Question

Prove the following identities and give the values of \(x\) for which they are true. $$\sin \left(2 \sin ^{-1} x\right)=2 x \sqrt{1-x^{2}}$$

Short answer

Question: Prove the identity \(\sin(2\sin^{-1}(x)) = 2x\sqrt{1-x^2}\) for \(-1 \leq x \leq 1\). Answer: We first express the double angle sine using sine and cosine: \(\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)\). Then, we replace \(\alpha\) with \(\sin^{-1}(x)\) in the identity and simplify the sine and cosine expressions. After plugging the simplified expressions back into the double angle sine formula, we obtain the result \(\sin(2\sin^{-1}(x)) = 2x\sqrt{1-x^2}\). This identity holds true for all \(x\) in the domain \(-1 \leq x \leq 1\).

Step-by-step solution

(Step 1: Express double angle of sine using sine and cosine)

(Recall the identity for double angle sine: \(\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)\). We will use this identity with \(\alpha=\sin^{-1}(x)\).)

(Step 2: Replace \(\alpha\) with \(\sin^{-1}(x)\) in the identity)

(Using the double angle sine identity, the expression \(\sin(2\sin^{-1}(x))\) can be written as: $$\sin(2\sin^{-1}(x)) = 2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))$$)

(Step 3: Simplify the sine and cosine expressions)

(We know that \(\sin(\sin^{-1}(x))=x\) because the sine and its inverse cancel each other out. To find \(\cos(\sin^{-1}(x))\), we can use the Pythagorean identity:$$\cos^2(\alpha) = 1 - \sin^2(\alpha)$$With \(\alpha=\sin^{-1}(x)\), we get $$\cos^2(\sin^{-1}(x)) = 1 - x^2$$Taking square root on both sides, we obtain $$\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$$ for \(-1 \leq x \leq 1\) as sine and cosine functions are real valued in this domain.)

(Step 4: Plug the simplified expressions back into the double angle sine formula)

(Substituting the simplified expressions of \(\sin(\sin^{-1}(x))\) and \(\cos(\sin^{-1}(x))\) back into the formula, we get$$\sin(2\sin^{-1}(x)) = 2x\sqrt{1-x^2}$$)

(Step 5: Determine the values of \(x\) where the identity holds true)

(The domain of the inverse sine function is \(-1 \leq x \leq 1\). Therefore, the given identity is true for all \(x\) in this domain:$$-1 \leq x \leq 1$$)

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us determine angle measures when given a trigonometric ratio. For instance, the inverse sine function, denoted as \( \sin^{-1}(x) \), gives you the angle whose sine is \( x \). This is particularly useful in trigonometry for solving equations and determining specific angle values.
A key property of inverse trigonometric functions is that they reverse the effect of the original function. So, applying \( \sin^{-1}(x) \) to \( x \) will produce an angle \( \theta \), such that \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \), where the sine of \( \theta \) is \( x \).

• For \( \sin(\sin^{-1}(x)) = x \), showing the action of sine and its inverse canceling each other.
• Within a certain range, specifically \(-1 \leq x \leq 1\), the inverse sine function outputs real angles.

Due to this behavior, understanding inverse trigonometric functions greatly enhances your ability to solve trigonometric identities and equations as seen in this exercise.

Double Angle Formulas

Double angle formulas are essential tools in trigonometry that allow you to express trigonometric functions of double angles (like \(2\theta\)) in terms of functions of \(\theta\). For instance, the double angle formula for sine is: \[ \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \] In the context of the given problem, using \(\alpha = \sin^{-1}(x)\), the expression \( \sin(2\sin^{-1}(x)) \) translates to \( 2x\sqrt{1-x^2} \). This manipulation uses both the identity of multiplying sine and cosine by 2 and the understanding of inverse trigonometric functions.
Here’s why it works:

• First, calculate \( \sin(\alpha) \) and \( \cos(\alpha) \) leveraging inverse trigonometric function properties.
• Substitute into the double angle formula to solve or simplify trigonometric expressions efficiently.

By mastering these formulas, trigonometric equations become more manageable and can often be transformed into expressions or values that are easier to interpret.

Pythagorean Identity

The Pythagorean identities in trigonometry are foundational tools that describe the inherent relationships between sine and cosine functions. One of the most well-known forms is \( \sin^2(\theta) + \cos^2(\theta) = 1 \). This formula stems directly from the Pythagorean Theorem and helps solve for unknown trigonometric functions.
In exercises involving inverse trig functions, such as proving the identity \( \sin(2\sin^{-1}(x)) = 2x\sqrt{1-x^2} \), we can rely on the Pythagorean identity to solve for \( \cos(\sin^{-1}(x)) \). Applying \( \cos^2(\alpha) = 1 - \sin^2(\alpha) \), and letting \( \alpha = \sin^{-1}(x)\), yields:

• \( \cos^2(\sin^{-1}(x)) = 1 - x^2 \)
• Simplifying to get \( \cos(\sin^{-1}(x)) = \sqrt{1-x^2} \).

This manipulation is crucial for solving the problem, ensuring that the domain restrictions are adhered to and the formulas used are accurate. Understanding the Pythagorean identity allows for the breakdown of components to determine exact trigonometric expressions critical for tests and practical applications.