Question

sin x+sin y=a & cos x+cos y=b
prove that sin(x+y)=2ab/a^2+b^2​

Answer 1

$\underline{\textbf{\textsf{Solution \: :-}}}$

$\dagger \;\underline{\underline{\small\purple{\rm Given\: :-}}}$

• $\sf \small sin(x) + sin(y) = a$
• $\sf \small cos(x) + cos(y) = b$

$\dagger \;\underline{\underline{\small\purple{\rm Formulas\;to\; be \; used \: :-}}}$

• $\sf \small sin(2x) = 2sin(x)cos(x)$
• $\sf \small sin(x+y) = sin(x)cos(y) + cos(x)sin(y)$
• $\sf \small sin(x)+sin(y) = 2sin\Bigg(\dfrac{x+y}{2}\Bigg)cos\Bigg(\dfrac{x+y}{2}\Bigg)$
• $\sf\small sin\theta cos\theta = \dfrac{sin2\theta}{2} (or) ^1/_2 sin2\theta$

◉ $\underline{\rm \small\orange{Finding\; a\times b}}$

$\longrightarrow \small \sf ab = \big[sin(x)+sin(y)\big]\big[cos(x)+cos(y)\big]$

$\longrightarrow \small \sf ab = sin(x)cos(x) + \big[ sin(x)cos(y) + cos(x)sin(y)\big] + sin(y)cos(y)$

$\longrightarrow \small \sf ab = ^1/_2 sin2x +sin(x+y) + ^1\!/\!_2 sin2y$

$\longrightarrow \small \sf ab = ^1/_2\big(sin2x+sin2y\big) + sin(x+y)$

$\longrightarrow \small \sf ab = ^1/_2\Bigg[2 sin\Bigg(\dfrac{2x +2y}{2}\Bigg) cos\Bigg(\dfrac{2x-2y}{2y}\Bigg)\Bigg] + sin(x+y)$

$\longrightarrow\small\sf ab = sin(x+y)cos(x-y) + sin(x+y)$

$\longrightarrow\small\sf ab = sin(x+y)\big[1 + cos(x-y)\big]$

◍ $\underline{\sf Multiplying\; with \; 2 \; on \; both \; sides \; :-}$

$\longrightarrow\small \sf 2ab = 2sin(x+y)\big[1 + cos(x-y)\big]...eq(1)$

◉ $\underline{\rm\small\orange{Finding\;a^2\:\&\: b^2}}$

$\dashrightarrow \small \sf a^2 = \big[sin(x) + sin(y)\big]^2 = sin^2(x)+sin^2(y)+2sin(x)sin(y)$

$\dashrightarrow\small \sf b^2 = \big[cos(x)+cos(y)\big]^2 = cos^2(x) + cos^2(y) + 2cos(x)cos(y)$

◍ $\underline{\sf a^2 + b^2 \; :-}$

$\longmapsto \small \sf a^2+b^2 = sin^2(x) + sin^2(y) + 2sin(x)sin(y) + cos^2(x)+cos^2(y) + 2cos(x)sin(y)$

$\longmapsto \small \sf a^2 + b^2 = 1 + 1 + 2\big[sin(x)sin(y)+cos(x)cos(y)\big]$

$\longmapsto \small \sf a^2 + b^2 = 2 + 2 \big[ cos(x-y)\big]$

$\longmapsto \small \sf a^2 + b^2 = 2\big[ 1 + cos(x-y) \big]...eq(2)$

◉ $\underline{\rm\small \orange{Eq(1) \div Eq(2)}}$

$\hookrightarrow \small\sf \dfrac{2ab}{a^2+b^2} = \dfrac{2sin(x+y)\bigg[1+cos(x-y)\bigg]}{2\bigg[1+cos(x-y)\bigg]}$

$\hookrightarrow \underline{\boxed{\bf sin(x+y)=\dfrac{2ab}{a^2+b^2}}}$

$\sf Hence \; proved \;\green{\checkmark}$