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 \({\left| \color{#cc00ff}{MN} \right|^2} = {\left| \color{#0099ff}{AP} \right|^2}:\)

Cosine Addition Formula

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

Sine Addition Formula

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.

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

\[{\cos \frac{{5\pi }}{{12}} }={ \cos \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) }={ \cos \frac{\pi }{4}\cos \frac{\pi }{6} – \sin \frac{\pi }{4}\sin \frac{\pi }{6} }={ \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 3 }}{2} – \frac{{\sqrt 2 }}{2} \cdot \frac{1}{2} }={ \frac{{\sqrt 6 }}{4} – \frac{{\sqrt 2 }}{4} }={ \frac{{\sqrt 6 – \sqrt 2 }}{4}.}\]

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

\[{\sin \frac{\pi }{{12}} }={ \cos \left( {\frac{\pi }{2} – \frac{\pi }{{12}}} \right) }={ \cos \frac{{6\pi – \pi }}{{12}} }={ \cos \frac{{5\pi }}{{12}} }={ \frac{{\sqrt 6 – \sqrt 2 }}{4}.}\]

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:

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

Solution.

We write the angle \(105^\circ\) as the sum of two angles:

\[{105^\circ} = {60^\circ} + {45^\circ}.\]

Therefore,

\[{\cos {105^\circ} = \cos \left( {{{60}^\circ} + {{45}^\circ}} \right) }={ \cos {60^\circ}\cos {45^\circ} – \sin {60^\circ}\sin {45^\circ} }={ \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} – \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} }={ \frac{{\sqrt 2 – \sqrt 6 }}{4} }={ – \frac{{\sqrt 6 – \sqrt 2 }}{4}.}\]

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.

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

\[{\sin \left( {\frac{\pi }{4} – \beta } \right) }={ \sin \frac{\pi }{4}\cos \beta – \cos \frac{\pi }{4}\sin \beta }={ \frac{{\sqrt 2 }}{2} \cdot \left( { – \frac{1}{2}} \right) – \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 3 }}{2} }={ – \frac{{\sqrt 2 – \sqrt 6 }}{4} }={ \frac{{\sqrt 6 – \sqrt 2 }}{4}.}\]

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

\[\require{cancel}{LHS }={ {\cos ^2}\alpha \left( {1 – {{\sin }^2}\beta } \right) }-{ \left( {1 – {{\cos }^2}\alpha } \right){\sin ^2}\beta }={ {\cos ^2}\alpha – \cancel{{\cos ^2}\alpha \,{\sin ^2}\beta }}-{ {\sin ^2}\beta + \cancel{{\cos ^2}\alpha \,{\sin ^2}\beta} }={ {\cos ^2}\alpha – {\sin ^2}\beta }={ RHS.}\]

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

\[{LHS }={ {\sin ^2}\alpha \left( {1 – {{\sin }^2}\beta } \right) }-{ \left( {1 – {{\sin }^2}\alpha } \right){\sin ^2}\beta }={ {\sin ^2}\alpha – \cancel{{\sin ^2}\alpha \,{\sin ^2}\beta} }-{ {\sin ^2}\beta + \cancel{{\sin ^2}\alpha \,{\sin ^2}\beta} }={ {\sin ^2}\alpha – {\sin ^2}\beta }={ RHS.}\]

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 \(-\sqrt{2}.\)

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\(-2.\)