Precalculus

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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: \(A\left( 0 \right),\) \(M\left( \alpha \right),\) \(N\left( \beta \right),\) \(P\left( {\alpha - \beta } \right).\)

Deriving the difference formula for cosine.
Figure 1.

The coordinates of these points are

\[A = A\left( {1,0} \right),\;\;M = M\left( {\cos \alpha ,\sin \alpha } \right),\;\;N = N\left( {\cos \beta ,\sin \beta } \right),\;\;P = P\left( {\cos \left( {\alpha - \beta } \right),\sin \left( {\alpha - \beta } \right)} \right).\]

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 subtraction formula

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.

Sine is odd, cosine is even.
Figure 2.

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:

Cosine and sine of a negative angle.

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

Cosine addition formula

Special Cases

Substitute \(\alpha = \frac{\pi }{2}\) 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,

Cofunction identity cos(pi/2-beta)=sin(beta)

Replace \(\beta \to \frac{\pi }{2} - \beta\) 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

Cofunction identity sin(pi/2-beta)=cos(beta)

Similarly, take the cosine addition formula and substitute \(\alpha = \frac{\pi }{2}:\)

\[\cos \left( {\frac{\pi }{2} + \beta } \right) = \cos \frac{\pi }{2}\cos \beta - \sin \frac{\pi }{2}\sin \beta = 0 \cdot \cos \beta - 1 \cdot \sin \beta = - \sin \beta .\]

We got the following identity:

Trig identity cos(pi/2+beta)=-sin(beta)

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

\[\sin \left( {\frac{\pi }{2} + \beta } \right) = \cos \left( { - \beta } \right) = \cos \beta ,\]

that is,

Trig identity sin(pi/2+beta) = cos(beta)

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 subtraction formula

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,

Sine addition formula

See solved problems on Page 2.

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