# Sinx + siny =a and coax +cos y =b show that( sinx +y) = 2ab /a^2+b^2

Ajay
209 Points
5 years ago
Please find the solution below. Also tell us what you tried at your end............................................................................
$Given\quad sinx\quad +\quad siny\quad =\quad a\quad .........................(1)\quad and\quad \\ \quad \quad \quad \quad \quad \quad cosx\quad +cosy\quad =\quad b...........................(2)\\ \\ squaring\quad and\quad adding\quad equations\quad 1\quad and\quad 2\\ \\ { (sinx\quad +\quad siny })^{ 2 }\quad +\quad { (cosx\quad +cosy) }^{ 2 }\quad =\quad { a }^{ 2 }+{ b }^{ 2 }\\ \\ on\quad simplifying\quad ths\quad becomes\\ 2\quad +\quad 2(sinxsiny\quad +\quad cosxcosy)\quad =\quad { a }^{ 2 }+{ b }^{ 2 }\\ 1\quad +\quad cos(x-y)\quad =\quad \frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } ...............................(3)$
To be continued

Ajay
209 Points
5 years ago
Posting remaining solution as the equation editor cant post whole solution in single answer.......................................................................................
$Now\quad mulliply\quad 1\quad and\quad 2\\ { (sinx\quad +\quad siny })*cosx\quad +cosy\quad =\quad ab\\ \\ sinx.cosx\quad +\quad siny.cosy\quad +\quad (siny.cosx\quad +\quad sinx.cosy)\quad =\quad ab\\ 1/2\quad sin2x\quad \quad +\quad 1/2sin2y\quad \quad +\quad sin(x+y)\quad =\quad ab\\ sin(x+y)cos(x-y)\quad +\quad sin(x+y)\quad =\quad ab\\ sin(x+y)(cos(x-y)\quad +1)\quad =\quad ab\\ substituting\quad from\quad eq\quad 3\\ sin(x+y).\frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } \quad =\quad ab\\ sin(x+y)\quad =\quad \frac { 2ab }{ { a }^{ 2 }+{ b }^{ 2 } } \quad Hence\quad Proved.$