# Addition and Subtraction Formulas for Sine and Cosine

### Cosine Subtraction Formula

Let $$\alpha, \beta$$ be two angles such that $$\alpha \gt \beta.$$ Take the following points on a unit circle: $$A\left( 0 \right),$$ $$M\left( \alpha \right),$$ $$N\left( \beta \right),$$ $$P\left( {\alpha – \beta } \right).$$

The coordinates of these points are

${A = A\left( {1,0} \right),\;\;}\kern0pt{M = M\left( {\cos \alpha ,\sin \alpha } \right),\;\;}\kern0pt{N = N\left( {\cos \beta ,\sin \beta } \right),\;\;}\kern0pt{P = P\left( {\cos \left( {\alpha – \beta } \right),\sin \left( {\alpha – \beta } \right)} \right).}$

Since $$\angle MON = \angle POA = \alpha – \beta,$$ the line segments $$\color{#cc00ff}{MN}$$ and $$\color{#0099ff}{AP}$$ have the same length:

$\left| \color{#cc00ff}{MN} \right| = \left| \color{#0099ff}{AP} \right|.$

The distance between two points on a plane is given by the formula

$d = \sqrt {{{\left( {{x_1} – {x_2}} \right)}^2} + {{\left( {{y_1} – {y_2}} \right)}^2}} .$

Therefore,

${{\left| \color{#cc00ff}{MN} \right|^2} }={ {\left( {{x_M} – {x_N}} \right)^2} + {\left( {{y_M} – {y_N}} \right)^2} }={ {\left( {\cos \alpha – \cos \beta } \right)^2} + {\left( {\sin \alpha – \sin \beta } \right)^2} }={ {\cos ^2}\alpha – 2\cos \alpha \cos \beta + {\cos ^2}\beta }+{ {\sin ^2}\alpha – 2\sin \alpha \sin \beta + {\sin ^2}\beta }={ \underbrace {{{\cos }^2}\alpha + {{\sin }^2}\alpha }_1 + \underbrace {{{\cos }^2}\beta + {{\sin }^2}\beta }_1 }-{ 2\left( {\cos \alpha \cos \beta }+{ \sin \alpha \sin \beta } \right) }={ 2 – 2\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right);}$

Similarly we find the distance $$\left| \color{#0099ff}{AP} \right|$$ squared:

${{\left| \color{#0099ff}{AP} \right|^2} }={ {\left( {{x_A} – {x_P}} \right)^2} + {\left( {{y_A} – {y_P}} \right)^2} }={ {\left( {1 – \cos \left( {\alpha – \beta } \right)} \right)^2} }+{ {\left( {0 – \sin \left( {\alpha – \beta } \right)} \right)^2} }={ 1 – 2\cos \left( {\alpha – \beta } \right) }+{ \underbrace {{{\cos }^2}\left( {\alpha – \beta } \right) + {{\sin }^2}\left( {\alpha – \beta } \right)}_1 }={ 2 – 2\cos \left( {\alpha – \beta } \right).}$

The cosine subtraction formula follows from the equality $${\left| \color{#cc00ff}{MN} \right|^2} = {\left| \color{#0099ff}{AP} \right|^2}:$$

Consider two points $$N\left( \beta \right)$$ and $$L\left( { – \beta } \right)$$ lying on the terminal sides of the angles $$\beta$$ and $$-\beta,$$ respectively.

These points are symmetric with respect to the $$x-$$axis. Therefore, they have the same $$x-$$coordinates. Their $$y-$$coordinates are equal in magnitude but opposite in sign. In other words, the cosine function is even and the sine function is odd:

Now, let’s go back to the cosine subtraction formula and replace $$\beta \to -\beta:$$

${\cos \left( {\alpha + \beta } \right) }={ \cos \alpha \cos \left( { – \beta } \right) }+{ \sin \alpha \sin \left( { – \beta } \right).}$

Since cosine is even and sine is odd, we get the cosine addition identity in the form

### Special Cases

Substitute $$\alpha = \large{\frac{\pi }{2}}\normalsize$$ in the cosine subtraction formula:

${\cos \left( {\frac{\pi }{2} – \beta } \right) }={ \cos \frac{\pi }{2}\cos \beta }+{ \sin \frac{\pi }{2}\sin \beta }={ 0 \cdot \cos \beta }+{ 1 \cdot \sin \beta }={ \sin \beta ,}$

that is,

Replace $$\beta \to \large{\frac{\pi }{2}}\normalsize – \beta$$ in the last expression:

${\sin \left( {\frac{\pi }{2} – \beta } \right) }={ \cos \left( {\frac{\pi }{2} – \left( {\frac{\pi }{2} – \beta } \right)} \right) }={ \cos \beta ,}$

or

Similarly, take the cosine addition formula and substitute $$\alpha = \large{\frac{\pi }{2}}\normalsize:$$

${\cos \left( {\frac{\pi }{2} + \beta } \right) }={ \cos \frac{\pi }{2}\cos \beta – \sin \frac{\pi }{2}\sin \beta }={ 0 \cdot \cos \beta – 1 \cdot \sin \beta }={ – \sin \beta .}$

We got the following identity:

By changing $$\beta \to -\beta$$ in the identity $$\sin \left( {\large{\frac{\pi }{2}}\normalsize – \beta } \right) = \cos \beta ,$$ we obtain

${\sin \left( {\frac{\pi }{2} + \beta } \right) }={ \cos \left( { – \beta } \right) }={ \cos \beta ,}$

that is,

### Sine Subtraction Formula

Using cofunction identities from the previous section, we derive the sine subtraction formula:

${\sin \left( {\alpha – \beta } \right) }={ \cos \left( {\frac{\pi }{2} – \left( {\alpha – \beta } \right)} \right) }={ \cos \left( {\left( {\frac{\pi }{2} – \alpha } \right) + \beta } \right) }={ \cos \left( {\frac{\pi }{2} – \alpha } \right)\cos \beta }-{ \sin \left( {\frac{\pi }{2} – \alpha } \right)\sin \beta }={ \sin \alpha \cos \beta – \cos \alpha \sin \beta .}$

Thus, we have

If we replace $$\beta \to -\beta$$ in this formula, we get the sine addition identity:

${\sin \left( {\alpha + \beta } \right) }={ \sin \alpha \cos \left( { – \beta } \right) }-{ \cos \alpha \sin \left( { – \beta } \right) }={ \sin \alpha \cos \beta + \cos \alpha \sin \beta .}$

Hence,

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate $$\cos \large{\frac{{5\pi }}{{12}}}\normalsize.$$

### Example 2

Calculate $$\sin \large{\frac{{\pi }}{{12}}}\normalsize.$$

### Example 3

Find the exact value of $$\sin \large{\frac{{7\pi }}{{5}}}\normalsize.$$

### Example 4

Find the exact value of $$\cos 105^\circ.$$

### Example 5

Determine the value of $$\cos \left( {\large{\frac{\pi }{3}}\normalsize + \alpha } \right)$$ if $$\sin \alpha = \large{\frac{1}{{\sqrt 3 }}}\normalsize$$ and the angle $$\alpha$$ lies in the $$1\text{st}$$ quadrant.

### Example 6

Determine the value of $$\sin \left( {\large{\frac{\pi }{4}}\normalsize – \beta } \right)$$ if $$\cos \beta = -\large{\frac{1}{2}}\normalsize$$ 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 $$\cos \large{\frac{{5\pi }}{{12}}}\normalsize.$$

Solution.

We represent the angle $$\large{\frac{{5\pi }}{{12}}}\normalsize$$ 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 $$\sin \large{\frac{{\pi }}{{12}}}\normalsize.$$

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 $$\sin \large{\frac{{7\pi }}{{5}}}\normalsize.$$

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 $$\cos \left( {\large{\frac{\pi }{3}}\normalsize + \alpha } \right)$$ if $$\sin \alpha = \large{\frac{1}{{\sqrt 3 }}}\normalsize$$ 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 $$\sin \left( {\large{\frac{\pi }{4}}\normalsize – \beta } \right)$$ if $$\cos \beta = -\large{\frac{1}{2}}\normalsize$$ 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:

${LHS }={ \cos \left( {\alpha + \beta } \right)\cos \left( {\alpha – \beta } \right) }={ \left( {\cos \alpha \cos \beta – \sin \alpha \sin \beta } \right)}\kern0pt{\cdot\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right) }={ {\cos ^2}\alpha \,{\cos ^2}\beta – {\sin ^2}\alpha\,{\sin ^2}\beta .}$

Recall that

${{\cos ^2}\beta = 1 – {\sin ^2}\beta \;\;\text{and}\;\;}\kern0pt{{\sin ^2}\alpha = 1 – {\cos ^2}\alpha .}$

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

${LHS }={ \sin \left( {\alpha + \beta } \right)\sin \left( {\alpha – \beta } \right) }={ \left( {\sin \alpha \cos \beta + \cos \alpha \sin \beta } \right)}\kern0pt{\cdot\left( {\sin \alpha \cos \beta – \cos \alpha \sin \beta } \right) }={ {\sin ^2}\alpha\, {\cos ^2}\beta – {\cos ^2}\alpha \,{\sin ^2}\beta .}$

Using the Pythagorean trig identities

${{\cos ^2}\alpha = 1 – {\sin ^2}\alpha ,\;\;}\kern0pt{{\cos ^2}\beta = 1 – {\sin ^2}\beta ,}$

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 $$\sin \left( {\alpha + \large{\frac{\pi }{4}}\normalsize} \right)$$ 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$$-2.$$