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Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.

Attempt:

I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.

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A proof (almost) without words: enter image description here

The sum between the red angle and the blue angle is the complementary angle of $\frac{\pi}{4}$, since in the figure we have a right isosceles triangle.

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    $\begingroup$ Coolest thing I've seen all day. $\endgroup$ Sep 29 '15 at 18:00
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$$\tan[\tan^{-1}(1/2)+\tan^{-1}(1/3)]=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}$$

Can you continue?

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Sine addition identity: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$

Cosine addition identity: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$

From the above, we obtain the tangent addition identity: $$\begin{align*} \tan (\alpha + \beta) &= \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \\ &= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \\ &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}. \end{align*}$$

From this, we now let $x = \tan \alpha$, $y = \tan \beta$, or equivalently, $\alpha = \tan^{-1} x$, $\beta = \tan^{-1} y$, and substitute: $$\tan(\tan^{-1} x + \tan^{-1} y) = \frac{x+y}{1-xy}.$$ Now taking the inverse tangent of both sides, we obtain the inverse tangent identity: $$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}.$$ Now let $x = 1/2$, $y = 1/3$, and simplify the right hand side.

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Let $$\displaystyle \tan^{-1}\left(\frac{1}{2}\right)=\alpha\Rightarrow \tan \alpha = \frac{1}{2}$$ and $$\displaystyle \tan^{-1}\left(\frac{1}{3}\right)=\beta\Rightarrow \tan \beta = \frac{1}{3}$$

So $$\displaystyle \tan(\alpha+\beta) = \frac{\tan \alpha+\tan \beta}{1-\tan \alpha\cdot \tan \beta} = \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}=1$$

So we get $$\displaystyle \tan(\alpha+\beta)=1=\tan \frac{\pi}{4}\Rightarrow \alpha+\beta = n\pi+\frac{\pi}{4}\;,n\in \mathbb{N}$$

But above $\displaystyle 0<\tan^{-1}\left(\frac{1}{2}\right)<\frac{\pi}{6}$ and $\displaystyle 0<\tan^{-1}\left(\frac{1}{3}\right)<\frac{\pi}{6}$

So we get $\displaystyle \alpha+\beta = \frac{\pi}{4}$

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  • $\begingroup$ Thank you. I'm confused a bit though. How is tan π/4 = 1? $\endgroup$
    – user274246
    Sep 29 '15 at 18:00
  • $\begingroup$ $\pi=180^0\Rightarrow \frac{\pi}{4} = \frac{180^0}{4}=45^0$ $\endgroup$
    – juantheron
    Sep 29 '15 at 18:03
  • $\begingroup$ Okay. I understand now. Thanks. $\endgroup$
    – user274246
    Sep 29 '15 at 18:11

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