Addition and Subtraction Formulas for Sine and Cosine
Solved Problems
Example 1.
Calculate \(\cos \frac{{5\pi }}{{12}}.\)
Solution.
We represent the angle \(\frac{{5\pi }}{{12}}\) as the sum of two angles:
\[\frac{{5\pi }}{{12}} = \frac{{3\pi + 2\pi }}{{12}} = \frac{{3\pi }}{{12}} + \frac{{2\pi }}{{12}} = \frac{\pi }{4} + \frac{\pi }{6}.\]
Using the cosine addition formula, we have
\[\cos \frac{{5\pi }}{{12}} = \cos \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) = \cos \frac{\pi }{4}\cos \frac{\pi }{6} - \sin \frac{\pi }{4}\sin \frac{\pi }{6} = \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2} \cdot \frac{1}{2} = \frac{{\sqrt 6 }}{4} - \frac{{\sqrt 2 }}{4} = \frac{{\sqrt 6 - \sqrt 2 }}{4}.\]
Example 2.
Calculate \(\sin \frac{{\pi }}{{12}}.\)
Solution.
Since
\[\frac{\pi }{{12}} = \frac{{4\pi - 3\pi }}{{12}} = \frac{{4\pi }}{{12}} - \frac{{3\pi }}{{12}} = \frac{\pi }{3} - \frac{\pi }{4},\]
then by the sine subtraction formula:
\[\sin \frac{\pi }{{12}} = \sin \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right) = \sin \frac{\pi }{3}\cos \frac{\pi }{4} - \cos \frac{\pi }{3}\sin \frac{\pi }{4} = \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 }}{4} - \frac{{\sqrt 2 }}{4} = \frac{{\sqrt 6 - \sqrt 2 }}{4}.\]
Notice that
\[\sin \frac{\pi }{{12}} = \cos \left( {\frac{\pi }{2} - \frac{\pi }{{12}}} \right) = \cos \frac{{6\pi - \pi }}{{12}} = \cos \frac{{5\pi }}{{12}} = \frac{{\sqrt 6 - \sqrt 2 }}{4}.\]
Example 3.
Find the exact value of \(\sin \frac{{7\pi }}{{5}}.\)
Solution.
We write the given angle in the form
\[\frac{{7\pi }}{{12}} = \frac{{4\pi + 3\pi }}{{12}} = \frac{{4\pi }}{{12}} + \frac{{3\pi }}{{12}} = \frac{\pi }{3} + \frac{\pi }{4}.\]
Use the sine addition identity:
\[\sin \frac{{7\pi }}{{12}} = \sin \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right) = \sin \frac{\pi }{3}\cos \frac{\pi }{4} + \cos \frac{\pi }{3}\sin \frac{\pi }{4} = \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} + \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 }}{4} + \frac{{\sqrt 2 }}{4} = \frac{{\sqrt 6 + \sqrt 2 }}{4}.\]
Example 4.
Find the exact value of \(\cos 105^\circ.\)
Solution.
We write the angle \(105^\circ\) as the sum of two angles:
\[{105^\circ} = {60^\circ} + {45^\circ}.\]
Therefore,
\[\cos {105^\circ} = \cos \left( {{{60}^\circ} + {{45}^\circ}} \right) = \cos {60^\circ}\cos {45^\circ} - \sin {60^\circ}\sin {45^\circ} = \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 - \sqrt 6 }}{4} = - \frac{{\sqrt 6 - \sqrt 2 }}{4}.\]
Example 5.
Determine the value of \(\cos \left( {\frac{\pi }{3} + \alpha } \right)\) if \(\sin \alpha = \frac{1}{{\sqrt 3 }}\) and the angle \(\alpha\) lies in the \(1\text{st}\) quadrant.
Solution.
The cosine function is positive in the \(1\text{st}\) quadrant. Therefore
\[\cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } = \sqrt {1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}} = \sqrt {1 - \frac{1}{3}} = \sqrt {\frac{2}{3}} = \frac{{\sqrt 2 }}{{\sqrt 3 }}.\]
Now we use the cosine addition formula:
\[\cos \left( {\frac{\pi }{3} + \alpha } \right) = \cos \frac{\pi }{3}\cos \alpha - \sin \frac{\pi }{3}\sin \alpha = \frac{1}{2} \cdot \frac{{\sqrt 2 }}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2} \cdot \frac{1}{{\sqrt 3 }} = \frac{{\sqrt 2 - \sqrt 3 }}{{2\sqrt 3 }} = \frac{{\sqrt 6 - 3}}{6}.\]
We can note that the cosine of this angle is negative.
Example 6.
Determine the value of \(\sin \left( {\frac{\pi }{4} - \beta } \right)\) if \(\cos \beta = -\frac{1}{2}\) and the angle \(\beta\) is in the \(2\text{nd}\) quadrant.
Solution.
The sine is positive in the \(2\text{nd}\) quadrant. Hence,
\[\sin \beta = \sqrt {1 - {{\sin }^2}\beta } = \sqrt {1 - {{\left( { - \frac{1}{2}} \right)}^2}} = \sqrt {1 - \frac{1}{4}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}.\]
Using the sine difference identity, we obtain
\[\sin \left( {\frac{\pi }{4} - \beta } \right) = \sin \frac{\pi }{4}\cos \beta - \cos \frac{\pi }{4}\sin \beta = \frac{{\sqrt 2 }}{2} \cdot \left( { - \frac{1}{2}} \right) - \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 3 }}{2} = - \frac{{\sqrt 2 - \sqrt 6 }}{4} = \frac{{\sqrt 6 - \sqrt 2 }}{4}.\]
Example 7.
Prove the identity
\(\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)\) \( = {\cos ^2}\alpha - {\sin ^2}\beta.\)
Solution.
Simplify the left-hand side \(LHS\) using the cosine addition and subtraction formulas:
\[LHS = \cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right) = \left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right) \left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right) = {\cos ^2}\alpha \,{\cos ^2}\beta - {\sin ^2}\alpha\,{\sin ^2}\beta .\]
Recall that
\[{\cos ^2}\beta = 1 - {\sin ^2}\beta \;\;\text{and}\;\;{\sin ^2}\alpha = 1 - {\cos ^2}\alpha .\]
Then
\[\require{cancel} LHS = {\cos ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) - \left( {1 - {{\cos }^2}\alpha } \right){\sin ^2}\beta = {\cos ^2}\alpha - \cancel{{\cos ^2}\alpha \,{\sin ^2}\beta } - {\sin ^2}\beta + \cancel{{\cos ^2}\alpha \,{\sin ^2}\beta} = {\cos ^2}\alpha - {\sin ^2}\beta = RHS.\]
Example 8.
Prove the identity
\(\sin \left( {\alpha + \beta } \right) \sin \left( {\alpha - \beta } \right)\) \(= {\sin ^2}\alpha - {\sin ^2}\beta.\)
Solution.
We write the left-hand side in the form
\[LHS = \sin \left( {\alpha + \beta } \right)\sin \left( {\alpha - \beta } \right) = \left( {\sin \alpha \cos \beta + \cos \alpha \sin \beta } \right) \left( {\sin \alpha \cos \beta - \cos \alpha \sin \beta } \right) = {\sin ^2}\alpha\, {\cos ^2}\beta - {\cos ^2}\alpha \,{\sin ^2}\beta .\]
Using the Pythagorean trig identities
\[{\cos ^2}\alpha = 1 - {\sin ^2}\alpha ,\;\;{\cos ^2}\beta = 1 - {\sin ^2}\beta ,\]
we get
\[LHS = {\sin ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) - \left( {1 - {{\sin }^2}\alpha } \right){\sin ^2}\beta = {\sin ^2}\alpha - \cancel{{\sin ^2}\alpha \,{\sin ^2}\beta} - {\sin ^2}\beta + \cancel{{\sin ^2}\alpha \,{\sin ^2}\beta} = {\sin ^2}\alpha - {\sin ^2}\beta = RHS.\]
Example 9.
Find the greatest and least value of \(\sin\alpha + \cos\alpha.\)
Solution.
We denote this expression by \(A.\) Using the auxiliary angle \(\frac{\pi }{4},\) we have
\[\frac{{A\sqrt 2 }}{2} = \sin \alpha \frac{{\sqrt 2 }}{2} + \cos \alpha \frac{{\sqrt 2 }}{2} = \sin \alpha \cos \frac{\pi }{4} + \cos \alpha \sin \frac{\pi }{4} = \sin \left( {\alpha + \frac{\pi }{4}} \right).\]
Hence,
\[A = \sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right).\]
The function \(\sin \left( {\alpha + \frac{\pi }{4}} \right)\) ranges from \(-1\) to \(1.\) Therefore, the maximum value of \(A\) is \(\sqrt{2},\) and the minimum value is \(-\sqrt{2}.\)
Example 10.
Find the greatest and least value of \(\sin\beta - \sqrt{3}\cos\beta.\)
Solution.
Let the value of this expression be denoted by \(B.\) We can represent it in the following form:
\[\frac{B}{2} = \frac{1}{2}\sin \beta - \frac{{\sqrt 3 }}{2}\cos \beta = \sin \frac{\pi }{6}\sin \beta - \cos \frac{\pi }{6}\cos \beta = - \left( {\cos \frac{\pi }{6}\cos \beta - \sin \frac{\pi }{6}\sin \beta } \right) = - \cos \left( {\frac{\pi }{6} + \beta } \right).\]
Then
\[B = - 2\cos \left( {\frac{\pi }{6} + \beta } \right).\]
The maximum value of cosine is \(1,\) and the minimum value is \(-1.\) Respectively, the greatest value of \(B\) is \(2,\) and the least value is \(-2.\)
