Trigonometric Functions and Identities
6.0 Transformation Formulae
6.0 Transformation Formulae
1. $\cos A + \cos B = 2\cos \frac{{A + B}}{2}\cos \frac{{A - B}}{2}$
Proof: Using identity $(1)$ and $(2)$ i.e., $$\cos (A + B) = \cos A\cos B - \sin A\sin B...(1)$$ and $$\cos (A - B) = \cos A\cos B + \sin A\sin B...(2)$$ Add $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} \cos (A + B) + \cos (A - B) = \cos A\cos B - \sin A\sin B + \cos A\cos B + \sin A\sin B \\ \cos (A + B) + \cos (A - B) = 2\cos A\cos B...(3) \\\end{aligned} \end{equation} $$ Put $A + B = \theta $ and $A - B = \phi $. Solving them, we get $$A = \frac{{\theta + \phi }}{2},B = \frac{{\theta - \phi }}{2}$$ Substitute the values of $A$ and $B$ in $(3)$, we get $$\begin{equation} \begin{aligned} \cos \left( {\frac{{\theta + \phi }}{2} + \frac{{\theta - \phi }}{2}} \right) + \cos \left( {\frac{{\theta + \phi }}{2} - \frac{{\theta - \phi }}{2}} \right) = 2\cos \left( {\frac{{\theta + \phi }}{2}} \right)\cos \left( {\frac{{\theta - \phi }}{2}} \right) \\ \cos (\theta ) + \cos (\phi ) = 2\cos \left( {\frac{{\theta + \phi }}{2}} \right)\cos \left( {\frac{{\theta - \phi }}{2}} \right) \\\end{aligned} \end{equation} $$ Since $\theta$ and $\phi$ can take any real values, replace $\theta$ by $A$ and $\phi$ by $B$, we get $$\cos A + \cos B = 2\cos \frac{{A + B}}{2}\cos \frac{{A - B}}{2}$$
2. $\cos A - \cos B = -2\sin \frac{{A + B}}{2}\sin \frac{{A - B}}{2}$
Proof: Using identity $(1)$ and $(2)$ i.e., $$\cos (A + B) = \cos A\cos B - \sin A\sin B...(1)$$ and $$\cos (A - B) = \cos A\cos B + \sin A\sin B...(2)$$ Subtracting $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} \cos (A + B) - \cos (A - B) = \cos A\cos B - \sin A\sin B - \cos A\cos B - \sin A\sin B \\ \cos (A + B) - \cos (A - B) = -2\sin A\sin B...(3) \\\end{aligned} \end{equation} $$ Put $A + B = \theta $ and $A - B = \phi $. Solving them, we get $$A = \frac{{\theta + \phi }}{2},B = \frac{{\theta - \phi }}{2}$$ Substitute the values of $A$ and $B$ in $(3)$, we get $$\begin{equation} \begin{aligned} \cos \left( {\frac{{\theta + \phi }}{2} + \frac{{\theta - \phi }}{2}} \right) - \cos \left( {\frac{{\theta + \phi }}{2} - \frac{{\theta - \phi }}{2}} \right) = -2\sin \left( {\frac{{\theta + \phi }}{2}} \right)\sin \left( {\frac{{\theta - \phi }}{2}} \right) \\ \cos (\theta ) - \cos (\phi ) = -2\sin \left( {\frac{{\theta + \phi }}{2}} \right)\sin \left( {\frac{{\theta - \phi }}{2}} \right) \\\end{aligned} \end{equation} $$ Since $\theta$ and $\phi$ can take any real values, replace $\theta$ by $A$ and $\phi$ by $B$, we get $$\cos A - \cos B = -2\sin \frac{{A + B}}{2}\sin \frac{{A - B}}{2}$$
3. $\sin A + \sin B = 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2}$
Proof: Using identity $(5)$ and $(6)$ i.e., $$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B...(5)$$ and $$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B...(6)$$ Add $(5)$ and $(6)$, we get $$\begin{equation} \begin{aligned} \sin (A + B) + \sin (A - B) = \sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B \\ \sin (A + B) + \sin (A - B) = 2\sin A\cos B...(7) \\\end{aligned} \end{equation} $$ Put $A + B = \theta $ and $A - B = \phi $. Solving them, we get $$A = \frac{{\theta + \phi }}{2},B = \frac{{\theta - \phi }}{2}$$ Substitute the values of $A$ and $B$ in $(7)$, we get $$\begin{equation} \begin{aligned} \sin \left( {\frac{{\theta + \phi }}{2} + \frac{{\theta - \phi }}{2}} \right) + \sin \left( {\frac{{\theta + \phi }}{2} - \frac{{\theta - \phi }}{2}} \right) = 2\sin \left( {\frac{{\theta + \phi }}{2}} \right)\cos \left( {\frac{{\theta - \phi }}{2}} \right) \\ \sin (\theta ) + \sin (\phi ) = 2\sin \left( {\frac{{\theta + \phi }}{2}} \right)\cos \left( {\frac{{\theta - \phi }}{2}} \right) \\\end{aligned} \end{equation} $$ Since $\theta$ and $\phi$ can take any real values, replace $\theta$ by $A$ and $\phi$ by $B$, we get $$\sin A + \sin B = 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2}$$
4. $\sin A - \sin B = 2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}$
Proof: Using identity $(5)$ and $(6)$ i.e., $$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B...(5)$$ and $$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B...(6)$$ Subtracting $(5)$ and $(6)$, we get $$\begin{equation} \begin{aligned} \sin (A + B) - \sin (A - B) = \sin A\cos B + \cos A\sin B - \sin A\cos B + \cos A\sin B \\ \sin (A + B) - \sin (A - B) = 2\cos A\sin B...(7) \\\end{aligned} \end{equation} $$ Put $A + B = \theta $ and $A - B = \phi $. Solving them, we get $$A = \frac{{\theta + \phi }}{2},B = \frac{{\theta - \phi }}{2}$$ Substitute the values of $A$ and $B$ in $(7)$, we get $$\begin{equation} \begin{aligned} \sin \left( {\frac{{\theta + \phi }}{2} + \frac{{\theta - \phi }}{2}} \right) - \sin \left( {\frac{{\theta + \phi }}{2} - \frac{{\theta - \phi }}{2}} \right) = 2\cos \left( {\frac{{\theta + \phi }}{2}} \right)\sin \left( {\frac{{\theta - \phi }}{2}} \right) \\ \sin (\theta ) - \sin (\phi ) = 2\cos \left( {\frac{{\theta + \phi }}{2}} \right)\sin \left( {\frac{{\theta - \phi }}{2}} \right) \\\end{aligned} \end{equation} $$ Since $\theta$ and $\phi$ can take any real values, replace $\theta$ by $A$ and $\phi$ by $B$, we get $$\sin A - \sin B = 2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}$$
By putting the values of $cos(A+B)$, $cos(A-B)$, $sin(A+B)$ and $sin(A-B)$, from the above explained identities, we get the following results
- $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
- $-2\sin A\sin B=\cos(A+B)-\cos(A-B)$
- $2\sin A\cos B=\sin(A+B)+\sin(A-B)$
- $2\cos A\sin B=\sin(A+B)-\sin(A-B)$
