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Given $$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$ $$a,b \in (0,\frac{\pi}{2})$$ Show that $$a+2b=\frac{\pi}{4}$$

Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$?

Thank You!

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    $\begingroup$ You need some additional conditions (such as $a$ and $b$ are both in the first quadrant), otherwise the implication is false. $\endgroup$ – mathguy Aug 11 '16 at 20:08
  • $\begingroup$ @mathguy Oh, soooorry! I really forgot to write that. Really sorry! Edited. $\endgroup$ – MM PP Aug 11 '16 at 20:16
  • $\begingroup$ Probably not, cause that's pretty fast $\endgroup$ – Hagen von Eitzen Aug 11 '16 at 20:26
  • $\begingroup$ I assume you can do something with a right triangle with sides $1$ and $7$. One angle would be $a$ and the other would be $2b$. You'd need the triangle bisector theorem, but I can't imagine that it won't work. $\endgroup$ – Arthur Aug 11 '16 at 20:33
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$\sin b = \dfrac{1}{\sqrt{10}}$, $b\in(0,\pi/2)$ $\;\Rightarrow\;$ $b\in(0,\pi/6)$, i.e. $2b\in (0,\pi/3)$, since $1/\sqrt{10}<1/2$, and

$$\sin 2b = 2\sin b \cos b = 2 \dfrac{1}{\sqrt{10}} \dfrac{3}{\sqrt{10}} = \dfrac{3}{5}.$$

Then

$$ \tan 2b = \dfrac{\sin 2b}{\cos 2b} = \dfrac{3/5}{4/5} = \dfrac{3}{4}. $$

Now use formula $$ \tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}, $$

$$ \tan(a+2b) = \dfrac{\tan a + \tan 2b}{1-\tan a \tan 2b} = \dfrac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7}\cdot\frac{3}{4}} = \dfrac{4+21}{28-3}=\dfrac{25}{25}=1. $$

Then , in general, $a+2b = \pi/4+\pi k$, $k \in \mathbb{Z}$.
But since $a$ and $2b$ are acute, then $a+2b=\pi/4$.

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Just for fun... Adding a picture for the first time, so bear with me.

enter image description here

$\sin b = 1/\sqrt{10} \implies \cos b = 3/\sqrt{10}$ and $\sin 2b = 3/5$.

Now consider an isosceles right triangle $ABC$ as in the picture, with $AB = AC = 4$. Pick $D$ on $AC$ so that $AD = 3$, and $DE$ perpendicular to $BC$. $BD = 5$ by the Pythagorean theorem.

$\sin \angle DBA = 3/5$, $\angle ABC = \pi/4$, so it suffices to show $a = \angle DBE$. Now the area of $\triangle ABC$ is $8$ and that of $\triangle ABD$ is $6$, so the area of $\triangle DBC$ is $2$, and $BC = 4\sqrt 2$. From the formula for the area, $DE = 1/\sqrt 2$ , and $EB = 7/\sqrt2$ (by the Pythagorean theorem), $\tan \angle DBE = 1/7$, and the conclusion follows.

ADDED: $DE = 1/\sqrt2$ follows, more directly, from the fact that $\triangle ECD$ is isosceles and right-angled - no need for all the nonsense about areas.

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So, $\cos b=+\sqrt{1-\left(\dfrac1{\sqrt{10}}\right)^2}=\dfrac3{\sqrt{10}}$

$\implies\tan b=\dfrac{\dfrac1{\sqrt{10}}}{\dfrac3{\sqrt{10}}}\implies b=\arctan \dfrac13$

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$2b=\arctan\dfrac{2\cdot\dfrac13}{1-\left(\dfrac13\right)^2}=\arctan\dfrac34$$

Again as $a=\arctan\dfrac17$

Again Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$, $a+2b=\arctan\dfrac17+\arctan\dfrac34=\arctan\dfrac{\dfrac17+\dfrac34}{1-\dfrac17\cdot\dfrac34}=\arctan1=\dfrac\pi4$

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