Calculating sin ( 0.5 arcsinx)

[Dethrone]

I did this question using two different trig identities, each of which gave me a different answer when I graphed them on Wolfram Alpha.

1. First identity: sinx = 2sin(0.5x)cos(0.5x)
isolating for sin(0.5x):

sin(0.5x) = (sinx) / (2cos(0.5x))

The question wants sin (0.5 arcsinx):

Applying the formula, we get the answer to be
\(\displaystyle \frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}\)

2. Second identity: sin²x = 0.5 (1-cos2x),
or sin²0.5x = 0.5(1-cosx)

Applying that formula, we get the answer to be
\(\displaystyle \sqrt{\frac{1}{2}(1-\sqrt{1-x^2})}\)

Why are the two answers different? Is my algebra wrong somewhere, or is this an issue of domain or something?

 

[Amer]

I did this question using two different trig identities, each of which gave me a different answer when I graphed them on Wolfram Alpha.

1. First identity: sinx = 2sin(0.5x)cos(0.5x)
isolating for sin(0.5x):

sin(0.5x) = (sinx) / (2cos(0.5x))

The question wants sin (0.5 arcsinx):

Applying the formula, we get the answer to be
\(\displaystyle \frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}\)

2. Second identity: sin²x = 0.5 (1-cos2x),
or sin²0.5x = 0.5(1-cosx)

Applying that formula, we get the answer to be
\(\displaystyle \sqrt{\frac{1}{2}(1-\sqrt{1-x^2})}\)

Why are the two answers different? Is my algebra wrong somewhere, or is this an issue of domain or something?

They are the same
\(\displaystyle \displaystyle \frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}= \frac{x \sqrt{1 -\sqrt{ 1-x^2}}}{\sqrt{2(1+\sqrt{1-x^2})}\sqrt{1 - \sqrt{1-x^2}}}= \frac{ x \sqrt{ 1 - \sqrt{ 1 -x^2 }} }{ \sqrt{ 2 ( 1 - (1 - x^2 ))} }= \sqrt{ \frac{1}{2} ( 1 - \sqrt{1 - x^2 } ) } \)

Note that $ \sqrt{x^2} = \mid x \mid $

 

[Dethrone]

Why then are their graphs different?

sqrt'('0.5'('1-sqrt'('1-x'^'2')'')' - Wolfram|Alpha

x '/' '(' 2 sqrt'('0.5'('1'+'sqrt'('1-x'^'2')'')'')'')' - Wolfram|Alpha

Actually, only one side it different, so does this have to do with absolute value or something?

 

[MarkFL]

It has been my experience that W|A does not handle imaginary numbers correctly with respect to graphing or odd roots of real negative values. Try graphing \(\displaystyle y=x^{\frac{1}{3}}\) and you get garbage for negative values of $x$.

 

[Dethrone]

Above, I used sin²0.5x = 0.5(1-cosx), but isolating \(\displaystyle \sin\left({x}\right) \)would give us |\(\displaystyle \sin\left({x}\right)\)|. And |\(\displaystyle \sin\left({x}\right)\)| is only equal to \(\displaystyle \sin\left({x}\right)\) on intervals such as \(\displaystyle (0, \pi)\) and \(\displaystyle (2\pi, 3\pi)\). Why, then, do they yield the same answer?