Question

If
$\sin \alpha + \sin \beta = a \\
\cos \alpha + \cos \beta = b \\ $
Then prove that:
$\left( i \right)\cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}} \\
\left( {ii} \right)\sin \left( {\alpha + \beta } \right) = \dfrac{{2ab}}{{{a^2} + {b^2}}} \\ $

Answer 1

Hint: In the first part of the question try proving it by first finding the numerator of the question and then the denominator and then divide both of them. In the second part of the question you can start with a different route to avoid more calculation. In this process you can use trigonometric identities and formulas.

Complete step by step solution:
We will start with proving the first part. We have been given that $\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$. We will have to use this given information along with trigonometric identities to prove $\left( i \right)\cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}$.
To start with our solution we will first start with what is given to us.
$\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$
Since in the question ${a^2}$ and ${b^2}$is given to us we will square the above two equations.
Therefore we get ${\left( {\sin \alpha + \sin \beta } \right)^2} = {a^2} - - - \left( 1 \right)$
And ${\left( {\cos \alpha + \cos \beta } \right)^2} = {b^2} - - - \left( 2 \right)$
We can also see ${a^2} + {b^2}$ in the given question, hence we will add equation (1) and (2)
Therefore we get ${a^2} + {b^2} = {\left( {\sin \alpha + \sin \beta } \right)^2} + {\left( {\cos \alpha + \cos \beta } \right)^2}$
Simplifying the above equation we get
${a^2} + {b^2} = \left( {{{\sin }^2}\alpha + 2\sin \alpha \sin \beta + {{\sin }^2}\beta } \right) + \left( {{{\cos }^2}\alpha + 2\cos \alpha \cos \beta + {{\cos }^2}\beta } \right)$
Solving further we get,
${a^2} + {b^2} = {\sin ^2}\alpha + 2\sin \alpha \sin \beta + {\sin ^2}\beta + {\cos ^2}\alpha + 2\cos \alpha \cos \beta + {\cos ^2}\beta $
$ \Rightarrow {a^2} + {b^2} = \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) + \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right)$
Using the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we get,
${a^2} + {b^2} = 1 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) + 1 \\
\Rightarrow {a^2} + {b^2} = 2 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) \\ $
Using the identity $\sin a\sin b + \cos a\cos b = \cos \left( {a - b} \right)$ we get,
${a^2} + {b^2} = 2 + 2\cos \left( {\alpha - \beta } \right) - - - \left( 3 \right)$
Now Subtracting equation (1) from equation (2)
${b^2} - {a^2} = {\left( {\cos \alpha + \cos \beta } \right)^2} - {\left( {\sin \alpha + \sin \beta } \right)^2}$
On further simplifying
\[{b^2} - {a^2} = \left( {{{\cos }^2}\alpha + 2\cos \alpha \cos \beta + {{\cos }^2}\beta } \right) - \left( {{{\sin }^2}\alpha + 2\sin \alpha \sin \beta + {{\sin }^2}\beta } \right)\]
\[ \Rightarrow {b^2} - {a^2} = {\cos ^2}\alpha + 2\cos \alpha \cos \beta + {\cos ^2}\beta - {\sin ^2}\alpha - 2\sin \alpha \sin \beta - {\sin ^2}\beta \]
$ \Rightarrow {b^2} - {a^2} = \left( {{{\cos }^2}\alpha - {{\sin }^2}\beta } \right) + 2\left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right) + \left( {{{\cos }^2}\beta - {{\sin }^2}\alpha } \right)$
$ \Rightarrow {b^2} - {a^2} = \left( {{{\cos }^2}\alpha - {{\sin }^2}\beta } \right) + 2\left( {\cos \left( {\alpha + \beta } \right)} \right) + \left( {{{\cos }^2}\beta - {{\sin }^2}\alpha } \right)$
On further simplifying,
$\Rightarrow {b^2} - {a^2} = 2\left( {\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)} \right) + 2\left( {\cos \left( {\alpha + \beta } \right)} \right) \\
\Rightarrow {b^2} - {a^2} = \cos \left( {\alpha + \beta } \right)\left( {2\cos \left( {\alpha - \beta } \right) + 2} \right) - - - \left( 4 \right) \\ $
Solving equation (3) and (4) we get
\[{b^2} - {a^2} = \cos \left( {\alpha + \beta } \right)\left( {{a^2} + {b^2}} \right) \\
\therefore\cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{\left( {{a^2} + {b^2}} \right)}} \\ \]
Hence \[\cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}\] is proved.

$(ii)\,\sin \left( {\alpha + \beta } \right) = \dfrac{{2ab}}{{{a^2} + {b^2}}}$
We will start with $\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$
$a = \sin \alpha + \sin \beta \\
a= 2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) - - - \left( 5 \right) \\ $
using $\sin \left( {a + b} \right) = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)$
Similarly,
$b = \cos \alpha + \cos \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) - - - \left( 6 \right)$
Dividing equation (5) and (6) we get
$\dfrac{a}{b} = \dfrac{{2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)}} = \tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)$
$ \Rightarrow \dfrac{a}{b} = \tan \left( {\dfrac{{\alpha + \beta }}{2}} \right) - - - \left( 7 \right)$
Also we know that
\[\sin \left( {\alpha + \beta } \right) = \dfrac{{2\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)}}\]
Using equation (7) we get
\[\sin \left( {\alpha + \beta } \right) = \dfrac{{2\left( {\dfrac{a}{b}} \right)}}{{1 + \left( {\dfrac{{{a^2}}}{{{b^2}}}} \right)}} \\
\therefore\sin \left( {\alpha + \beta } \right) = \dfrac{{2ab}}{{{a^2} + {b^2}}} \]
Hence we get our required result.

This proves $\sin \left( {\alpha + \beta } \right) = \dfrac{{2ab}}{{{a^2} + {b^2}}}$.

Note: If you notice carefully, both the parts are solved differently. The second part can also be solved in a somewhat similar process that is used in the first part but to avoid long calculations and trigonometric formulas, identities are always handy in this kind of situation. You should be able to find the correct spot to apply them.