# If (a^2-b^2)sintheta + 2ab costheta = a^2+b^2, then what will the value of sintheta be?

Nov 27, 2016

Given

$\left({a}^{2} - {b}^{2}\right) \sin \theta + 2 a b \cos \theta = {a}^{2} + {b}^{2}$

$\implies \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} \sin \theta + \frac{2 a b}{{a}^{2} + {b}^{2}} \cos \theta = 1. \ldots . \left(1\right)$

Now if $\frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} = \cos \alpha$

then

$\sin \alpha = \sqrt{1 - {\cos}^{2} \alpha}$

=sqrt(1- ((a^2-b^2)/(a^2+b^2))^2

$= \frac{2 a b}{{a}^{2} + {b}^{2}}$

So the relation (1) becomes

$\cos \alpha \sin \theta + \sin \alpha \cos \theta = 1$

$\implies \sin \left(\theta + \alpha\right) = \sin \left(\frac{\pi}{2}\right)$

$\implies \alpha = \frac{\pi}{2} - \theta$

$\implies \cos \alpha = \cos \left(\frac{\pi}{2} - \theta\right)$

$\implies \cos \alpha = \sin \theta$

$\therefore \sin \theta = \cos \alpha = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$

Alternative-1

Given

$\left({a}^{2} - {b}^{2}\right) \sin \theta + 2 a b \cos \theta = {a}^{2} + {b}^{2}$

Transferring to RHS we get

${a}^{2} \left(1 - \sin \theta\right) + {b}^{2} \left(1 + \sin \theta\right) - 2 a b \cos \theta = 0$

$\implies {\left(a \sqrt{1 - \sin \theta}\right)}^{2} + {\left(b \sqrt{1 + \sin \theta}\right)}^{2} - 2 a b \sqrt{1 - {\sin}^{2} \theta} = 0$

$\implies {\left[\left(a \sqrt{1 - \sin \theta}\right) - \left(b \sqrt{1 + \sin \theta}\right)\right]}^{2} = 0$

$\implies a \sqrt{1 - \sin \theta} = b \sqrt{1 + \sin \theta}$

$\implies {a}^{2} \left(1 - \sin \theta\right) = {b}^{2} \left(1 + \sin \theta\right)$

$\implies \sin \theta = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Alternative-2

Putting $b = a \tan \phi$ in the relation (1) we get

$\implies \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} \sin \theta + \frac{2 a b}{{a}^{2} + {b}^{2}} \cos \theta = 1$

$\implies \frac{{a}^{2} - {a}^{2} {\tan}^{\phi}}{{a}^{2} + {a}^{2} {\tan}^{2} \phi} \sin \theta + \frac{2 a \cdot a \tan \phi}{{a}^{2} + {a}^{2} {\tan}^{2} \phi} \cos \theta = 1$

$\implies \cos 2 \phi \sin \theta + \sin 2 \phi \cos \theta = 1$

$\implies \sin \left(\theta + 2 \phi\right) = \sin \left(\frac{\pi}{2}\right)$

$= \theta = \frac{\pi}{2} - 2 \phi$

$= \sin \theta = \sin \left(\frac{\pi}{2} - 2 \phi\right) = \cos 2 \phi$

$\implies \sin \theta = \frac{1 - {\tan}^{2} \phi}{1 + {\tan}^{2} \phi}$

$\implies \sin \theta = \frac{1 - {\left(\frac{b}{a}\right)}^{2}}{1 + {\left(\frac{b}{a}\right)}^{2}} = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$

Nov 27, 2016

$\sin \theta = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$

#### Explanation:

After reducing to

$\frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} \sin \theta + \frac{2 a b}{{a}^{2} + {b}^{2}} \cos \theta = 1$

Now

${\left(\frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}\right)}^{2} + {\left(\frac{2 a b}{{a}^{2} + {b}^{2}}\right)}^{2} = \frac{{a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4}}{{a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4}} = 1$ so there exists $\phi$ such that

$\sin \phi = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$ and
$\cos \phi = \frac{2 a b}{{a}^{2} + {b}^{2}}$

note that

$- 1 \le \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} \le 1$
$- 1 \le \frac{2 a b}{{a}^{2} + {b}^{2}} \le 1$

so

$\sin \theta \sin \phi + \cos \theta \cos \phi = 1$

and $\theta = \phi$ because ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ then

$\sin \theta = \sin \phi = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}$