# If sinA+sinB=x and cosA+cosB=y then show that tan((A-B)/2)=+-sqrt((4-x^2-y^2)/(x^2+y^2))?

Feb 23, 2018

#### Explanation:

• $\sin A + \sin B = x$
$\implies 2 \cdot \sin \left(\frac{A + B}{2}\right) \cdot \cos \left(\frac{A - B}{2}\right) = x \text{ } \ldots \left(1\right)$

• $\cos A + \cos B = y$
$\implies 2 \cdot \cos \left(\frac{A + B}{2}\right) \cdot \cos \left(\frac{A - B}{2}\right) = y \text{ } \ldots \left(2\right)$

Now, we have to remove the term of sine and cosine value of " "((A+B)/2)" "to get the value of $\tan \left(\frac{A - B}{2}\right)$.

From 1st equation,

$\implies 4 \cdot {\sin}^{2} \left(\frac{A + B}{2}\right) \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) = {x}^{2}$
$\implies 4 \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) \cdot \left\{1 - {\cos}^{2} \left(\frac{A + B}{2}\right)\right\} = {x}^{2} \text{ } \ldots \left(3\right)$

Similarly, from 2nd equation,

$\implies 4 \cdot {\cos}^{2} \left(\frac{A + B}{2}\right) \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) = {y}^{2}$
$\implies 4 \cdot {\cos}^{2} \left(\frac{A + B}{2}\right) \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) = {y}^{2} \text{ } \ldots \left(4\right)$

From 3rd and 4th equation, we get

$\implies 4 \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) \cdot \left\{1 - {y}^{2} / \left(4 \cdot {\cos}^{2} \left(\frac{A - B}{2}\right)\right\}\right\} = {x}^{2}$

$\implies 4 \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) - {y}^{2} = {x}^{2}$

$\implies 4 \cdot {\cos}^{2} \left(\frac{A - B}{2}\right) = {x}^{2} + {y}^{2}$

$\implies {\cos}^{2} \left(\frac{A - B}{2}\right) = \frac{{x}^{2} + {y}^{2}}{4}$

$\implies {\sec}^{2} \left(\frac{A - B}{2}\right) = \frac{4}{{x}^{2} + {y}^{2}}$

$\implies 1 + {\tan}^{2} \left(\frac{A - B}{2}\right) = \frac{4}{{x}^{2} + {y}^{2}}$

$\implies {\tan}^{2} \left(\frac{A - B}{2}\right) = \frac{4}{{x}^{2} + {y}^{2}} - 1$

$\implies {\tan}^{2} \left(\frac{A - B}{2}\right) = \frac{4 - {x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$

=>tan((A-B)/2)=+-sqrt((4-x^2-y^2)/(x^2+y^2)

Hope it helps..
Thank you...

Feb 23, 2018

Proof is done considering squaring both the sides, and verified

#### Explanation:

Given:
$\sin A + \sin B = x$
$\cos A + \cos B = y$

To show that:

$\tan \left(\frac{A - B}{2}\right) = \pm \sqrt{\frac{4 - {x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}}$

Squaring both sides,

${\tan}^{2} \left(\frac{A - B}{2}\right) = \frac{4 - {x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$

RHS

$= \frac{4 - {\left(\sin A + \sin B\right)}^{2} - {\left(\cos A + \cos B\right)}^{2}}{{\left(\sin A + \sin B\right)}^{2} + {\left(\cos A + \cos B\right)}^{2}}$

=(4-(sin^2A+2sinAsinB+sin^2B)-(cos^2A+2cosAcosB+cos^2B))/(sin^2A+2sinAsinB+sin^2B+cos^2A+2cosAcosB+cos^2B

$= \frac{4 - \left({\sin}^{2} A + {\cos}^{2} A + {\sin}^{2} B + {\cos}^{2} B + 2 \left(\cos A \cos B + \sin A \sin B\right)\right)}{{\sin}^{2} A + {\cos}^{2} A + {\sin}^{2} B + {\cos}^{2} B + 2 \left(\cos A \cos B + \sin A \sin B\right)}$

${\sin}^{2} A + {\cos}^{2} A = 1$
${\sin}^{2} B + {\cos}^{2} B = 1$
$\cos A \cos B + \sin A \sin B = \cos \left(A - B\right)$

$= \frac{4 - \left(1 + 1 + 2 \cos \left(A - B\right)\right)}{1 + 1 + 2 \cos \left(A - B\right)}$

=(4-1-1-2cos(A-B))/(1+1+2cos(A-B)

$\frac{2 - 2 \cos \left(A - B\right)}{2 + 2 \cos \left(A - B\right)}$

$\frac{2 \left(1 - \cos \left(A - B\right)\right)}{2 \left(1 + \cos \left(A - B\right)\right)}$

$= \frac{1 - \cos \left(A - B\right)}{1 + \cos \left(A - B\right)}$

$= \frac{2 {\sin}^{2} \left(\frac{A - B}{2}\right)}{2 {\cos}^{2} \left(\frac{A - B}{2}\right)}$

$= \frac{{\sin}^{2} \left(\frac{A - B}{2}\right)}{{\cos}^{2} \left(\frac{A - B}{2}\right)}$

$= {\tan}^{2} \left(\frac{A - B}{2}\right)$

=LHS

Hence, proved