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

sin(x+y) =
`(2AB)/((A^(2)+B^(2))  )`

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((x+y)/(2))sin((y-x)/(2) )) + 2B(cos((x+y)/(2) )sin((x-y)/(2) )) =0\\\\`

Cancelling 2 and
`sin((x-y)/(2) )`,

`Asin((x+y)/(2) ) = Bcos((x+y)/(2) )\\\\tan((x+y)/(2) ) = (B)/(A)`

We know
`sin(\alpha) = (2tan(\alpha /2))/(1+tan^(2)(\alpha /2) )`

`sin(x+y) = (2tan((x+y)/(2) ))/(1+tan^(2)((x+y)/(2) ) ) \\\\`

`sin(x+y) = (2(B)/(A) )/(1+(B^(2) )/(A^(2) )) \\\\sin(x+y) = (2AB)/(A^(2)+B^(2)) \\`