Question

if acosalpha+bsinbeta=c ;
asinalpha+bcosbeta=c then sin(alpha+beta)=?
​

Answer 1

Step-by-step explanation:

Given that: if

$acos \alpha + bsin \beta = c\:\:...eq1 \\ asin \alpha + bcos \beta = c\:\:...eq2$

To find:

$sin( \alpha + \beta ) = ?$

Solution: Square both given equations 1 and 2

$(acos \alpha + bsin \beta)^{2} = {c}^{2} \\ \\ {a}^{2} {cos}^{2} \alpha + {b}^{2} {sin}^{2} \beta + 2abcos \alpha sin \beta = {c}^{2} \:\:...eq3 \\ \\( asin \alpha + bcos \beta)^{2} = {c}^{2} \\ \\ {a}^{2} {sin}^{2} \alpha + {b}^{2} {cos}^{2} \beta + 2absin \alpha cos \beta = {c}^{2}\:\:...eq4 \\ \\ add \: both \: equations3\:and\:4 \: \\ \\ {a}^{2} ( {sin}^{2} \alpha + {cos}^{2} \alpha ) + {b}^{2} ( {sin}^{2} \beta + {cos}^{2} \beta ) \\+ 2ab(sin \alpha cos \beta + cos \alpha sin \beta ) = 2 {c}^{2} \\\\ \because\:({sin}^{2} \theta + {cos}^{2} \theta ) =1\\\\and\\\\ (sin\:A cos\:B + cos\:A sin\:B ) =sin(A+B)\\\\ {a}^{2} + {b}^{2} + 2absin( \alpha + \beta ) = 2 {c}^{2} \\ \\ 2absin( \alpha + \beta ) = 2 {c}^{2} - {a}^{2} - {b}^{2} \\ \\ \bold{sin( \alpha + \beta ) = \frac{2 {c}^{2} - {a}^{2} - {b}^{2}}{2ab} }$

Hope it helps you.