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

2 Answers
Oct 10, 2016

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

Explanation:

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#

Oct 10, 2016

#x=pm1# and of course #x = 0#

Explanation:

Calling #alpha = (ax)/c# and #beta = (bx)/c# we have

#sin(alpha+beta) = sin alpha cos beta + cos alpha sin beta# so
#sin(alpha+beta)=((ax)/c)sqrt(1-((b x)/c)^2)+((bx)/c)sqrt(1-((ax)/c)^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]=abx^2#

squaring again

#c^4 - (a^2 + b^2) c^2 x^2=c^4(1-x^2)=0# so finally

#x = pm1# and #x=0# (formerly cancelled).