# 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

$\frac{{7\pi }}{{12}} = \frac{{4\pi + 3\pi }}{{12}} = \frac{{4\pi }}{{12}} + \frac{{3\pi }}{{12}} = \frac{\pi }{3} + \frac{\pi }{4}.$

$\sin \frac{{7\pi }}{{12}} = \sin \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right) = \sin \frac{\pi }{3}\cos \frac{\pi }{4} + \cos \frac{\pi }{3}\sin \frac{\pi }{4} = \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} + \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 }}{4} + \frac{{\sqrt 2 }}{4} = \frac{{\sqrt 6 + \sqrt 2 }}{4}.$

### 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 $$-2.$$