Question

if x and y are two distinct angles satisfying the equation Acos(z)+Bsin(z)=C
show that
sin(x+y)=2AB/(A²+B²)​

Answer 1

Answer:

sin(x+y) =$\frac{2AB}{(A^{2}+B^{2})  }$

Step-by-step explanation:

As x and y are angles satisfying the given equations ,

$Acos(x) + Bsin(x)= C\\Acos(y) + Bsin(y)= C$

Subtracting second equation from first,

$A(cos(x)-cos(y)) + B(sin(x)-sin(y)) =0$

Applying trigonometric formulas,

$2A(sin(\frac{x+y}{2})sin(\frac{y-x}{2} )) + 2B(cos(\frac{x+y}{2} )sin(\frac{x-y}{2} )) =0$

Cancelling 2 and $sin(\frac{x-y}{2} )$,

$Asin(\frac{x+y}{2} ) = Bcos(\frac{x+y}{2} )\\\\tan(\frac{x+y}{2} ) = \frac{B}{A}$

We know $sin(\alpha) = \frac{2tan(\alpha /2)}{1+tan^{2}(\alpha /2) }$

$sin(x+y) = \frac{2tan(\frac{x+y}{2} )}{1+tan^{2}(\frac{x+y}{2} ) }$

$sin(x+y) = \frac{2\frac{B}{A} }{1+\frac{B^{2} }{A^{2} }} \\\\sin(x+y) = \frac{2AB}{A^{2}+B^{2}}$