📂Functions

Sum and Difference Identities for Trigonometric Functions

Formulas

• Composition into sine

  $$A \cos \theta + B \sin \theta = C\sin(\theta + \phi)$$

  Here, $C = \sqrt{A^{2} + B^{2}}$, $\phi = \sin^{-1} \left( \dfrac{A}{\sqrt{A^{2} + B^{2}}} \right) = \cos^{-1} \left( \dfrac{B}{\sqrt{A^{2} + B^{2}}} \right)$ are given.

• Composition into cosine

  $$A \cos \theta + B \sin \theta = C\cos(\theta - \phi)$$

  Here, $C = \sqrt{A^{2} + B^{2}}$, $\phi = \sin^{-1} \left( \dfrac{B}{\sqrt{A^{2} + B^{2}}} \right) = \cos^{-1} \left( \dfrac{A}{\sqrt{A^{2} + B^{2}}} \right)$ are given.

Proof

Grouping two terms $A \cos \theta + B \sin \theta$ into $\sqrt{A^{2} + B^{2}}$,

$$A \cos \theta + B \sin \theta = \sqrt{A^{2} + B^{2}} \left( \dfrac{A}{\sqrt{A^{2} + B^{2}}}\cos \theta + \dfrac{B}{\sqrt{A^{2} + B^{2}}}\sin \theta \right)$$

Where $-1 \lt \dfrac{A}{\sqrt{A^{2} + B^{2}}} \lt 1$, let’s denote this value as $\sin \phi$.

$$\sin \phi = \dfrac{A}{\sqrt{A^{2} + B^{2}}}$$

Then, as $\sin^{2} \phi - 1 = \cos^{2} \phi$,

$$\sin^{2} \phi - 1 = \dfrac{A^{2}}{A^{2} + B^{2}} - \dfrac{A^{2} + B^{2}}{A^{2} + B^{2}} = \dfrac{B^{2}}{A^{2} + B^{2}} = \cos \phi$$

$$\implies \dfrac{B}{\sqrt{A^{2} + B^{2}}} = \cos \phi$$

Now, if we set $C = \sqrt{A^{2} + B^{2}}$, according to the addition formula of trigonometric functions,

$$A \cos \theta + B \sin \theta = C \left( \sin\phi \cos\theta + \cos\phi \sin\theta \right) = C\sin(\theta + \phi)$$