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How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?

I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.

For instance: $\sin(2x) = 2 \sin(x) \cos(x)$

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    $\begingroup$ $4sin^2 cos^2 = (24/25)^2$, $ cos^2 =1 - sin^2 $, solve the previous for $sin$. $\endgroup$ – Rodrigo Fontana May 18 '18 at 21:54
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$$\cos ^2 {2x} = 1-\sin ^2 {2x} = \frac {49}{625}$$ $$ \cos 2x = \frac {7}{25}$$ $$ \sin ^ 2x = \frac {1-\cos 2x}{2} = 9/25 $$ $$ \sin x = \frac {3}{5} $$

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  • $\begingroup$ Check the case of $\frac{-3}{5}$. If $\sin(x)=\frac{-3}{5}$, then $\sin(2x)=-\frac{24}{25}$ $\endgroup$ – Rhys Hughes May 18 '18 at 22:32
  • $\begingroup$ @RhysHughes Good comment, thanks, $\endgroup$ – Mohammad Riazi-Kermani May 18 '18 at 23:48
  • $\begingroup$ In your first line, you should have $\cos^2(2x) = 1 - \sin^2(2x) = \frac{49}{625}$. $\endgroup$ – N. F. Taussig May 19 '18 at 10:15
  • $\begingroup$ @N.F.Taussig Thanks, I fixed it. $\endgroup$ – Mohammad Riazi-Kermani May 19 '18 at 12:50
  • $\begingroup$ That covers only one of four solutions due to two $y^2=z^2\to y=z$ fallacies. $\endgroup$ – J.G. May 20 '18 at 21:12
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Since $\cos 2x=\pm\frac{7}{25}$, $\tan x=\frac{\sin 2x}{1+\cos 2x}\in\left\{\frac{24}{32},\,\frac{24}{18}\right\}=\left\{\frac{3}{4},\,\frac{4}{3}\right\}$ so $\sin x=\pm\frac{\tan x}{\sqrt{1+\tan^2 x}}\in\pm\left\{\frac{3}{5},\,\frac{4}{5}\right\}=\left\{-\frac{4}{5},\,-\frac{3}{5},\,\frac{3}{5},\,\frac{4}{5}\right\}$.

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$$|\sin(2x)|=|2\sin(x)\cos(x)|=|2\sin(x)\sqrt{1-\sin(x)^2}|$$ so $$\left(\frac{24}{25}\right)^2=4\sin(x)^2(1-\sin(x)^2)$$ solve this quadratic for $\sin(x)^2$ and get $|\sin(x)|$. Both the positive and negative values will be possible (think of why).

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  • $\begingroup$ Square should be inside paranthesis not outside.. $\endgroup$ – Isham May 18 '18 at 22:11
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$tanx=\frac{sin2x}{1+cos2x}$. You have $\sin2x$ and you can calculate $\cos2x$ with the Pythagorean theorem. Once you have $\ tanx$, you can use SOHCAHTOA. Don't forget the quadrants though

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    $\begingroup$ Use a backslash before trig functions like \sin x so they render as $\sin x$ and not $sin x$ $\endgroup$ – Chase Ryan Taylor May 18 '18 at 23:27
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Refer to the diagram below.

$AD$ is the angle bisector of the right-angled triangle $\Delta ABC$.

Given $BC=24$ and $AC=25$.

Let $\angle DAB = x$.

From $\Delta ABC$ we see that $\sin(2x) = \dfrac{24}{25}$.

Now, by angle bisector theorem, $BD:DC = 7:25$.

Therefore, $BD = \dfrac{7}{7+25} \times 24 = \dfrac{21}4$.

Observing that $AB:BD = 4:3$, we see that $AD = \dfrac{35}4$

Therefore, $\sin x = \dfrac {BD} {DA} = \dfrac 3 5$.

The other possible $x$ would be $x+180^\circ$, since the period of sine is $360^\circ$, making $\sin x = -\dfrac35$.

In conclusion, $\sin x = \pm \dfrac 35$.

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  • $\begingroup$ Be careful $\sin^{-1}(-\frac 35)\approx -0.6435$ then $\sin(2\cdot-0.6435)=-\frac{24}{25}$ $\endgroup$ – Rhys Hughes May 18 '18 at 22:29
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$$\sin(2x)=\frac{24}{25}$$ The identities you want are: $$\sin(2x)=2\sin(x)\cos(x) \space [I]$$

and $$\sin^2(x)+\cos^2(x)=1\space[II]$$

Square $[I]$ to get: $$4\sin^2(x)\cos^2(x)=\frac{576}{625}\to\sin^2x(1-\sin^2x)=\frac{144}{625}$$ $$\to\sin^4(x)-\sin^2x+\frac{144}{625}=0$$ Let $\sin^2(x)=\sigma$. Your equation becomes: $\sigma^2-\sigma+\frac{144}{625}=0$

Solve via Quadratic Formula: $$\sigma=\frac{1\pm\sqrt{1-\frac{576}{625}}}{2}=\frac 12\pm\frac{\sqrt{\frac{49}{625}}}{2}=\frac{9}{25}, \frac{16}{25}.$$

Since $\sin^2(x)=\frac{9}{25}, \frac{16}{25}, \sin(x)=\pm \frac 35, \pm \frac 45$

Check these values with the function $f(S)=\sin(2\cdot\sin^{-1}(S))$, (where $S$ is the solutions i.e. $\pm\frac 35, \pm \frac 45$) to see the ones which provide the correct solution, and note the values which work are $\frac 35, \frac 45$.

We can see them graphed here:

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