Question #b42d1

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Answer 1 · 2016-10-31T07:48:04

Ans : $x=0,+1 and -1$

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

$sin^-1((ax)/c) + sin^-1( (bx)/c) = sin^-1( x )$

$=>sin^-1{((ax)/c)sqrt(1-( (bx)/c)^2)+( (bx)/c)sqrt(1-( (ax)/c)^2)} = sin^-1( x )$

$=>{((ax)/c)sqrt(1-( (bx)/c)^2)+( (bx)/c)sqrt(1-( (ax)/c)^2)}^2 = x^2$

$=>((ax)/c)^2(1-( (bx)/c)^2)+( (bx)/c)^2(1-( (ax)/c)^2)+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2$

$=>((a^2+b^2))/c^2x^2-(2a^2b^2x^4)/c^4+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2$

$=>c^2/c^2x^2-(2a^2b^2x^4)/c^4+2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = x^2$

$=>2((ax)/c)sqrt(1-( (bx)/c)^2) ((bx)/c)sqrt(1-( (ax)/c)^2) = (2a^2b^2x^4)/c^4$

$=>x^2sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2) - (abx^4)/c^2=0$

$=>x^2[sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2) - (abx^2)/c^2]=0$

so $x^2=0=>x=0$

And

$=>sqrt(1-( (bx)/c)^2) sqrt(1-( (ax)/c)^2)= (abx^2)/c^2$

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

$=>x^2-1=0$

$=>x=+-1$

Ans : $x=0,+1 and -1$