Question

If $\sin \left(A\right)-\sin \left(B\right)=c$ and $\cos \left(A\right)-\cos \left(B\right)=d$ then find $cose{c}^2\left(\frac{A+B}{2}\right)$.

Answer 1

Step 1: Simplify $c$ and $d$.

In the question, an equation $\sin \left(A\right)-\sin \left(B\right)=c$ is given.

$$\begin{gathered} \sin \left(A\right)-\sin \left(B\right)=c \\ \Rightarrow 2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)=c\left[\because \sin \left(x\right)-\sin \left(y\right)=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right]...1 \end{gathered}$$

Another equation $\cos \left(A\right)-\cos \left(B\right)=d$ is also given.

$$\begin{gathered} \cos \left(A\right)-\cos \left(B\right)=d \\ \Rightarrow 2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)=d\left[\because \cos \left(C\right)-\cos \left(D\right)=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{D-C}{2}\right)\right]....2 \end{gathered}$$

Step 2: Divide both equations.

Dividing equation $1$ and equation $2$.

$$\begin{gathered} \frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)}=\frac{c}{d} \\ \Rightarrow \frac{-\cos \left({\displaystyle \frac{A+B}{2}}\right)}{\sin \left({\displaystyle \frac{A+B}{2}}\right)}=\frac{c}{d} \\ \Rightarrow \mathrm{co}t\left(\frac{A+B}{2}\right)=-\frac{c}{d}\left[\because \frac{\cos \left(x\right)}{\sin \left(x\right)}=cot\left(x\right)\right]....3 \end{gathered}$$

Step 3: Find the required value.

Using the identity ${\cot }^2\left(x\right)+1=cose{c}^2\left(x\right)$

$$\begin{gathered} \mathrm{co}{t}^2\left(\frac{A+B}{2}\right)+1=cose{c}^2\left(\frac{A+B}{2}\right) \\ \Rightarrow {\left(\frac{-c}{d}\right)}^2+1=cose{c}^2\left(\frac{A+B}{2}\right)\left[\text{From equation 3}\right] \\ \Rightarrow cose{c}^2\left(\frac{A+B}{2}\right)=1+\frac{c^2}{d^2} \end{gathered}$$

Therefore, the value of the given expression is $1+\frac{c^2}{d^2}$.

Hence, the value of the given expression $cose{c}^2\left(\frac{A+B}{2}\right)$ is $1+\frac{c^2}{d^2}$.