Inverse Trigonometric Functions – JEE Main MathematicsPractice Questions & PYQs

The sine inverse function $\sin^{-1}(u)$ is defined only when its input argument $u$ lies within the closed interval $[-1, 1]$. Consequently, to determine the domain of the function $f(x)=\sin^{-1}\left(\frac{x^{2-}3x+2}{x^{2+}2x+7}\right)$, we must satisfy the condition $-1 \leq \frac{x^{2-}3x+2}{x^{2+}2x+7} \leq 1$.

Because the denominator $x^{2+}2x+7$ is always positive for all real $x$-confirmed by its negative discriminant-we can multiply throughout the inequalities without changing the sign. First, solving $\frac{x^{2-}3x+2}{x^{2+}2x+7} \leq 1$ yields $x^{2-}3x+2 \leq x^{2+}2x+7$, which simplifies to $-3x+2 \leq 2x+7$, resulting in $5x \geq -5$, or $x \geq -1$. Next, solving the lower bound condition $\frac{x^{2-}3x+2}{x^{2+}2x+7} \geq -1$ leads to $x^{2-}3x+2 \geq -x^{2-}2x-7$, which rearranges to $2x^{2-}x+9 \geq 0$. Since this quadratic also has a negative discriminant, it is always positive for all real numbers. Taking the intersection of these two conditions, $x \geq -1$ and $x \in \mathbb{R}$, the final domain is $[-1, \infty )$.

The evaluation of the internal expression relies on simplifying the trigonometric ratios, where $\cos\left( \frac{15\pi}{4} \right)$ is equivalent to $\cos\left( \frac{7\pi}{4} \right)$, which yields $\frac{1}{\sqrt{2}}$. Substituting this along with $\sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$ transforms the fraction into $\frac{ \frac{1}{\sqrt{2}} - 1 }{ \frac{1}{\sqrt{2}} }$, which simplifies directly to $1 - \sqrt{2}$. Recognizing that $\tan\left( \frac{\pi}{8} \right) = \sqrt{2} - 1$ implies that the inverse tangent of $1 - \sqrt{2}$ is equivalent to $-\frac{\pi}{8}$.

The inner function $g(x)=\sin x-\cos x$ can be rewritten using the harmonic identity as $\sqrt{2}\sin(x-\frac{\pi}{4})$. As $x$ ranges from $0$ to $\pi$, the argument $(x-\frac{\pi}{4})$ spans the interval $[-\frac{\pi}{4}, \frac{3\pi}{4}]$, causing $g(x)$ to traverse the values from $-1$ to $\sqrt{2}$. Because $\tan^{-1}$ is a strictly increasing function, the range of $f(x)=\tan^{-1}(g(x))$ is $[\tan^{-1}(-1), \tan^{-1}(\sqrt{2})]$, which simplifies to $[-\frac{\pi}{4}, \tan^{-1}(\sqrt{2})]$. Adding these absolute minimum and maximum values yields $\tan^{-1}(\sqrt{2})-\frac{\pi}{4}$. By considering a right-angled triangle where $\tan \theta = \sqrt{2}$, one observes that the opposite side is $\sqrt{2}$ and the adjacent side is $1$, resulting in a hypotenuse of $\sqrt{3}$ and therefore $\cos \theta = \frac{1}{\sqrt{3}}$. This identity allows the expression to be written as $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}$.

To determine the range of values for $k$, recognize the fundamental relationship between inverse trigonometric functions, specifically that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$. By introducing a variable $t = \tan^{-1} x$, the expression is transformed into a single-variable function $f(t)$ defined over the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$:
$f(t) = t^3 + \left( \frac{\pi}{2} - t \right)^3$
Expanding this cubic expression results in a quadratic function, as the $t^3$ terms cancel out:
$f(t) = \frac{3\pi}{2} t^2 - \frac{3\pi^2}{4} t + \frac{\pi^3}{8}$
Because the coefficient of $t^2$ is positive, this represents an upward-opening parabola. The vertex of the parabola is located where the derivative $f'(t) = 3\pi t - \frac{3\pi^2}{4}$ equals zero, which occurs at $t = \frac{\pi}{4}$. Since this point lies within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, the function reaches its minimum value at $t = \frac{\pi}{4}$, which is $f(\frac{\pi}{4}) = \frac{\pi^3}{32}$. As $t$ approaches the lower boundary of $-\frac{\pi}{2}$, the expression approaches the value $f(-\frac{\pi}{2}) = -\frac{\pi^3}{8} + \pi^3 = \frac{7\pi^3}{8}$. Therefore, the sum takes on all values in the interval $[\frac{\pi^3}{32}, \frac{7\pi^3}{8})$, which corresponds to $k$ ranging from $\frac{1}{32}$ to $\frac{7}{8}$.

Converting the $\sin^{-1}(x)$ term into a $\cos^{-1}(x)$ format using the identity $\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)$ transforms the equation $\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$ into $\cos^{-1}(x) - 2(\frac{\pi}{2} - \cos^{-1}(x)) = \cos^{-1}(2x)$. This expression simplifies algebraically to $3\cos^{-1}(x) - \pi = \cos^{-1}(2x)$.

Applying the cosine function to both sides results in $\cos(3\cos^{-1}(x) - \pi) = \cos(\cos^{-1}(2x))$. Recalling that $\cos(\theta - \pi) = -\cos(\theta)$, the left side transforms into $-\cos(3\cos^{-1}(x))$, which can be expanded using the triple angle identity $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$ to yield $-(4x^3 - 3x)$. Setting this equal to the right side, which simplifies directly to $2x$, creates the equation $-4x^3 + 3x = 2x$, or $4x^3 - x = 0$.

Factoring this equation as $x(4x^2 - 1) = 0$ provides the solutions $x = 0$, $x = \frac{1}{2}$, and $x = -\frac{1}{2}$. All these candidates lie within the required domain of $x \in [-\frac{1}{2}, \frac{1}{2}]$, and their sum is $0 + \frac{1}{2} + (-\frac{1}{2}) = 0$.

The function is defined only when the input to the inverse cosine function lies within the interval $[-1, 1]$ and the argument of the logarithm is strictly positive while not equal to $1$. Simplifying the inverse cosine input $\frac{x^{2-}5x+6}{x^{2-}9}$ by factoring the numerator and denominator yields $\frac{x-2}{x+3}$ for $x \neq 3$, and solving the inequality $-1 \leq \frac{x-2}{x+3} \leq 1$ produces the set $\left[-\frac{1}{2}, \infty \right)$ excluding $x=3$. Simultaneously, the condition $x^{2-}3x+2 > 0$ restricts the variable to $(-\infty , 1) \cup (2, \infty )$, while the restriction $x^{2-}3x+2 \neq 1$ requires excluding the values $x = \frac{3 \pm \sqrt{5}}{2}$. Intersecting these requirements leads to the domain $\left[-\frac{1}{2}, 1\right) \cup (2, \infty ) - \left\{3, \frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right\}$.

Simplifying this expression requires applying inverse trigonometric identities and the tangent half-angle formula. Focusing on the first term, $50 \tan \left( 3 \tan^{-1} \left( \frac{1}{2} \right) + 2 \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \right)$, recognize that $\cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$ is equivalent to $\tan^{-1} 2$ because a right triangle with an adjacent side of 1 and an opposite side of 2 has a hypotenuse of $\sqrt{5}$. The argument then becomes $\tan^{-1} \left( \frac{1}{2} \right) + 2 \left( \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} 2 \right)$. Since $\tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} 2 = \frac{\pi}{2}$, the expression simplifies to $50 \tan \left( \tan^{-1} \left( \frac{1}{2} \right) + \pi \right)$, which equals $50 \times \frac{1}{2} = 25$.

For the second term, $4\sqrt{2} \tan \left( \frac{1}{2} \tan^{-1} \left( 2\sqrt{2} \right) \right)$, define $\theta = \tan^{-1} \left( 2\sqrt{2} \right)$, which implies $\tan \theta = 2\sqrt{2}$. Using the half-angle formula $\tan \left( \frac{\theta}{2} \right) = \frac{\sin \theta}{1 + \cos \theta}$ with $\sin \theta = \frac{2\sqrt{2}}{3}$ and $\cos \theta = \frac{1}{3}$, the value becomes $\frac{2\sqrt{2}/3}{1 + 1/3} = \frac{1}{\sqrt{2}}$. Multiplying this by $4\sqrt{2}$ yields 4. Combining both parts gives a final value of 29.

The inverse cosine function $\cos^{-1}(u)$ is defined only when the input $u$ falls within the closed interval $[-1, 1]$, requiring the condition $-1 \leq \frac{x^{2-}4x+2}{x^{2+}3} \leq 1$. The left inequality $\frac{x^{2-}4x+2}{x^{2+}3} \geq -1$ simplifies to $2x^{2-}4x+5 \geq 0$, which holds for all real $x$ because the quadratic has a negative discriminant and a positive leading coefficient. The right inequality $\frac{x^{2-}4x+2}{x^{2+}3} \leq 1$ simplifies to $x^{2-}4x+2 \leq x^{2+}3$, which further reduces to $-4x \leq 1$, meaning $x \geq -\frac{1}{4}$. These conditions overlap to define the domain as $[-\frac{1}{4}, \infty )$.

The fundamental identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2}$ simplifies the expression significantly because the function $f(x) = \sin^{-1} 2x + \sin 2x + \cos^{-1} 2x + \cos 2x$ contains both inverse trigonometric terms. Grouping these inverse terms, we find that the expression becomes $f(x) = \frac{\pi}{2} + \sin 2x + \cos 2x$. To determine the range of this function, we can rewrite the trigonometric component $\sin 2x + \cos 2x$ as $\sqrt{2} \sin(2x + \frac{\pi}{4})$ by applying the harmonic addition theorem.

Considering the given interval $x \in [0, \frac{\pi}{8}]$, the argument of the sine function, $2x + \frac{\pi}{4}$, ranges from $\frac{\pi}{4}$ to $\frac{\pi}{2}$. Since the sine function is monotonically increasing on this interval, the minimum value $m$ occurs at the lower bound where $2x + \frac{\pi}{4} = \frac{\pi}{4}$, and the maximum value $M$ occurs at the upper bound where $2x + \frac{\pi}{4} = \frac{\pi}{2}$. Evaluating these, the minimum is $m = \frac{\pi}{2} + \sqrt{2} \sin(\frac{\pi}{4}) = \frac{\pi}{2} + 1$, and the maximum is $M = \frac{\pi}{2} + \sqrt{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} + \sqrt{2}$. Summing these values gives $m + M = \pi + \sqrt{2} + 1$.

Evaluating composite trigonometric functions requires systematic reduction from the innermost term, while the relationship between quadratic roots and coefficients is governed by Vieta's formulas. Starting with the innermost expression $\sin^{-1} x = \theta$, we observe $\cos \theta = \sqrt{1-x^2}$. Expanding the next term $\tan^{-1}(\sqrt{1-x^2})$ yields an angle $\phi$ such that $\tan \phi = \sqrt{1-x^2}$, which implies $\cot \phi = \frac{1}{\sqrt{1-x^2}}$. This simplifies the argument of the outer cosine function to $\sin^{-1}(\frac{x}{\sqrt{1-x^2}})$. Applying the cosine identity leads to the equation:
$\cos\left(\sin^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right) = k$
Using the property $\cos(\sin^{-1} u) = \sqrt{1-u^2}$, the equation becomes $\sqrt{1 - \frac{x^2}{1-x^2}} = k$, which simplifies to $\frac{1-2x^2}{1-x^2} = k^2$.

Solving for $x^2$ gives $x^2 = \frac{k^{2-}1}{k^{2-}2}$. Because this quadratic form is $x^2 - C = 0$, the roots $\alpha$ and $\beta$ satisfy $\alpha = -\beta$, which means their ratio is $\frac{\alpha}{\beta} = -1$ and the sum of their reciprocal squares is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{2}{\alpha^2} = 2\left(\frac{k^{2-}2}{k^{2-}1}\right)$. The quadratic $x^2 - bx - 5 = 0$ is defined with roots equal to these values, so the product of the roots is:
$\left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) \cdot \left(\frac{\alpha}{\beta}\right) = -5$
Substituting $\frac{\alpha}{\beta} = -1$ into this product gives $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 5$. Equating this to the previous expression $2\left(\frac{k^{2-}2}{k^{2-}1}\right)$ yields $2k^2 - 4 = 5k^2 - 5$, or $k^2 = \frac{1}{3}$. Since the sum of the roots for $x^2 - bx - 5 = 0$ is $b$, we have $b = 5 + (-1) = 4$. Consequently, the value of $\frac{b}{k^2}$ is $\frac{4}{1/3} = 12$.

Rewriting the equation $\cos^{-1}\left(\frac{y}{2}\right) = \log_e\left(\frac{x}{5}\right)^5$ as $\cos^{-1}\left(\frac{y}{2}\right) = 5\ln\left(\frac{x}{5}\right)$ allows for differentiation with respect to $x$, giving $-\frac{1}{\sqrt{1 - \frac{y^2}{4}}} \cdot \frac{y'}{2} = \frac{5}{x}$. Because the square root simplifies to $\frac{\sqrt{4-y^2}}{2}$, the twos cancel out to produce $-\frac{y'}{\sqrt{4-y^2}} = \frac{5}{x}$. Cross-multiplying and squaring both sides yields $x^2(y')^2 = 25(4 - y^2)$, which simplifies to the expression $x^2(y')^2 = 100 - 25y^2$.

Differentiating this result with respect to $x$ gives $2x(y')^2 + x^2(2y'y'') = -50yy'$. Dividing the entire equation by $2y'$ isolates the second-order derivative and simplifies the expression to $x y' + x^2 y'' = -25y$. Rearranging this to group all terms on one side leads to the following equation:

$x^2 y'' + x y' + 25 y = 0$

Inverse trigonometric functions possess restricted principal value domains, meaning the angle inside must be brought into these specific ranges before the inverse operation can be applied. For $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$, since $\frac{2\pi}{3}$ exceeds the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we use the property $\sin(\pi - \theta) = \sin\theta$ to express it as $\sin(\pi - \frac{2\pi}{3}) = \sin\frac{\pi}{3}$, which results in $\frac{\pi}{3}$. The cosine term $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ must lie within $[0, \pi]$, so applying the symmetry $\cos(2\pi - \theta) = \cos\theta$ gives $\cos(2\pi - \frac{7\pi}{6}) = \cos\frac{5\pi}{6}$, resulting in $\frac{5\pi}{6}$. For the tangent component, $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$ requires the input to fall within the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is satisfied by shifting the angle using $\tan(\theta - \pi) = \tan\theta$, giving $\tan(\frac{3\pi}{4} - \pi) = \tan(-\frac{\pi}{4})$ and simplifying to $-\frac{\pi}{4}$. Summing these principal values yields the final result:

$\frac{\pi}{3} + \frac{5\pi}{6} - \frac{\pi}{4} = \frac{4\pi + 10\pi - 3\pi}{12} = \frac{11\pi}{12}$

The function $\cos^{-1}(u)$ is defined for inputs in the interval $[-1, 1]$. Applying this constraint to our expression, we must satisfy $-1 \leq \frac{2\sin^{-1}\left(\frac{1}{4x^{2-}1}\right)}{\pi} \leq 1$, which simplifies to $-\frac{\pi}{2} \leq \sin^{-1}\left(\frac{1}{4x^{2-}1}\right) \leq \frac{\pi}{2}$. Because the output of the inverse sine function is restricted by definition to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, this condition is inherently satisfied whenever the argument $\frac{1}{4x^{2-}1}$ is within the valid domain for $\sin^{-1}$, which is $[-1, 1]$.

This requirement leads to the inequality $-1 \leq \frac{1}{4x^{2-}1} \leq 1$. Considering the left side, $\frac{1}{4x^{2-}1} \geq -1$, we rearrange this to $\frac{1}{4x^{2-}1} + 1 \geq 0$, which yields $\frac{4x^2}{4x^{2-}1} \geq 0$. For this fraction to be non-negative, the denominator must be positive (since the numerator $4x^2$ is always non-negative), or the numerator must be zero. This requires $4x^{2-}1 > 0$ or $x=0$, which gives $x \in (-\infty , -\frac{1}{2}) \cup (\frac{1}{2}, \infty ) \cup \{0\}$.

Next, addressing the right side, $\frac{1}{4x^{2-}1} \leq 1$, we rearrange the inequality to $\frac{1}{4x^{2-}1} - 1 \leq 0$, leading to $\frac{2-4x^2}{4x^{2-}1} \leq 0$. Multiplying by $-1$ reverses the inequality to $\frac{4x^{2-}2}{4x^{2-}1} \geq 0$, or $\frac{2(2x^{2-}1)}{4x^{2-}1} \geq 0$. Analyzing the sign of this expression across the critical points $\pm\frac{1}{2}$ and $\pm\frac{1}{\sqrt{2}}$, we find that it holds true for $x \in (-\infty , -\frac{1}{\sqrt{2}}] \cup (-\frac{1}{2}, \frac{1}{2}) \cup [\frac{1}{\sqrt{2}}, \infty )$. Taking the intersection of these two solution sets, we obtain the valid domain as $\left(-\infty , -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty \right) \cup \{0\}$.

The core principle for solving this expression is to consolidate the inverse trigonometric terms using standard sum and double-angle identities to reduce the complexity before evaluating the tangent. We first isolate the terms $2 \tan^{-1} \frac{1}{5}$ and $2 \tan^{-1} \frac{1}{8}$ by factoring out the two and applying the sum formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$ to the arguments $1/5$ and $1/8$, which results in $2 \tan^{-1} \left( \frac{1/5+1/8}{1-(1/5)(1/8)} \right) = 2 \tan^{-1} \frac{1}{3}$. Applying the double-angle identity $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$ to this result yields $\tan^{-1} \left( \frac{2/3}{1-1/9} \right)$, which simplifies to $\tan^{-1} \frac{3}{4}$. We then convert the inverse secant term $\sec^{-1} \frac{\sqrt{5}}{2}$ into an equivalent inverse tangent by interpreting it as an angle whose secant is $\frac{\sqrt{5}}{2}$, which corresponds to a tangent of $\frac{1}{2}$ via the identity $\tan \theta = \sqrt{\sec^2 \theta - 1}$. Combining $\tan^{-1} \frac{3}{4}$ and $\tan^{-1} \frac{1}{2}$ using the addition formula gives $\tan^{-1} \left( \frac{3/4+1/2}{1-(3/4)(1/2)} \right)$, which simplifies to $\tan^{-1} 2$. Applying the tangent function to this inverse tangent value leaves us with the final result of 2.

The operation defined by $x \cdot y = x^{2} + y^{3}$ requires applying the rule sequentially to resolve the nested terms in the given equality. Evaluating the left side, $(x \cdot 1) \cdot 1$ first yields $(x^2 + 1^3) \cdot 1 = (x^2 + 1) \cdot 1$, which further becomes $(x^2 + 1)^2 + 1^3$, expanding to $x^4 + 2x^2 + 2$. On the right side, the inner part $1 \cdot 1$ evaluates to $1^2 + 1^3 = 2$, simplifying the expression to $x \cdot 2$, which is $x^2 + 2^3 = x^2 + 8$.

Equating these two results produces $x^4 + 2x^2 + 2 = x^2 + 8$, which simplifies to the quadratic form $x^4 + x^2 - 6 = 0$. Factoring this leads to $(x^2 + 3)(x^2 - 2) = 0$, where $x^2 = 2$ is the only valid solution because the square of a real number cannot be negative. Finally, substituting $x^2 = 2$ into the expression $2\sin^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)$ gives $2\sin^{-1}\left(\frac{4+2-2}{4+2+2}\right)$, which simplifies to $2\sin^{-1}\left(\frac{1}{2}\right)$, resulting in $2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

Translating inverse trigonometric expressions into algebraic forms provides the most direct path to solving this problem. The term $x$ corresponds to the standard double-angle identity for sine, where $x = \sin(2\tan^{-1}\alpha) = \frac{2\alpha}{1+\alpha^2}$. For the value of $y$, letting $\theta = \tan^{-1}\frac{4}{3}$ implies $\tan\theta = \frac{4}{3}$, which corresponds to a triangle with sides in a $3:4:5$ ratio, resulting in $\cos\theta = \frac{3}{5}$. Applying the half-angle identity $\sin(\frac{\theta}{2}) = \sqrt{\frac{1-\cos\theta}{2}}$, we find $y = \sqrt{\frac{1-3/5}{2}} = \frac{1}{\sqrt{5}}$, which means $y^2 = \frac{1}{5}$.

Substituting these expressions into the equation $y^2 = 1 - x$ provides the relationship
$\frac{1}{5} = 1 - \frac{2\alpha}{1+\alpha^2}$
Rearranging the terms yields $\frac{2\alpha}{1+\alpha^2} = \frac{4}{5}$, which simplifies to $10\alpha = 4(1+\alpha^2)$. This leads to the quadratic equation
$2\alpha^2 - 5\alpha + 2 = 0$
Factoring this quadratic gives $(2\alpha - 1)(\alpha - 2) = 0$, identifying the roots as $\alpha = \frac{1}{2}$ and $\alpha = 2$. Calculating the sum $16\alpha^3$ for these values gives $16(2^3) + 16(\frac{1}{2})^3$, which equals $16(8) + 16(\frac{1}{8}) = 128 + 2 = 130$.

To determine the domain of the function, we must ensure that each component part is well-defined within its specific mathematical constraints. The inverse sine function $\sin^{-1}(t)$ is defined only when its argument satisfies $-1 \leq t \leq 1$. Given the argument is the greatest integer function $[2x^{2-}3]$, the possible integer outputs are $-1, 0,$ and $1$. This requirement is satisfied when $-1 \leq 2x^{2-}3 < 2$, which rearranges to $2 \leq 2x^2 < 5$ or $1 \leq x^2 < \frac{5}{2}$. From this, $x$ must lie in the intervals $(-\sqrt{\frac{5}{2}}, -1] \cup [1, \sqrt{\frac{5}{2}})$.

Simultaneously, the logarithmic term $\log_2(\log_{\frac{1}{2}}(x^{2-}5x+5))$ requires the inner logarithmic expression to be strictly positive. Since the base $\frac{1}{2}$ is less than $1$, the inequality $\log_{\frac{1}{2}}(x^{2-}5x+5) > 0$ implies $0 < x^{2-}5x+5 < (\frac{1}{2})^0$, which simplifies to $0 < x^{2-}5x+5 < 1$. The quadratic inequality $x^{2-}5x+5 > 0$ is satisfied for $x \in (-\infty , \frac{5-\sqrt{5}}{2}) \cup (\frac{5+\sqrt{5}}{2}, \infty )$, while $x^{2-}5x+5 < 1$ reduces to $x^{2-}5x+4 < 0$, identifying the interval $(1, 4)$. Intersecting these two regions provides the valid range for this component as $(1, \frac{5-\sqrt{5}}{2})$. By taking the final intersection of this result with the condition derived from the inverse sine function, we find that the values must reside within the interval $(1, \frac{5-\sqrt{5}}{2})$.