# Question #c5d00

Jun 16, 2016

$\sin \left(20\right) = \sqrt{\frac{1 - \sqrt{1 - {X}^{2}}}{2}}$

#### Explanation:

Let us draw a figure to depict given and what is to calculated

$\sin A = \sin \left(40\right) = X$
$\implies \text{opposite"/ "hypotenuse} = \frac{X}{1}$
This is shown in the figure above.
From Pythagoras theorem
$A B = \sqrt{1 - {X}^{2}}$ .....(1)
To calculate $\sin \left(20\right)$ we need to consider $\Delta A D B$
In this triangle to find $A D \mathmr{and} D B$ we proceed as below

If $A D$ is the angular bisector of $\angle C A B$, by angle bisector theorem we have
$C D : D B : : A C : A B$
$\implies \frac{C D}{D B} : : \frac{A C}{A B}$, From (1)
$\implies \frac{1}{D B} : : \frac{1}{\sqrt{1 - {X}^{2}}}$
From ratio and proportion
$D B = X \times \frac{\sqrt{1 - {X}^{2}}}{1 + \sqrt{1 - {X}^{2}}}$ .......(2)

Also using Pythagoras theorem
$A D = \sqrt{D {B}^{2} + A {B}^{2}}$
from (1) and (2)
$A D = \sqrt{{\left(X \times \frac{\sqrt{1 - {X}^{2}}}{1 + \sqrt{1 - {X}^{2}}}\right)}^{2} + {\left(\sqrt{1 - {X}^{2}}\right)}^{2}}$
$\implies A D = \sqrt{1 - {X}^{2}} \sqrt{{\left(\frac{X}{1 + \sqrt{1 - {X}^{2}}}\right)}^{2} + 1}$
$\implies A D = \sqrt{1 - {X}^{2}} \sqrt{\frac{{X}^{2} + {\left(1 + \sqrt{1 - {X}^{2}}\right)}^{2}}{1 + \sqrt{1 - {X}^{2}}} ^ 2}$
$\implies A D = \frac{\sqrt{1 - {X}^{2}}}{1 + \sqrt{1 - {X}^{2}}} \sqrt{{X}^{2} + 1 + 2 \sqrt{1 - {X}^{2}} + 1 - {X}^{2}}$
$\implies A D = \sqrt{2} \frac{\sqrt{1 - {X}^{2}}}{1 + \sqrt{1 - {X}^{2}}} \sqrt{1 + \sqrt{1 - {X}^{2}}}$
$\implies A D = \sqrt{2} \frac{\sqrt{1 - {X}^{2}}}{\sqrt{1 + \sqrt{1 - {X}^{2}}}}$ .......(3)
Now $\sin \left(20\right) = \frac{D B}{A D}$
Inserting the values from (2) and (3) we obtain

$\sin \left(20\right) = \frac{X \times \frac{\sqrt{1 - {X}^{2}}}{1 + \sqrt{1 - {X}^{2}}}}{\sqrt{2} \frac{\sqrt{1 - {X}^{2}}}{\sqrt{1 + \sqrt{1 - {X}^{2}}}}}$
$\sin \left(20\right) = \frac{X}{\sqrt{2} \sqrt{1 + \sqrt{1 - {X}^{2}}}}$
Rationalizing the denominator by multiplying and dividing RHS with the conjugate of denominator we get
RHS$= \frac{X}{\sqrt{2} \sqrt{1 + \sqrt{1 - {X}^{2}}}} \times \frac{\sqrt{1 - \sqrt{1 - {X}^{2}}}}{\sqrt{1 - \sqrt{1 - {X}^{2}}}}$
$\implies$RHS$= X \frac{\sqrt{1 - \sqrt{1 - {X}^{2}}}}{\sqrt{2} \sqrt{1 - \left(1 - {X}^{2}\right)}}$
$\implies$RHS$= \sqrt{\frac{1 - \sqrt{1 - {X}^{2}}}{2}}$

Jun 17, 2016

Let $\sin \left(20\right) = t$, so that $\cos \left(20\right) = \sqrt{1 - {t}^{2}} .$

Now given that, $\sin \left(40\right) = X$

$\therefore \sin \left(20 + 20\right) = X .$

Now to expand $L H S$, we use the identity $\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right) .$

$\therefore \sin \left(20\right) \cos \left(20\right) + \cos \left(20\right) \sin \left(20\right) = X .$

$\therefore 2 \sin \left(20\right) \cos \left(20\right) = X .$

Putting the values of $\sin \left(20\right)$ & $\cos \left(20\right) ,$

$\therefore 2 t \cdot \sqrt{1 - {t}^{2}} = X .$

Squaring, $4 {t}^{2} \cdot \left(1 - {t}^{2}\right) = {X}^{2.}$

$\therefore 4 {t}^{2} - 4 {t}^{4} = {X}^{2.}$

$\therefore 4 {t}^{4} - 4 {t}^{2} = - {X}^{2.}$

Completing square,
$4 {t}^{4} - 4 {t}^{2} + 1 = 1 - {X}^{2.}$

$\therefore {\left(2 {t}^{2} - 1\right)}^{2} = 1 - {X}^{2.}$

$\therefore 2 {t}^{2} - 1 = \pm \sqrt{1 - {X}^{2}} ,$ giving,

$t = \sin \left(20\right) = \sqrt{\frac{1 \pm \sqrt{1 - {X}^{2}}}{2}}$

The $+$ solution is not good since $\sin {20}^{o}$ should be less than $\sin {30}^{o} = \frac{1}{2}$ (function $y = \sin x$ is monotonically increasing in the 1st quadrant).

So the final solution is
$t = \sin \left(20\right) = \sqrt{\frac{1 - \sqrt{1 - {X}^{2}}}{2}}$