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During helping a junior , I came to this term: $$ \sin\left(4\tan^{-1}\frac{1}{2}\right) $$

using calculator , you will find the value $\frac{24}{25}$ which is correct one. But when you expand it like this(which my junior did) : \begin{align} \sin\left(4\tan^{-1}\frac{1}{2}\right) &= \\ &= \sin\left(2\tan^{-1}\frac{1}{1-\frac{1}{4}}\right) & [2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}] \\ &=\sin\left(2\tan^{-1}\frac{4}{3}\right) \\ &=\sin\left(\tan^{-1}\frac{2\cdot\frac{4}{3}}{1-\frac{16}{9}}\right) \\ &=\sin\left(\tan^{-1}\left(-\frac{24}{7}\right)\right) \end{align}

which value is $-\frac{24}{25}$ . Why this expansion is giving wrong answer ? where is the error ?

Misha Lavrov
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Anklon
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3 Answers3

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The identity $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2\theta}$$ is always true, but it does not always imply that $$2 \arctan x = \arctan \frac{2x}{1-x^2}.$$ The problem is that multiple values of $\theta$ have the same value of $\tan \theta$, and by convention $\arctan x$ chooses the angle $\theta$ between $-\frac\pi2$ and $\frac\pi2$.

In your case, $2 \arctan \frac43 \approx 1.85$, which is bigger than $\frac\pi2$, and $\arctan -\frac{24}{7} \approx -1.29$: it is $2 \arctan \frac43 - \pi$. When you take the sine of both values, since $\sin (x + \pi) = - \sin x$, you will be off by a minus sign.

To deal with this, you can observe that if $x > 1$, then $\arctan x > \frac\pi4$, so $2 \arctan x > \frac\pi2$ and you'll get an answer off by $\pi$ when that happens. This tells you when to subtract a $\pi$ when working with this identity.

Misha Lavrov
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It should be $$\sin { \left( 4\tan ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } \right) } =\\ \tan ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } =t\\ \tan { t=\frac { 1 }{ 2 } } \Rightarrow \sin { t } =\frac { 1 }{ \sqrt { 5 } } ,\cos { t } =\frac { 2 }{ \sqrt { 5 } } \\ \sin { \left( 4t \right) } =2\sin { 2t\cos { 2t } } =4\sin { t\cos { t } \left( \cos ^{ 2 }{ t } -\sin ^{ 2 }{ t } \right) } =4\cdot \frac { 2 }{ \sqrt { 5 } } \cdot \frac { 1 }{ \sqrt { 5 } } \left( \frac { 4 }{ 5 } -\frac { 1 }{ 5 } \right) =\frac { 24 }{ 25 } $$

haqnatural
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  • If you follow this procedure , you can get the answer. But my question is not how can I get proper answer. I wanna know in my procedure , what is my mistake ? – Anklon Apr 01 '17 at 18:12
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When bigger angles are calculated we should check in which quadrant the angle lies after rotation.The tan in second quadrant is negative.So we need to change sign of $$ \tan^{-1}\dfrac{2 \dfrac{4}{3}}{1-\dfrac{16}{9}} $$

Else all OK.

Narasimham
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  • We don't change the sign of $\arctan$: we subtract $\pi$. (In this case, because we're taking $\sin$ afterwards, the two are equivalent, but if we were to apply $\cos$ on the outside instead, changing the sign would give you the wrong answer.) – Misha Lavrov Apr 01 '17 at 17:00
  • It was a typo, corrected. Since tan in second quadrant is < 0, and we are having a definite angle etc.... anyway – Narasimham Apr 01 '17 at 18:08