\(3=1\)? Please inform if there is any mistake or any contradiction in this

3sin⁡−1x=2sin⁡−1x+sin⁡−1x=sin⁡−1(2x1−x2)+sin⁡−1x=sin⁡−1(2x1−x21−x2+x1−(2x1−x2)2)=sin⁡−1(2x(1−x2)+x1−4x2(1−x2))=sin⁡−1(2x(1−x2)+x1−4x2+4x4))=sin⁡−1(2x−2x3+x(2x2−1)2)=sin⁡−1(2x−2x3+x(2x2−1))=sin⁡−1(2x−2x3+2x3−x)=sin⁡−1x \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 3sin⁡−1x=sin⁡−1x3\sin^{-1} x = \sin^{-1} x or 3=13=1?

#Geometry

Note by Darvin Rio
5 years, 3 months ago

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Comments

You're making two different mistakes.

The first one is that if x2≤12 x^2 \leq \frac{1}{2} , then (2x2−1)2=(1−2x2)≠(2x2−1) \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 ∣sin⁡−1(x)∣>π4 |\sin^{-1}(x)| > \frac{\pi}{4} . and 2∣sin⁡−1(x)∣≥π2 2|\sin^{-1}(x)| \geq \frac{\pi}{2}

Therefore it should be obvious that, 2∣sin⁡−1(x)∣≠∣sin⁡−1(2x1−x2)∣ 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 sin−1 sin^{-1} is [π2,−π2] [\frac{\pi}{2},-\frac{\pi}{2}] )

Siddhartha Srivastava - 5 years, 3 months ago

Isn't 3sin⁡−1x=sin⁡−1x3\sin^{-1} x=\sin^{-1} x satisfied only when sin⁡−1x=0\sin^{-1}x=0. If that is true then recalling basic identities of equations: ac=ad⇒c=dac=ad\Rightarrow c=d (Only when a is non zero).
However in our case c=3, d=1 while a=sin⁡−1x\sin^{-1} x which is 0!! Hence 3sin⁡−1x=sin⁡−1x⇏3=1 3\sin^{-1} x = \sin^{-1} x \not\Rightarrow 3=1

Rishabh Jain - 5 years, 3 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 - 5 years, 3 months ago

I think that there is another mistake because if everything is correct till the last step then, 2sin⁡−1x+sin⁡−1x=sin⁡−1x2 \sin^{-1}{x}+\sin^{-1}{x}=\sin^{-1}{x} or sin⁡−1x=0\sin^{-1}{x}=0.

A Former Brilliant Member - 5 years, 3 months ago

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

Rishabh Jain - 5 years, 3 months ago

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@Rishabh Jain – The problem is that the first four lines are only valid if x=0. If not sin−1xsin^{-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 - 5 years, 3 months ago
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