Question

If $\sin A + \sin B = x$ and $\cos A + \cos B = y$ then show $\tan \left( {\dfrac{{A - B}}{2}} \right) = \pm \sqrt {\dfrac{{4 - {x^2} - {y^2}}}{{{x^2} + {y^2}}}} $?

Answer 1

Hint: In this question, first we have to use the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ and $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ to convert the angles into half-angles. Then we have to use the relation between the sine function and cosine function of ${\sin ^2}x + {\cos ^2}x = 1$ to get to the final answer.

Complete step-by-step answer:
In the above question, it is given that
$\sin A + \sin B = x$
Now use the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ in the above equation.
$ \Rightarrow 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) = x................\left( 1 \right)$
Also,
$\cos A + \cos B = y$
Now, use the formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ in the above equation.
$ \Rightarrow 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) = y.....................\left( 2 \right)$
Now, we have to remove the term of sine and cosine value of $\left( {\dfrac{{A + B}}{2}} \right)$ to get the value of $\tan \left( {\dfrac{{A - B}}{2}} \right)$.
Now squaring both sides in equation $\left( 1 \right)$.
$ \Rightarrow 4{\sin ^2}\left( {\dfrac{{A + B}}{2}} \right){\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {x^2}$
Now use the identity ${\sin ^2}x + {\cos ^2}x = 1$in the above equation.
$ \Rightarrow 4\left\{ {1 - {{\cos }^2}\left( {\dfrac{{A + B}}{2}} \right)} \right\}{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {x^2}..........\left( 3 \right)$
Now, squaring both sides in equation $\left( 2 \right)$.
$ \Rightarrow 4{\cos ^2}\left( {\dfrac{{A + B}}{2}} \right){\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {y^2}$

$ \Rightarrow {\cos ^2}\left( {\dfrac{{A + B}}{2}} \right) = \dfrac{{{y^2}}}{{4{{\cos }^2}\left( {\dfrac{{A - B}}{2}} \right)}}.................\left( 4 \right)$
From $3rd\,and\,4th$ equation, we get
$ \Rightarrow 4\left\{ {1 - \dfrac{{{y^2}}}{{4{{\cos }^2}\left( {\dfrac{{A - B}}{2}} \right)}}} \right\}{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {x^2}$
$ \Rightarrow 4\left\{ {\dfrac{{4{{\cos }^2}\left( {\dfrac{{A - B}}{2}} \right) - {y^2}}}{{4{{\cos }^2}\left( {\dfrac{{A - B}}{2}} \right)}}} \right\}{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {x^2}$
Now, cancel $4{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right)$ from numerator and denominator in left-hand side
$ \Rightarrow 4{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) - {y^2} = {x^2}$
Transpose ${y^2}$ to RHS.
$ \Rightarrow 4{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = {x^2} + {y^2}$
Now, divide both sides by $4$.
$ \Rightarrow {\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{{x^2} + {y^2}}}{4}$
Now use the identity $\cos x = \dfrac{1}{{\sec x}}$
$ \Rightarrow \dfrac{1}{{{{\sec }^2}\left( {\dfrac{{A - B}}{2}} \right)}} = \dfrac{{{x^2} + {y^2}}}{4}$
Now using cross-multiplication
$ \Rightarrow {\sec ^2}\left( {\dfrac{{A - B}}{2}} \right) = \dfrac{4}{{{x^2} + {y^2}}}$
Now using the identity $1 + {\tan ^2}x = {\sec ^2}x$
$ \Rightarrow 1 + {\tan ^2}\left( {\dfrac{{A - B}}{2}} \right) = \dfrac{4}{{{x^2} + {y^2}}}$
Now, transpose $1$ to RHS
$ \Rightarrow {\tan ^2}\left( {\dfrac{{A - B}}{2}} \right) = \dfrac{4}{{{x^2} + {y^2}}} - 1$
Now taking LCM in right hand side
$ \Rightarrow {\tan ^2}\left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{4 - {x^2} - {y^2}}}{{{x^2} + {y^2}}}$
Now taking root both sides, we get
$ \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = \pm \sqrt {\dfrac{{4 - {x^2} - {y^2}}}{{{x^2} + {y^2}}}} $
Hence proved.

Note: Formulas are the most important thing in trigonometry. Using some half angle- formula we can convert an expression with exponents to one without exponents, and whose angles are multiples of the original angle. It is to note that we get half-angle formulas from double angle formulas. Both \[sin{\text{ }}\left( {2A} \right)\] and \[cos{\text{ }}\left( {2A} \right)\] are obtained from the double angle formula for the cosine.