A geometry problem by Nivedit Jain

Relevant wiki: Sum and Difference Formulas

If $\sin a + \sin b = \dfrac{1}{\sqrt{2}}$ so we can write as $2 \sin(\dfrac{a+b}{2}) \cos(\dfrac{a-b}{2}) = \dfrac{1}{\sqrt{2}}$ ... (I)

Now again in 2nd equation $\cos a + \cos b = \sqrt{\dfrac{3}{2}}$ or $2 \cos (\dfrac{a+b}{2}) \cos(\dfrac{a-b}{2}) = \sqrt{\dfrac{3}{2}}$ ...(II)

Now ,

$\dfrac{I}{II} = \tan(\dfrac{a+b}{2})$ = $\sqrt{\dfrac{1}{3}}$

So $\dfrac{a+b}{2} = 30^{o}$ or $a+b = 60^{o}$

So $\sin(a+b) = \sin 60^{o} = \dfrac{\sqrt{3}}{2}$ = $\boxed{0.866}$

If we square and add both the given equations, we get

$\large \sin^{2}a + \cos^{2}a + sin^{2}b + \cos^{2}b + 2(\sin a \cos a + \sin b \cos b) = 2$

$\Longrightarrow \large 2 + 2(\sin a \cos a + \sin b \cos b) = 2$

$\Longrightarrow \large \sin a \cos a + \sin b \cos b = 0$

If we multiply both the given equations, we get

$\large \sin a \cos b + \cos a \sin b + \sin a \cos a + \sin b \cos b$ = $\dfrac{\sqrt 3}{2}$ = $\large 0.866$

$\Longrightarrow \large \sin a \cos b + \cos a \sin b + 0$ = $\large 0.866$

$\Longrightarrow \large \sin (a+b) = 0.866$