Answer :
To write the expression \( 4 \sin \left(2 \sin^{-1} x \right) \) as an algebraic expression in \( x \) for \( x > 0 \), follow these steps:
1. Let \(\theta = \sin^{-1}(x)\):
Since \(\theta\) is the angle whose sine is \(x\), we have:
[tex]\[ \sin(\theta) = x \][/tex]
and
[tex]\[ \theta = \sin^{-1}(x) \][/tex]
2. Work with the double angle identity:
We need to express \(\sin(2\theta)\) where \(\theta = \sin^{-1}(x)\). The double angle identity for sine is:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
3. Substitute \(\sin(\theta) = x\):
Now we have:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x \cos(\theta) \][/tex]
4. Find \(\cos(\theta)\):
Given \(\sin(\theta) = x\), we can use the Pythagorean identity to find \(\cos(\theta)\). Recall that:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Since \(\sin^2(\theta) = x^2\), we get:
[tex]\[ x^2 + \cos^2(\theta) = 1 \][/tex]
Hence,
[tex]\[ \cos^2(\theta) = 1 - x^2 \][/tex]
Taking the positive square root (since \( x > 0 \) and we work within the principal range of arcsine \(0 \leq \theta \leq \frac{\pi}{2}\)):
[tex]\[ \cos(\theta) = \sqrt{1 - x^2} \][/tex]
5. Substitute \(\cos(\theta)\) back into the expression for \(\sin(2\theta)\):
[tex]\[ \sin(2\theta) = 2x \cos(\theta) = 2x \sqrt{1 - x^2} \][/tex]
6. Multiply by 4:
Finally, substitute \(\sin(2 \sin^{-1}(x))\) back into the original expression and multiply by 4:
[tex]\[ 4 \sin(2\sin^{-1}(x)) = 4 \cdot 2x \sqrt{1 - x^2} \][/tex]
Simplifying, we obtain:
[tex]\[ 4 \sin(2\sin^{-1}(x)) = 8x \sqrt{1 - x^2} \][/tex]
Therefore, the expression \( 4 \sin \left( 2 \sin^{-1} x \right) \) written as an algebraic expression in \( x \) for \( x > 0 \) is:
[tex]\[ 8x \sqrt{1 - x^2} \][/tex]
1. Let \(\theta = \sin^{-1}(x)\):
Since \(\theta\) is the angle whose sine is \(x\), we have:
[tex]\[ \sin(\theta) = x \][/tex]
and
[tex]\[ \theta = \sin^{-1}(x) \][/tex]
2. Work with the double angle identity:
We need to express \(\sin(2\theta)\) where \(\theta = \sin^{-1}(x)\). The double angle identity for sine is:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
3. Substitute \(\sin(\theta) = x\):
Now we have:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x \cos(\theta) \][/tex]
4. Find \(\cos(\theta)\):
Given \(\sin(\theta) = x\), we can use the Pythagorean identity to find \(\cos(\theta)\). Recall that:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Since \(\sin^2(\theta) = x^2\), we get:
[tex]\[ x^2 + \cos^2(\theta) = 1 \][/tex]
Hence,
[tex]\[ \cos^2(\theta) = 1 - x^2 \][/tex]
Taking the positive square root (since \( x > 0 \) and we work within the principal range of arcsine \(0 \leq \theta \leq \frac{\pi}{2}\)):
[tex]\[ \cos(\theta) = \sqrt{1 - x^2} \][/tex]
5. Substitute \(\cos(\theta)\) back into the expression for \(\sin(2\theta)\):
[tex]\[ \sin(2\theta) = 2x \cos(\theta) = 2x \sqrt{1 - x^2} \][/tex]
6. Multiply by 4:
Finally, substitute \(\sin(2 \sin^{-1}(x))\) back into the original expression and multiply by 4:
[tex]\[ 4 \sin(2\sin^{-1}(x)) = 4 \cdot 2x \sqrt{1 - x^2} \][/tex]
Simplifying, we obtain:
[tex]\[ 4 \sin(2\sin^{-1}(x)) = 8x \sqrt{1 - x^2} \][/tex]
Therefore, the expression \( 4 \sin \left( 2 \sin^{-1} x \right) \) written as an algebraic expression in \( x \) for \( x > 0 \) is:
[tex]\[ 8x \sqrt{1 - x^2} \][/tex]
