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

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

## Comments

Sort by:

TopNewestYou'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 \( [\frac{\pi}{2},-\frac{\pi}{2}] \)) – Siddhartha Srivastava · 11 months, 1 week ago

Log in to reply

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 \[ 3\sin^{-1} x = \sin^{-1} x \not\Rightarrow 3=1\] – Rishabh Cool · 11 months, 1 week ago

Log in to reply

– Darvin Rio · 11 months ago

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

– Svatejas Shivakumar · 11 months, 1 week 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 \(\sin^{-1}{x}=0\).Log in to reply

– Rishabh Cool · 11 months, 1 week ago

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

– Maximilian Wackenhuth · 11 months, 1 week 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 \(\{0\}\)Log in to reply