Considering the principal values of the inverse trigonometric functions, $\sin^{-1}\bigl(\tfrac{\sqrt{3}}{2}x + \tfrac{1}{2}\sqrt{1-x^2}\bigr), -\tfrac{1}{2} < x < \tfrac{1}{\sqrt{2}}$ is equal to:

Step 1: Substitute x with a trigonometric function

To simplify the expression, we can use a trigonometric substitution. Let $x = \sin\theta$. This substitution is valid because the domain of $x$ is given as $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$, which corresponds to a valid range for $\sin\theta$.

Step 2: Rewrite the expression using the substitution

Substitute $x = \sin\theta$ into the given expression. We also know that $\sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta|$. Since $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$, we have $-\frac{1}{2} < \sin\theta < \frac{1}{\sqrt{2}}$. This implies $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$. In this interval, $\cos\theta$ is positive, so $\sqrt{1-\sin^2\theta} = \cos\theta$.

Step 3: Apply trigonometric identity

We recognize that $\frac{\sqrt{3}}{2} = \sin\frac{\pi}{3}$ and $\frac{1}{2} = \cos\frac{\pi}{3}$. Substituting these values, the expression becomes $\sin^{-1}\left(\sin\frac{\pi}{3}\sin\theta + \cos\frac{\pi}{3}\cos\theta\right)$. This is in the form of the cosine addition formula.

Step 4: Simplify using cosine addition formula

Using the cosine addition formula, $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we can simplify the expression inside the inverse sine function to $\cos\left(\theta - \frac{\pi}{3}\right)$.

Step 5: Convert cosine to sine

To further simplify, we use the identity $\cos x = \sin\left(\frac{\pi}{2} - x\right)$. Applying this, we get $\sin^{-1}\left(\sin\left(\frac{\pi}{2} - \left(\theta - \frac{\pi}{3}\right)\right)\right)$. Simplifying the argument gives $\sin^{-1}\left(\sin\left(\frac{5\pi}{6} - \theta\right)\right)$.

Step 6: Final simplification

Since $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$, we have $-\frac{\pi}{4} < -\theta < \frac{\pi}{6}$. Adding $\frac{5\pi}{6}$ to the inequality, we get $\frac{5\pi}{6} - \frac{\pi}{4} < \frac{5\pi}{6} - \theta < \frac{5\pi}{6} + \frac{\pi}{6}$, which simplifies to $\frac{7\pi}{12} < \frac{5\pi}{6} - \theta < \pi$. This range is within the principal value branch of $\sin^{-1}y$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. However, the range $\frac{7\pi}{12} < \frac{5\pi}{6} - \theta < \pi$ is not within the principal value branch. Let's re-evaluate the identity. Instead of $\cos x = \sin(\frac{\pi}{2} - x)$, we can use $\cos x = \sin(\frac{\pi}{2} + x)$ if the argument is in the correct range. Let's check the range of $\theta - \frac{\pi}{3}$. Since $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$, then $-\frac{\pi}{6} - \frac{\pi}{3} < \theta - \frac{\pi}{3} < \frac{\pi}{4} - \frac{\pi}{3}$, which means $-\frac{\pi}{2} < \theta - \frac{\pi}{3} < -\frac{\pi}{12}$. In this range, $\cos(\theta - \frac{\pi}{3})$ is positive. We need to ensure that the argument of $\sin^{-1}(\sin A)$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Let $A = \theta - \frac{\pi}{3}$. Then $\cos A = \sin(\frac{\pi}{2} - A)$. So, $\sin^{-1}(\cos(\theta - \frac{\pi}{3})) = \sin^{-1}(\sin(\frac{\pi}{2} - (\theta - \frac{\pi}{3})))$. $\frac{\pi}{2} - (\theta - \frac{\pi}{3}) = \frac{\pi}{2} - \theta + \frac{\pi}{3} = \frac{5\pi}{6} - \theta$. The range of $\frac{5\pi}{6} - \theta$: Since $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$, then $-\frac{\pi}{4} < -\theta < \frac{\pi}{6}$. Adding $\frac{5\pi}{6}$: $\frac{5\pi}{6} - \frac{\pi}{4} < \frac{5\pi}{6} - \theta < \frac{5\pi}{6} + \frac{\pi}{6}$. $\frac{10\pi - 3\pi}{12} < \frac{5\pi}{6} - \theta < \frac{6\pi}{6}$. $\frac{7\pi}{12} < \frac{5\pi}{6} - \theta < \pi$. This range is not in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We know that $\sin^{-1}(\sin x) = x$ if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. If $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$, then $\sin^{-1}(\sin x) = \pi - x$. Since $\frac{7\pi}{12} < \frac{5\pi}{6} - \theta < \pi$, the argument is in the interval $(\frac{\pi}{2}, \pi)$. Therefore, $\sin^{-1}\left(\sin\left(\frac{5\pi}{6} - \theta\right)\right) = \pi - \left(\frac{5\pi}{6} - \theta\right) = \pi - \frac{5\pi}{6} + \theta = \frac{\pi}{6} + \theta$. Substitute back $\theta = \sin^{-1}x$.

More from this chapter

• cos ( sin^(-1) (3)/(8) + sin^(-1) (5)/(13) + sin^(-1) (33)/(65)) is equal to:
• If α > β > > 0, then the expression cot^(-1) β + ((1 + β^5))/((α - β)) + cot^(-1) + ((1 + ²))/((β -)) + cot^(-1) α + ((1 + α²))/(( - α)) is equal to:
• The value of cot^(-1) ( (√(1 + tan²(2)) - 1)/(tan(2))) - cot^(-1) ( (√(1 + tan²((1)/(2))) + 1)/(tan((1)/(2)))) is equal to:
• Solve for x: sin⁻¹(x) + sin⁻¹(√(1-x²)) = π/2 for x ∈ [0,1].
• Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec^(-1) x)² + (csc^(-1) x)²) is:
• If for some α, β; α ≤ β, α + β = 8 and sec²(tan^(-1) α) + csc²(cot^(-1) β) = 36, then α² + β is
• If y = cos((π)/(3) + cos^(-1)(x)/(2)), then (x - y)² + 3y² is equal to:
• If (π)/(2) ≤ x ≤ (3π)/(4), then cos^(-1) ( (12)/(13) cos x + (5)/(13) sin x) is equal to

All Inverse Trigonometric Functions solutions →