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The question is to find $x\in\left(0,\frac{\pi}{2}\right)$:

$$\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $$

What I did was to take the $\cos x$ fraction to the right and try to simplify ;

But it looked very messy and trying to write $\sin x$ in terms of $\cos x$ didn't help.

Is there a more simple (elegant) way to do this.

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    $\begingroup$ divide both sides by $2\sqrt{2}$ and use the fact that $\sin(\frac{\pi}{6}) = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$ $\endgroup$
    – dezdichado
    Commented Mar 20, 2016 at 8:56
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    $\begingroup$ $\sin \frac \pi6$ is $\frac 12$, did you mean $\frac \pi{12}$? $\endgroup$
    – user26977
    Commented Mar 20, 2016 at 8:59

4 Answers 4

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$\sin(\frac{\pi}{12}) = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\cos(\frac{\pi}{12}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$. Plugging them into your equation yields $\sin(x+\frac{\pi}{12}) = \sin(2x)$. So $x = \frac{\pi}{12}$

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  • $\begingroup$ It $\pi/12$ not $\pi/6$ $\endgroup$ Commented Mar 20, 2016 at 9:00
  • $\begingroup$ yes edited accordingly $\endgroup$
    – dezdichado
    Commented Mar 20, 2016 at 9:01
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Another approach is to write the equation as

$(\sqrt{3} - 1) \cos x +(\sqrt{3} + 1) \sin x = 4 \sqrt2 \sin x \cos x$

then rearranging gives

$\frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x +\frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x = 2 \sin x \cos x$.

Now note that $(\frac{\sqrt{3} - 1}{2 \sqrt{2}})^2 +(\frac{\sqrt{3} + 1}{2 \sqrt{2}})^2 = \frac{1}{8} ((3 -2 \sqrt{3} + 1) + (3 +2 \sqrt{3} + 1) = 1$ so that we can write $\frac{\sqrt{3} - 1}{2 \sqrt{2}}$ as the $\sin$ of some angle, say $\alpha$, and identify $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$ as $\cos \alpha$.

Then you have to solve $\sin(\alpha + x) = \sin(2x)$, with $\tan \alpha = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.

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Replace $\sqrt3=\tan60^\circ,1=\tan45^\circ$

to find $$\sin(15^\circ+x)=2\sin x\cos x=\sin2x$$

Now if $\sin y=\sin A, y=n180^\circ+(-1)^nA$ where $n$ is any integer

As $\sqrt3-1=\tan60^\circ-\tan45^\circ=\dfrac{\sin(60-45)^\circ}{\cos60^\circ\cos45^\circ}=2\sqrt2\sin15^\circ$

Similarly, $\sqrt3+1=2\sqrt2\sin75^\circ=2\sqrt2\cos15^\circ$

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  • $\begingroup$ @user307178, Please follow the last two lines $\endgroup$ Commented Mar 20, 2016 at 9:05
  • $\begingroup$ Thanks got it...My reputation is less so I can't upvote $\endgroup$
    – user307178
    Commented Mar 20, 2016 at 9:07
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    $\begingroup$ @user307178, I actually tried to derive the sine/cosine value of $$\dfrac{\sqrt3\pm1}{2\sqrt2}$$ without assuming them. $\endgroup$ Commented Mar 20, 2016 at 9:15
  • $\begingroup$ And I have learned a useful method to derive them! Because I always forget it. Thanks! $\endgroup$
    – user307178
    Commented Mar 20, 2016 at 9:25
  • $\begingroup$ Do you know a similar trick to get sin18°? That would be awesome $\endgroup$
    – user307178
    Commented Mar 20, 2016 at 9:29
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We could obtain the following: $$(\sqrt{3}-1)\cos{x} + (\sqrt{3} + 1)\sin{x}=2\sqrt{2} \sin{2x}.$$ The LHS could be transformed into the following form: $$\text{LHS}=2\sqrt{2}\sin{(x+\alpha)},$$ where $\alpha = \arcsin{\frac{\sqrt{6}-\sqrt{2}}{4}}.$ Now what is $\alpha$? We could also see that $\alpha = \arctan{(2-\sqrt{3})}$, which inspires us to guess that $\alpha$ has something to do with $\frac{\pi}{6}$. Indeed, we verify that $\tan{\frac{\pi}{12}}=2-\sqrt{3}$. Therefore, the equation becomes $$\sin{(x+\frac{\pi}{12})}=\sin{2x}.$$ Thus, we obtain the solutions $$x_1=\frac{\pi}{12},\text{ }x_2=\frac{11}{36}\pi.$$

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