MHB Calculating sin ( 0.5 arcsinx)

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Two different trigonometric identities were used to calculate sin(0.5 arcsinx), resulting in two distinct expressions. The first identity yielded the result $$\frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}$$, while the second gave $$\sqrt{\frac{1}{2}(1-\sqrt{1-x^2})}$$. Despite appearing different, further algebraic manipulation shows these expressions are equivalent, although their graphs differ due to the handling of absolute values and domain restrictions. The discussion highlights potential issues with graphing tools like Wolfram Alpha, particularly regarding imaginary numbers and odd roots of negative values. Ultimately, both expressions represent the same mathematical relationship under specific conditions.
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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
$$\frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}$$

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

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

Why are the two answers different? Is my algebra wrong somewhere, or is this an issue of domain or something?
 
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Rido12 said:
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
$$\frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}$$

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

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

Why are the two answers different? Is my algebra wrong somewhere, or is this an issue of domain or something?
They are the same
$$\displaystyle \frac{x}{\sqrt{2(1+\sqrt{1-x^2})}}= \frac{x \sqrt{1 -\sqrt{ 1-x^2}}}{\sqrt{2(1+\sqrt{1-x^2})}\sqrt{1 - \sqrt{1-x^2}}}= \frac{ x \sqrt{ 1 - \sqrt{ 1 -x^2 }} }{ \sqrt{ 2 ( 1 - (1 - x^2 ))} }= \sqrt{ \frac{1}{2} ( 1 - \sqrt{1 - x^2 } ) } $$

Note that $ \sqrt{x^2} = \mid x \mid $
 
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 $$y=x^{\frac{1}{3}}$$ and you get garbage for negative values of $x$.
 
Above, I used sin20.5x = 0.5(1-cosx), but isolating $$\sin\left({x}\right) $$would give us |$$\sin\left({x}\right)$$|. And |$$\sin\left({x}\right)$$| is only equal to $$\sin\left({x}\right)$$ on intervals such as $$(0, \pi)$$ and $$(2\pi, 3\pi)$$. Why, then, do they yield the same answer?
 
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