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

$\begin{eqnarray} 3\sin^{-1} x &=& 2\sin^{-1} x + \sin^{-1} x \\ &=& \sin^{-1} (2x \sqrt{1-x^2} ) + \sin^{-1} x \\ &=& \sin^{-1} \left( 2x \sqrt{1-x^2} \sqrt{1-x^2} + x \sqrt{1 - (2x\sqrt{1-x^2} )^2} \right) \\ &=& \sin^{-1} \left( 2x(1-x^2) + x \sqrt{1 - 4x^2(1-x^2)} \right) \\ &=& \sin^{-1} \left( 2x(1-x^2) + x \sqrt{1 - 4x^2 + 4x^4)} \right) \\ &=& \sin^{-1} \left( 2x - 2x^3 + x \sqrt{(2x^2-1)^2} \right) \\ &=& \sin^{-1} (2x - 2x^3 + x(2x^2 - 1)) \\ &=&\require{cancel} \sin^{-1} (2x - \cancel{2x^3} + \cancel{2x^3} - x) \\ &=& \sin^{-1} x \end{eqnarray}$

Thus $$3\sin^{-1} x = \sin^{-1} x$$ or $$3=1$$?

Note by Darvin Rio
2 years, 6 months ago

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You're making two different mistakes.

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

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

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

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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

- 2 years, 6 months ago

The problem is that the first four lines are only valid if x=0. If not $$sin^{-1} x$$ 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 $$\{0\}$$

- 2 years, 6 months ago