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If

$${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$$

then find the value of $\sin(4x)$.

The options given are:

a) $x$

b) $4x$

c) $2x$

I tried applying many trigo identities but none of them is working and the radicals are posing a big problem...need help...Thanks!!

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SirXYZ
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    You've asked 22 questions, answered 14 and being a member of this site for 45 days. I think it is about time you learn the easy directions to properly write mathematics in this site. – DonAntonio Nov 17 '16 at 18:31
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    @DonAntonio You should check out this user...45 days seem like nothing compared to the linked user who never typesets math properly. I personally feel like there should be some sort of ban after a time, but I highly doubt the community would go for that. Shameful nonetheless. – Daniel W. Farlow Nov 17 '16 at 18:35
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    Yeah... I completely agree with you guys. Also thanks for those stats @DonAntonio. I appreciate it. – SirXYZ Nov 17 '16 at 18:59
  • I'd be surprised if this had a closed form answer. – dxiv Nov 17 '16 at 21:20
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    Using $\sin 4x=\frac{2u}{1+u^2}$, with $u=\tan 2x$ and then $u=\frac{1+t^2}{1-t^2}$ with $t=\tan x$ gives you the answer... which is not so nice. – Paul Enta Nov 17 '16 at 21:31
  • @PaulEnta The given equation has one real solution $x \approx 0.6186 \approx 0.1969 ,\pi;$ so $\sin(4x)$ is one real value, not an expression in $x$. – dxiv Nov 17 '16 at 21:42
  • @dxiv You're right, sorry about that! – Paul Enta Nov 17 '16 at 21:44
  • Check out the edited question... the given answer is $x$. – SirXYZ Nov 18 '16 at 01:10
  • The answer is x(calculator checks out taking $x=0.61864419684245725314110752758288$). I would try to simplify the expression obtained by @PaulEnta above. – Shraddheya Shendre Nov 18 '16 at 01:31
  • @SirJMP Next time please post the entire question upfront. It's not as much a matter of which of a/b/c is correct, but the fact that the problem was asking for an expression in $x$ i.e. not the actual value of $\sin(4x)$. – dxiv Nov 18 '16 at 01:46
  • @PaulEnta It looks like you guessed right what the question meant to ask after all. – dxiv Nov 18 '16 at 01:53
  • @DonAntonio Where could I learn more about MathJax? – Myungjin Hyun Oct 16 '17 at 12:03

2 Answers2

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Let: $$t = \tan(x) = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$$

Using the double-angle sine and half-angle tangent formulas:

$$ \sin(4x) \;=\; 2\;\sin(2x)\;\cos(2x) \;=\; 2 \cdot \frac{2 t}{1 + t^2} \cdot \frac{1-t^2}{1+t^2} \;=\; \frac{4t(1-t^2)}{(1+t^2)^2} $$

Substituting back $t$ in terms of $x$ and simplifying:

$$ \require{cancel} \begin{align} \sin(4x) & =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(\sqrt{1-x}+\sqrt{1+x}+2)^2} \\ & =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(1-x) + (1+x) + 4 + 2 \sqrt{(1-x)(1+x)} + 4 \sqrt{1-x} + 4 \sqrt{1+x} } \\ & =\; \frac{\cancel{2} x \bcancel{(\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}}{\cancel{2}\bcancel{(3 + \sqrt{1-x^2} + \sqrt{1-x}+\sqrt{1+x})} } \\ & =\; x \end{align} $$

dxiv
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For convenience, write $$ s := \sin x \qquad c := \cos x \qquad m := 1 +\sqrt{1+x} \qquad n := 1 + \sqrt{1-x}$$

The initial equation can then be rewritten as $$m s = n c \qquad\to\qquad m^2 s^2 = n^2 c^2 \qquad \to\qquad c^2 = \frac{m^2}{m^2+n^2} \qquad s^2 = \frac{n^2}{m^2 + n^2}$$ Note that $m$ and $n$ are both strictly positive. Since (presumably real) $x$ must lie between $1$ and $-1$, we know that $\cos x \geq 0$; since $\tan x$ must be positive in the original equation, $\sin x$ is, too. Thus, we have ... $$c = \frac{m}{\sqrt{m^2+n^2}} \qquad s = \frac{n}{\sqrt{m^2+n^2}}$$

Now ...

$$\begin{align} \sin 4x &= 2 \sin 2x \cos 2x \\[4pt] &= 4 sc ( 2 c^2 - 1 ) \\[4pt] &= \frac{4mn}{m^2 + n^2} \left(\frac{2 m^2}{m^2 + n^2} - 1\right) \\[4pt] &= \frac{4mn(m^2-n^2)}{(m^2 + n^2)^2} \\[4pt] &= \frac{8 x (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})}{8 (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})} \\[4pt] &= x \end{align}$$

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