Answer :
To solve the expression \( 2 \sin \left(2 \sin^{-1} x\right) \), we can utilize trigonometric identities and properties of inverse trigonometric functions. Here’s the step-by-step solution:
1. Understanding the Inner Function: Let’s first consider the inner function \( \sin^{-1}(x) \). Suppose \( \theta = \sin^{-1}(x) \), which implies \( \sin(\theta) = x \) and \( \theta \) is an angle.
2. Double Angle Identity: We now need to evaluate \( \sin(2\theta) \) where \( \theta = \sin^{-1}(x) \). Using the double angle identity for sine, we have:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
3. Substitution for \( \sin(\theta) \): From our initial substitution, we know \( \sin(\theta) = x \).
4. Finding \( \cos(\theta) \): To find \( \cos(\theta) \), we use the Pythagorean identity. Since \( \sin(\theta) = x \), we have:
[tex]\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2} \][/tex]
5. Putting it All Together: Using the information above in the double angle identity, we get:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x \sqrt{1 - x^2} \][/tex]
6. Scaling by 2: Finally, the original expression was \( 2 \sin (2 \sin^{-1} x) \):
[tex]\[ 2 \sin(2 \theta) = 2(2 x \sqrt{1 - x^2}) = 4 x \sqrt{1 - x^2} \][/tex]
Therefore, the expression \( 2 \sin \left(2 \sin^{-1} x \right) \) can be written as the algebraic expression:
[tex]\[ 4 x \sqrt{1 - x^2} \][/tex]
1. Understanding the Inner Function: Let’s first consider the inner function \( \sin^{-1}(x) \). Suppose \( \theta = \sin^{-1}(x) \), which implies \( \sin(\theta) = x \) and \( \theta \) is an angle.
2. Double Angle Identity: We now need to evaluate \( \sin(2\theta) \) where \( \theta = \sin^{-1}(x) \). Using the double angle identity for sine, we have:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
3. Substitution for \( \sin(\theta) \): From our initial substitution, we know \( \sin(\theta) = x \).
4. Finding \( \cos(\theta) \): To find \( \cos(\theta) \), we use the Pythagorean identity. Since \( \sin(\theta) = x \), we have:
[tex]\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2} \][/tex]
5. Putting it All Together: Using the information above in the double angle identity, we get:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x \sqrt{1 - x^2} \][/tex]
6. Scaling by 2: Finally, the original expression was \( 2 \sin (2 \sin^{-1} x) \):
[tex]\[ 2 \sin(2 \theta) = 2(2 x \sqrt{1 - x^2}) = 4 x \sqrt{1 - x^2} \][/tex]
Therefore, the expression \( 2 \sin \left(2 \sin^{-1} x \right) \) can be written as the algebraic expression:
[tex]\[ 4 x \sqrt{1 - x^2} \][/tex]
