3=13=1? Please inform if there is any mistake or any contradiction in this

3sin1x=2sin1x+sin1x=sin1(2x1x2)+sin1x=sin1(2x1x21x2+x1(2x1x2)2)=sin1(2x(1x2)+x14x2(1x2))=sin1(2x(1x2)+x14x2+4x4))=sin1(2x2x3+x(2x21)2)=sin1(2x2x3+x(2x21))=sin1(2x2x3+2x3x)=sin1x \begin{aligned} 3\sin^{-1} x &=& 2\sin^{-1} x + \sin^{-1} x \\ &=& \sin^{-1} (2x \sqrt{1-x^2} ) + \sin^{-1} x \\ &=& \sin^{-1} \left( 2x \sqrt{1-x^2} \sqrt{1-x^2} + x \sqrt{1 - (2x\sqrt{1-x^2} )^2} \right) \\ &=& \sin^{-1} \left( 2x(1-x^2) + x \sqrt{1 - 4x^2(1-x^2)} \right) \\ &=& \sin^{-1} \left( 2x(1-x^2) + x \sqrt{1 - 4x^2 + 4x^4)} \right) \\ &=& \sin^{-1} \left( 2x - 2x^3 + x \sqrt{(2x^2-1)^2} \right) \\ &=& \sin^{-1} (2x - 2x^3 + x(2x^2 - 1)) \\ &=& \sin^{-1} (2x - \cancel{2x^3} + \cancel{2x^3} - x) \\ &=& \sin^{-1} x \end{aligned}

Thus 3sin1x=sin1x3\sin^{-1} x = \sin^{-1} x or 3=13=1?

Note by Darvin Rio
3 years, 8 months ago

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You're making two different mistakes.

The first one is that if x212 x^2 \leq \frac{1}{2} , then (2x21)2=(12x2)(2x21) \sqrt{(2x^2 -1)^2} = (1 - 2x^2) \not = (2x^2 -1)

The second one is that if x2>12 x^2 > \frac{1}{2} then sin1(x)>π4 |\sin^{-1}(x)| > \frac{\pi}{4} . and 2sin1(x)π2 2|\sin^{-1}(x)| \geq \frac{\pi}{2}

Therefore it should be obvious that, 2sin1(x)sin1(2x1x2) 2|\sin^{-1}(x)| \not = |\sin^{-1}(2x\sqrt{1-x^2})| , since the LHS >π2 > \frac{\pi}{2} and the RHS π2 \leq \frac{\pi}{2} (Range of sin1 sin^{-1} is [π2,π2] [\frac{\pi}{2},-\frac{\pi}{2}] )

Siddhartha Srivastava - 3 years, 8 months ago

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Isn't 3sin1x=sin1x3\sin^{-1} x=\sin^{-1} x satisfied only when sin1x=0\sin^{-1}x=0. If that is true then recalling basic identities of equations: ac=adc=dac=ad\Rightarrow c=d (Only when a is non zero).
However in our case c=3, d=1 while a=sin1x\sin^{-1} x which is 0!! Hence 3sin1x=sin1x⇏3=1 3\sin^{-1} x = \sin^{-1} x \not\Rightarrow 3=1

Rishabh Jain - 3 years, 8 months ago

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i tried deriving a standard derivation......so universally doesnt it mean that sin inverse x universally is 0!!!!

Darvin Rio - 3 years, 8 months ago

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I think that there is another mistake because if everything is correct till the last step then, 2sin1x+sin1x=sin1x2 \sin^{-1}{x}+\sin^{-1}{x}=\sin^{-1}{x} or sin1x=0\sin^{-1}{x}=0.

A Former Brilliant Member - 3 years, 8 months ago

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Exactly sin1x=0\sin^{-1} x=0 that's why in last step we cannot cancel sin1x\sin^{-1} x from both sides to claim 1=31=3.

Rishabh Jain - 3 years, 8 months ago

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@Rishabh Jain The problem is that the first four lines are only valid if x=0. If not sin1xsin^{-1} x would be 0 for al x, by this argumentation, but the first 4 lines make implicit assumtions that x lies in certain ranges and the intersection of those is {0}\{0\}

Maximilian Wackenhuth - 3 years, 8 months ago

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