Question

if sinA+sinB=x and cosA+cosB=y then show that sin (A+B)=2xy/x^2+y^2​

Answer 1

Answer:

Step-by-step explanation:

Given,

1) sinA + sinB = x

2) cosA + cosB = y

To Prove :-

sin(A + B) = 2xy/x² + y²

Formula Required :-

1) sinC + sinD = 2sin(C+D/2)cos(C-D/2)

2)  cosC + cosD = 2cos(C+D/2)cos(C-D/2)

3) sinC/cosC = tanC

4) sin(A + B) = 2sin(A+B/2)cos(A+B/2)

5) (hypotenuse side)^2 = (ooposite side)^2 + (adjacent side)^2

Solution :-

sinA + sinB = x

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

[ ∴ sinC + sinD = 2sin(C+D/2)cos(C-D/2) ]

[ Let it be equation - 1]

cosA + cosB = y

$2cos\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)=y$

[ ∴ cosC + cosD = 2cos(C+D/2)cos(C-D/2) ]

[Let it be equation - 2]

Equation - 1 divided by equation - 2 :-

$\dfrac{2sin\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)}{2cos\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)}=\dfrac{x}{y}$

Cancelling the common terms :-

$\dfrac{sin\left(\dfrac{A+B}{2}\right)}{cos\left(\dfrac{A+B}{2}\right)}=\dfrac{x}{y}$

$tan\left(\dfrac{A+B}{2}\right)=\dfrac{x}{y}$

→ opposite side/adjacent side = x/y

opposite side = x , adjacent side = y

After applying Pythagoras theorem we get :-

$Hypotenuse side = \sqrt{x^2+y^2}$

→ sin(A+B/2) = opposite side/hypotenuse side

$sin\left(\dfrac{A+B}{2}\right)=\dfrac{x}{\sqrt{x^2+y^2}}$

cos(A+B/2) = adjacent side/hypotenuse side

$cos\left(\dfrac{A+B}{2}\right)=\dfrac{y}{x^2+y^2}$

sin(A + B) = 2sin(A+B/2)cos(A+B/2)

[ ∴ sin2A = 2sinAcosA ]

Substituting those values :-

$=2\times \dfrac{x}{\sqrt{x^2+y^2}}\times \dfrac{y}{\sqrt{x^2+y^2}}$

$=\dfrac{2xy}{(\sqrt{x^2+y^2})^2}$

$=\dfrac{2xy}{x^2+y^2}$

∴ sin(A + B) = 2xy/x² + y²

Hence proved.