Question

Find the value of $\sin \left( \alpha +\beta \right)$ if $\sin \alpha +\sin \beta =a$ and $\cos \alpha +\cos \beta =b$.
A) $\dfrac{2ab}{\left( {{a}^{2}}-{{b}^{2}} \right)}$
B) $\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}$
C) $\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}$
D) None of these

Answer 1

Hint: For solving this problem, first we square and add the given two equations to obtain a relationship in terms of $\left( \alpha -\beta \right)$ by applying suitable expansion of $\sin \theta \text{ and }\cos \theta$. Now, we add both the given equations and then square the whole value to obtain another relationship in terms of $\left( \alpha -\beta \right)$ by applying suitable expansion of $\sin \theta \text{ and }\cos \theta$. Now, we replace the value obtained in the first operation into the value obtained in the second operation to get the final result in terms of $\left( \alpha +\beta \right)$.

Complete step-by-step answer:
According to the problem statement, we are given two equations as:
$\begin{aligned} & \Rightarrow a=\sin \alpha +\sin \beta ...(1) \\\ & \Rightarrow b=\cos \alpha +\cos \beta ...(2) \\\ \end{aligned}$
Squaring both sides of both the equation (1) and equation (2) and then add both the equations with each other, we get

& \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( \sin \alpha +\sin \beta \right)}^{2}}+{{\left( \cos \alpha +\cos \beta \right)}^{2}} \\\ & \Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\left( \sin \alpha \right)\left( \sin \beta \right)+{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\left( \cos \alpha \right)\left( \cos \beta \right) \\\ \end{aligned}$$ Now, by using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get $$\begin{aligned} & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta +2\left( \sin \alpha \right)\left( \sin \beta \right)+2\left( \cos \alpha \right)\left( \cos \beta \right) \\\ & \Rightarrow 2+2\left( \sin \alpha \right)\left( \sin \beta \right)+2\left( \cos \alpha \right)\left( \cos \beta \right) \\\ & \Rightarrow 2\left[ 1+\left( \sin \alpha \right)\left( \sin \beta \right)+\left( \cos \alpha \right)\left( \cos \beta \right) \right] \\\ \end{aligned}$$ Now, by using the identity $$\left( \sin \alpha \right)\left( \sin \beta \right)+\left( \cos \alpha \right)\left( \cos \beta \right)=\cos \left( \alpha -\beta \right)$$, we get $${{a}^{2}}+{{b}^{2}}=2\left( 1+\cos \left( \alpha -\beta \right) \right)$$ Again, by using the identity $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$, we get $\Rightarrow {{a}^{2}}+{{b}^{2}}=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)$ Now, add both the equation (1) and (2) and then squaring both sides, $\begin{aligned} & \Rightarrow a+b=\sin \alpha +\sin \beta +\cos \alpha +\cos \beta \\\ & \Rightarrow {{\left( a+b \right)}^{2}}={{\left( \sin \alpha +\sin \beta +\cos \alpha +\cos \beta \right)}^{2}} \\\ \end{aligned}$ Now, let $A=\sin \alpha +\sin \beta \text{ and }B=\cos \alpha +\cos \beta $. By using the identity ${{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2AB$, we get $\Rightarrow {{\left( a+b \right)}^{2}}={{\left( \sin \alpha +\sin \beta \right)}^{2}}+{{\left( \cos \alpha +\cos \beta \right)}^{2}}+2\left( \sin \alpha +\sin \beta \right)\left( \cos \alpha +\cos \beta \right)$ Now, by using the previous result, we get $\begin{aligned} & \Rightarrow {{\left( \sin \alpha +\sin \beta \right)}^{2}}+{{\left( \cos \alpha +\cos \beta \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right) \\\ & \Rightarrow 4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)+2\left( \sin \alpha +\sin \beta \right)\left( \cos \alpha +\cos \beta \right)\ldots (3) \\\ \end{aligned}$ By using the identity $\begin{aligned} & \sin \alpha +\sin \beta =2\left[ \sin \left( \dfrac{\alpha +\beta }{2} \right) \right]\left[ \cos \left( \dfrac{\alpha -\beta }{2} \right) \right]\text{and }\cos \alpha +\cos \beta =2\left[ \cos \left( \dfrac{\alpha +\beta }{2} \right) \right]\left[ \cos \left( \dfrac{\alpha -\beta }{2} \right) \right] \\\ & 2\left( \sin \alpha +\sin \beta \right)\left( \cos \alpha +\cos \beta \right)=8{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right)\times \sin \left( \dfrac{\alpha +\beta }{2} \right) \\\ \end{aligned}$ Now, replacing the above value in equation (3), we get $\Rightarrow {{\left( a+b \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)+8{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)\times \sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right)$ Then, use the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand the left-hand side: $\Rightarrow {{a}^{2}}+{{b}^{2}}+2ab=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)+8{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)\times \sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right)$ The value of ${{a}^{2}}+{{b}^{2}}=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)$ as calculated above. Now, putting in the left-hand side, we get $\Rightarrow 4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)+2ab=4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)+8{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)\times \sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right)$ Cancel out $4{{\cos }^{2}}\dfrac{\alpha -\beta }{2}$ from both the sides, $\Rightarrow 2ab=2\times 4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)\times \sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right)$ Replacing, the value of $4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)=\left( {{a}^{2}}+{{b}^{2}} \right)$ in the above expression, we get $$\begin{aligned} & \Rightarrow 2ab=\left( {{a}^{2}}+{{b}^{2}} \right)\times 2\sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right) \\\ & \Rightarrow \dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}=2\sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right) \\\ & \Rightarrow \dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}=\sin \left( \alpha +\beta \right)\text{ }\left[ \because \sin \left( \alpha +\beta \right)=2\sin \left( \dfrac{\alpha +\beta }{2} \right)\times \cos \left( \dfrac{\alpha +\beta }{2} \right) \right] \\\ \end{aligned}$$ Hence, the value of $\sin \left( \alpha +\beta \right)=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}$. Therefore, the correct option is (C). Note: This problem can be alternatively solved by using the triangle property. First expand $\sin \alpha +\sin \beta =a\text{ and }\cos \alpha +\cos \beta =b$ by using the angle to multiplication property$\left[ \sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\text{ and cos}\alpha \text{+cos}\beta \text{=}2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right) \right]$. Now, take the ratio of both the quantities to obtain $\tan \left( \dfrac{\alpha +\beta }{2} \right)$. Now, by constructing a triangle having an angle $\left( \dfrac{\alpha +\beta }{2} \right)$, we can easily evaluate the value of $\sin \left( \alpha +\beta \right)$.