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Question: If \(\sin \alpha + \sin \beta = a \\\ \cos \alpha + \cos \beta = b \\\ \) Then prove that: ...

If
sin⁡α+sin⁡β=a cos⁡α+cos⁡β=b \sin \alpha + \sin \beta = a \\\ \cos \alpha + \cos \beta = b \\\
Then prove that:
(i)cos⁡(α+β)=b2−a2b2+a2 (ii)sin⁡(α+β)=2aba2+b2 \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}}} \\\

Explanation

Solution

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⁡α+sin⁡β=a\sin \alpha + \sin \beta = a and cos⁡α+cos⁡β=b\cos \alpha + \cos \beta = b. We will have to use this given information along with trigonometric identities to prove (i)cos⁡(α+β)=b2−a2b2+a2\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⁡α+sin⁡β=a\sin \alpha + \sin \beta = a and cos⁡α+cos⁡β=b\cos \alpha + \cos \beta = b
Since in the question a2{a^2} and b2{b^2}is given to us we will square the above two equations.
Therefore we get (sin⁡α+sin⁡β)2=a2−−−(1){\left( {\sin \alpha + \sin \beta } \right)^2} = {a^2} - - - \left( 1 \right)
And (cos⁡α+cos⁡β)2=b2−−−(2){\left( {\cos \alpha + \cos \beta } \right)^2} = {b^2} - - - \left( 2 \right)
We can also see a2+b2{a^2} + {b^2} in the given question, hence we will add equation (1) and (2)
Therefore we get a2+b2=(sin⁡α+sin⁡β)2+(cos⁡α+cos⁡β)2{a^2} + {b^2} = {\left( {\sin \alpha + \sin \beta } \right)^2} + {\left( {\cos \alpha + \cos \beta } \right)^2}
Simplifying the above equation we get
a2+b2=(sin⁡2α+2sin⁡αsin⁡β+sin⁡2β)+(cos⁡2α+2cos⁡αcos⁡β+cos⁡2β){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,
a2+b2=sin⁡2α+2sin⁡αsin⁡β+sin⁡2β+cos⁡2α+2cos⁡αcos⁡β+cos⁡2β{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
⇒a2+b2=(sin⁡2α+cos⁡2α)+2(sin⁡αsin⁡β+cos⁡αcos⁡β)+(sin⁡2β+cos⁡2β)\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we get,
a2+b2=1+2(sin⁡αsin⁡β+cos⁡αcos⁡β)+1 ⇒a2+b2=2+2(sin⁡αsin⁡β+cos⁡αcos⁡β) {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⁡asin⁡b+cos⁡acos⁡b=cos⁡(a−b)\sin a\sin b + \cos a\cos b = \cos \left( {a - b} \right) we get,
a2+b2=2+2cos⁡(α−β)−−−(3){a^2} + {b^2} = 2 + 2\cos \left( {\alpha - \beta } \right) - - - \left( 3 \right)
Now Subtracting equation (1) from equation (2)
b2−a2=(cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2{b^2} - {a^2} = {\left( {\cos \alpha + \cos \beta } \right)^2} - {\left( {\sin \alpha + \sin \beta } \right)^2}
On further simplifying
b2−a2=(cos⁡2α+2cos⁡αcos⁡β+cos⁡2β)−(sin⁡2α+2sin⁡αsin⁡β+sin⁡2β){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)
⇒b2−a2=cos⁡2α+2cos⁡αcos⁡β+cos⁡2β−sin⁡2α−2sin⁡αsin⁡β−sin⁡2β\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
⇒b2−a2=(cos⁡2α−sin⁡2β)+2(cos⁡αcos⁡β−sin⁡αsin⁡β)+(cos⁡2β−sin⁡2α)\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)
⇒b2−a2=(cos⁡2α−sin⁡2β)+2(cos⁡(α+β))+(cos⁡2β−sin⁡2α)\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,
⇒b2−a2=2(cos⁡(α+β)cos⁡(α−β))+2(cos⁡(α+β)) ⇒b2−a2=cos⁡(α+β)(2cos⁡(α−β)+2)−−−(4) \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

\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.