Question

If $\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$ , show that
$\cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}$

Answer 1

In this question use given equations and apply the division operation and also remember to use trigonometric identities $\sin C + \sin D = 2\sin \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right)$, use this information to approach the solution of the problem.

Complete step-by-step answer :
Given that: $\sin \alpha + \sin \beta = a$---- (1)
And $\cos \alpha + \cos \beta = b$ ----- (2)
Let us divide equation (1) by equation (2)
$\Rightarrow \dfrac{{\sin \alpha + \sin \beta }}{{\cos \alpha + \cos \beta }} = \dfrac{a}{b}$----- (3)
We know the identities of for addition of trigonometric terms which are:
$\sin C + \sin D = 2\sin \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right) \\\ \cos C + \cos D = 2\cos \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right) \\\$
Using the identities in equation (3) we get:
$\because \dfrac{{\sin \alpha + \sin \beta }}{{\cos \alpha + \cos \beta }} = \dfrac{a}{b} \\\ \Rightarrow \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)}} = \dfrac{a}{b} \\\$
Let us further simplify the result by cancelling the common terms in numerator and denominator
$\Rightarrow \dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}} = \dfrac{a}{b} \\\ \Rightarrow \tan \left( {\dfrac{{\alpha + \beta }}{2}} \right) = \dfrac{a}{b}........{\text{(4) }}\left[ {\because \dfrac{{\sin x}}{{\cos x}} = \tan x} \right] \\\$
In the RHS we have terms containing a and b but we need terms containing ${a^2}\& {b^2}$ , so we will use the identities to double angle.
As we know that:
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
So, in the above identity substituting the value of$\theta$ as $\left( {\dfrac{{\alpha + \beta }}{2}} \right)$
So, we get:
$\cos 2\left( {\dfrac{{\alpha + \beta }}{2}} \right) = \dfrac{{1 - {{\tan }^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)}}$----- (5)
Substituting the value in equation (5) from equation (4) we get:
$\Rightarrow \cos \left( {\alpha + \beta } \right) = \dfrac{{1 - {{\left( {\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)} \right)}^2}}}{{1 + {{\left( {\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)} \right)}^2}}} \\\ \Rightarrow \cos \left( {\alpha + \beta } \right) = \dfrac{{1 - {{\left( {\dfrac{a}{b}} \right)}^2}}}{{1 + {{\left( {\dfrac{a}{b}} \right)}^2}}} \\\$
Now let us simplify the above term by taking LCM on RHS
$\Rightarrow \cos \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{{b^2} - {a^2}}}{{{b^2}}}}}{{\dfrac{{{b^2} + {a^2}}}{{{b^2}}}}} \\\ \Rightarrow \cos \left( {\alpha + \beta } \right) = \dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}} \\\$
Hence, the given trigonometric equation is proved.

Note : In order to solve such complex trigonometric problems with difference in angles in the given terms and the problem statement. Students must first identify the trigonometric identity connecting both of the angles. In order to solve such types of problems students must remember trigonometric identities to find an easy solution and should manipulate different identities in order to fetch the answer.