# Solve sin^-1((ax)/c) + sin^-1( (bx)/c) = sin^-1( x )where a²+b²=c² and c≠0 ?

Oct 10, 2016

Ans : $x = 0 , + 1 \mathmr{and} - 1$

#### Explanation:

Solve
${\sin}^{-} 1 \left(\frac{a x}{c}\right) + {\sin}^{-} 1 \left(\frac{b x}{c}\right) = {\sin}^{-} 1 \left(x\right)$where a²+b²=c² and c≠0

${\sin}^{-} 1 \left(\frac{a x}{c}\right) + {\sin}^{-} 1 \left(\frac{b x}{c}\right) = {\sin}^{-} 1 \left(x\right)$

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

$\implies {\left\{\left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} + \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}}\right\}}^{2} = {x}^{2}$

$\implies {\left(\frac{a x}{c}\right)}^{2} \left(1 - {\left(\frac{b x}{c}\right)}^{2}\right) + {\left(\frac{b x}{c}\right)}^{2} \left(1 - {\left(\frac{a x}{c}\right)}^{2}\right) + 2 \left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = {x}^{2}$

$\implies \frac{\left({a}^{2} + {b}^{2}\right)}{c} ^ 2 {x}^{2} - \frac{2 {a}^{2} {b}^{2} {x}^{4}}{c} ^ 4 + 2 \left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = {x}^{2}$

$\implies {c}^{2} / {c}^{2} {x}^{2} - \frac{2 {a}^{2} {b}^{2} {x}^{4}}{c} ^ 4 + 2 \left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = {x}^{2}$

$\implies 2 \left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = \frac{2 {a}^{2} {b}^{2} {x}^{4}}{c} ^ 4$

$\implies {x}^{2} \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} - \frac{a b {x}^{4}}{c} ^ 2 = 0$

$\implies {x}^{2} \left[\sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} - \frac{a b {x}^{2}}{c} ^ 2\right] = 0$

so ${x}^{2} = 0 \implies x = 0$

And

$\implies \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = \frac{a b {x}^{2}}{c} ^ 2$

=>1-((a^2+b^2))/c^2x^2+cancel((a^2b^2x^4)/c^2)= cancel((a^2b^2x^4)/c^2

$\implies {x}^{2} - 1 = 0$

$\implies x = \pm 1$

Ans : $x = 0 , + 1 \mathmr{and} - 1$

Oct 10, 2016

$x = \pm 1$ and of course $x = 0$

#### Explanation:

Calling $\alpha = \frac{a x}{c}$ and $\beta = \frac{b x}{c}$ we have

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ so
$\sin \left(\alpha + \beta\right) = \left(\frac{a x}{c}\right) \sqrt{1 - {\left(\frac{b x}{c}\right)}^{2}} + \left(\frac{b x}{c}\right) \sqrt{1 - {\left(\frac{a x}{c}\right)}^{2}} = x$

Squaring both sides

 (a^2 x^2)/c^2 + (b^2 x^2)/c^2 - (2 a^2 b^2 x^4)/c^4 + ( 2 a b x^2 sqrt[1 - (a^2 x^2)/c^2] sqrt[1 - (b^2 x^2)/c^2])/c^2=x^2

Using the fact ${a}^{2} + {b}^{2} = {c}^{2}$ we have

$\sqrt{{c}^{2} - {a}^{2} {x}^{2}} \sqrt{{c}^{2} - {b}^{2} {x}^{2}} = a b {x}^{2}$

squaring again

${c}^{4} - \left({a}^{2} + {b}^{2}\right) {c}^{2} {x}^{2} = {c}^{4} \left(1 - {x}^{2}\right) = 0$ so finally

$x = \pm 1$ and $x = 0$ (formerly cancelled).