha = - {\cos ^2}\alpha - {\sin ^2}\alpha = - \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = - 1.\]
Example 8.
Simplify the expression \(\tan \left( {2\pi - \alpha } \right)\sin \left( {2\pi - \alpha } \right)\) \(- \cos \left( {\pi - \alpha } \right).\)
Solution.
We calculate each term separately. The tangent function is odd and has a period of \(\pi.\) Hence,
\[\tan \left( {2\pi - \alpha } \right) = \tan \left( { - \alpha } \right) = - \tan \alpha .\]
The sine function is odd and has a period 0f \(2\pi.\) Therefore, we can write
\[\sin \left( {2\pi - \alpha } \right) = \sin \left( { - \alpha } \right) = - \sin \alpha .\]
The cosine function \(\cos \left( {\pi - \alpha } \right)\) is represented as
\[\cos \left( {\pi - \alpha } \right) = - \cos \alpha .\]
Substituting these results, we have
\[\tan \left( {2\pi - \alpha } \right)\sin \left( {2\pi - \alpha } \right) - \cos \left( {\pi - \alpha } \right) = \left( { - \tan \alpha } \right) \cdot \left( { - \sin \alpha } \right) - \left( { - \cos \alpha } \right) = \frac{{\sin \alpha }}{{\cos \alpha }}\sin \alpha + \cos \alpha = \frac{{{{\sin }^2}\alpha }}{{\cos \alpha }} + \frac{{{{\cos }^2}\alpha }}{{\cos \alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha }} = \frac{1}{{\cos \alpha }} = \sec \alpha .\]
Example 9.
Prove that in any triangle \[\sin \left( {\alpha + \beta } \right) = \sin \gamma .\]
Solution.
The sum of all angles of a triangle is equal to \(180^\circ = \pi:\)
\[\alpha + \beta + \gamma = \pi ,\;\; \Rightarrow \alpha + \beta = \pi - \gamma .\]
Then
\[\sin \left( {\alpha + \beta } \right) = \sin \left( {\pi - \gamma } \right).\]
By the reduction formula, we find that
\[\sin \left( {\alpha + \beta } \right) = \sin \left( {\pi - \gamma } \right) = \sin \gamma .\]
Example 10.
Prove that in any triangle \[\tan \left( {\alpha + \beta } \right) = - \tan \gamma .\]
Solution.
Let \(\alpha, \beta, \gamma\) be the angles of a triangle. It is known that
\[\alpha + \beta + \gamma = \pi .\]
Hence,
\[\alpha + \beta = \pi - \gamma .\]
Using the reduction formula for tangent, we get
\[\tan \left( {\alpha + \beta } \right) = \tan \left( {\pi - \gamma } \right) = - \tan \gamma .\]
