Precalculus

Trigonometry

Trigonometry Logo

Addition and Subtraction Formulas for Sine and Cosine

Solved Problems

Click or tap a problem to see the solution.

Example 1

Calculate cos (5π/12).

Example 2

Calculate sin (π/12).

Example 3

Find the exact value of \(\sin \frac{{7\pi }}{{5}}.\)

Example 4

Find the exact value of \(\cos 105^\circ.\)

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.

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.

Example 7

Prove the identity

\(\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)\) \( = {\cos ^2}\alpha - {\sin ^2}\beta.\)

Example 8

Prove the identity

\(\sin \left( {\alpha + \beta } \right) \sin \left( {\alpha - \beta } \right)\) \(= {\sin ^2}\alpha - {\sin ^2}\beta.\)

Example 9

Find the greatest and least value of \(\sin\alpha + \cos\alpha.\)

Example 10

Find the greatest and least value of \(\sin\beta - \sqrt{3}\cos\beta.\)

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 \(-2.\)

Page 1 Page 2