Inverse Trig Functions — Free Trigonometry Review Games.

sin(pi/6) = 1/2, so arcsin(1/2) = pi/6.

arcsin returns values between -pi/2 and pi/2 inclusive.

cos(pi/2) = 0, so arccos(0) = pi/2.

arccos returns values between 0 and pi inclusive.

tan(pi/4) = 1, so arctan(1) = pi/4.

arcsin is only defined for inputs between -1 and 1.

cos(pi) = -1, so arccos(-1) = pi.

tan(0) = 0, so arctan(0) = 0.

sin and arcsin are inverse functions, so sin(arcsin(0.5)) = 0.5.

arctan returns values in the open interval (-pi/2, pi/2).

If sin(theta)=3/5, then cos(theta)=4/5 (in the range of arcsin, cos is positive).

If cos(theta)=5/13, opp=12, adj=5. tan=12/5.

sin(5*pi/6)=1/2. arcsin(1/2)=pi/6 (must be in [-pi/2, pi/2]).

Let theta=arctan(3/4). sin(theta)=3/5, cos(theta)=4/5. sin(2*theta)=2*(3/5)*(4/5)=24/25.

For \(x > 0\), \(\arctan(x) + \arctan(1/x) = \pi/2\).

Since \(\sin(0) = 0\) and \(0\) lies in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) of arcsine, \(\arcsin(0) = 0\). The choice \(\frac{\pi}{2}\) is wrong because \(\sin(\frac{\pi}{2}) = 1\), not \(0\). Always check that the angle you produce both satisfies the trig equation and lies within the restricted range of the inverse function.

The angle \(-\frac{\pi}{6}\) satisfies \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\) and lies within the arcsine range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\). The choice \(\frac{7\pi}{6}\) is incorrect because it falls outside that restricted range even though sine there is also \(-\frac{1}{2}\). Remember that arcsine always returns the unique output in \([-\frac{\pi}{2}, \frac{\pi}{2}]\), not just any coterminal angle.

Arcsine reaches its maximum output at \(\frac{\pi}{2}\) because \(\sin\left(\frac{\pi}{2}\right) = 1\) and this is the top of its restricted range. The choice \(\pi\) is wrong since \(\sin(\pi) = 0\), not \(1\). The endpoints of the arcsine range, \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), correspond to the input values \(-1\) and \(1\).

Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\frac{\pi}{3}\) lies in the arccosine range of \([0, \pi]\), this is the correct value. The choice \(-\frac{\pi}{3}\) is wrong because negative angles never appear in the output of arccosine. Arccosine always returns values between \(0\) and \(\pi\) inclusive, unlike arcsine which allows negative outputs.

Arccosine of \(1\) equals \(0\) because \(\cos(0) = 1\), the maximum value of cosine occurring at the start of its restricted domain. The choice \(\pi\) is incorrect because \(\cos(\pi) = -1\), not \(1\). The endpoints of the arccosine range, \(0\) and \(\pi\), correspond to inputs \(1\) and \(-1\) respectively.

Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\) and \(\frac{2\pi}{3}\) lies within \([0, \pi]\), this is the correct arccosine value. The choice \(-\frac{2\pi}{3}\) is wrong because arccosine never outputs negative angles by definition of its restricted range. Negative cosine inputs always produce angles in the second quadrant, between \(\frac{\pi}{2}\) and \(\pi\), when using arccosine.

The value \(-\frac{\pi}{4}\) works because \(\tan\left(-\frac{\pi}{4}\right) = -1\) and it lies within the arctangent range of \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). The choice \(\frac{3\pi}{4}\) is wrong since it falls outside that open interval even though its tangent also equals \(-1\). Arctangent always selects the unique angle in the open interval between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

Since \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\) and \(\frac{\pi}{6}\) lies in the arctangent range, this is the correct value. The choice \(\frac{\pi}{3}\) is wrong because \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\), the reciprocal of the given value. Memorizing the tangent values at \(30°\), \(45°\), and \(60°\) makes evaluating arctangent of common ratios much faster.

Arccosine only accepts inputs that a cosine function could actually output, which restricts the domain to \([-1, 1]\). The choice \([0, \pi]\) is incorrect because that interval describes the range of arccosine, not its domain. Every inverse trig function built from sine or cosine has domain \([-1, 1]\) since those parent functions never exceed those bounds.

Since tangent can output any real number, its inverse arctangent accepts every real number as input, making the domain all real numbers. The choice \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) is wrong because that interval is actually the range of arctangent, not its domain. Because tangent is unbounded, arctangent is the only common inverse trig function with an unrestricted domain.

Arcsine of \(-1\) equals \(-\frac{\pi}{2}\) since \(\sin\left(-\frac{\pi}{2}\right) = -1\), the minimum value at the bottom of its restricted range. The choice \(-\pi\) is wrong because \(\sin(-\pi) = 0\), not \(-1\). Recognizing that arcsine outputs lie strictly between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) inclusive helps eliminate many incorrect answer choices quickly.

Since \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\) and \(\frac{\pi}{4}\) lies within \([0, \pi]\), this is the correct arccosine value. The choice \(-\frac{\pi}{4}\) is wrong because arccosine never produces negative output angles by definition. Knowing the unit circle values for \(45°\), \(30°\), and \(60°\) speeds up evaluating common inverse cosine inputs.

Both arcsine and arctangent are odd functions because their graphs are symmetric about the origin, so \(\arcsin(-x) = -\arcsin(x)\) and \(\arctan(-x) = -\arctan(x)\). Arccosine fails this property since \(\arccos(-x) = \pi - \arccos(x)\), not simply \(-\arccos(x)\). Recognizing symmetry properties of inverse trig functions helps simplify expressions involving negative inputs on the exam.

The angle \(-\frac{\pi}{3}\) satisfies \(\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\) and lies within the arcsine range of \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). The choice \(-\frac{2\pi}{3}\) is wrong because it lies outside that restricted range even though its sine is also negative. Always search only within the restricted range when identifying the correct reference angle for arcsine.

Since \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) and \(\frac{5\pi}{6}\) lies in \([0, \pi]\), this is the correct value for arccosine. The choice \(\frac{\pi}{6}\) is wrong because \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), a positive value, not the negative one given. Negative cosine values always correspond to arccosine outputs in the second quadrant, between \(\frac{\pi}{2}\) and \(\pi\).

Since \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\) and \(\frac{\pi}{3}\) lies within the arctangent range, this is the correct output. The choice \(\frac{2\pi}{3}\) is wrong because it falls outside the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) even though it may share related tangent behavior. Knowing exact tangent values for special angles makes arctangent evaluation of common radicals straightforward.

The angle \(-\frac{\pi}{3}\) satisfies \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\) and lies within the arctangent's restricted range. The choice \(-\frac{2\pi}{3}\) is wrong because it is outside the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Because arctangent is odd, negating the input simply negates the output angle for standard reference values.

Letting \(\theta = \arccos(0.6)\) means \(\cos\theta = 0.6\), and using the Pythagorean identity \(\sin\theta = \sqrt{1 - 0.36} = 0.8\) since \(\theta\) is in the first quadrant. The choice \(0.6\) is wrong because that value is the cosine itself, not the sine, of the constructed angle. Building a right triangle from the given ratio is the standard technique for evaluating compositions of inverse and direct trig functions.

Since \(\arctan(1) = \frac{\pi}{4}\), evaluating cosine there gives \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). The choice \(1\) is wrong because it confuses the tangent value used to define the angle with the cosine of that angle. Always first identify the exact angle produced by the inner inverse function before applying the outer trig function.

Because \(\frac{\pi}{3}\) already lies within the restricted range of arcsine, \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), the composition simply returns \(\frac{\pi}{3}\). The choice \(\frac{2\pi}{3}\) is wrong because that angle falls outside arcsine's range even though its sine value matches. When the original angle lies within the inverse function's range, the composition acts as the identity and returns that same angle.

Since \(\frac{\pi}{4}\) already lies within arccosine's range of \([0, \pi]\), the composition returns the same angle, \(\frac{\pi}{4}\). The choice \(-\frac{\pi}{4}\) is wrong because arccosine never outputs a negative angle by definition of its restricted range. Whenever the original input angle already lies in the inverse function's range, the round-trip composition simplifies directly to that angle.

Because \(\frac{\pi}{6}\) lies within the arctangent range of \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the composition returns \(\frac{\pi}{6}\) directly. The choice \(\frac{5\pi}{6}\) is wrong because it lies outside that open interval even though it might share a related tangent value elsewhere. Checking whether the original angle already lies in the correct restricted range is the key first step in these composition problems.

Since \(\sin(\pi) = 0\), the expression becomes \(\arcsin(0)\), which equals \(0\) because that is the angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) with sine zero. The choice \(\pi\) is wrong because although \(\pi\) produces sine equal to \(0\), it lies outside the arcsine range, so it can never be the output. Always evaluate the inner trig function to a numeric value first, then apply the inverse function's range restriction.

Since \(\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\), evaluating cosine there gives \(\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), which is positive because cosine is an even function. The choice \(-\frac{\sqrt{3}}{2}\) is wrong because it incorrectly assumes cosine inherits the negative sign from the original sine input. Because arcsine outputs lie in quadrants I and IV where cosine is always nonnegative, this composition always yields a positive result.

Since \(\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\), evaluating tangent gives \(\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}\). The choice \(\sqrt{3}\) is wrong because it is the reciprocal of the correct tangent value, corresponding instead to \(\tan\left(\frac{\pi}{3}\right)\). Identifying the exact angle first, then applying the outer function, prevents mixing up reciprocal trig values.

Because \(\frac{2\pi}{3}\) already lies within arccosine's range of \([0, \pi]\), the composition returns the same angle, \(\frac{2\pi}{3}\). The choice \(\frac{\pi}{3}\) is wrong because it corresponds to a different cosine value entirely, not the reference angle process used here. Arccosine's range spans the entire interval \([0, \pi]\), making this identity check simpler than arcsine or arctangent compositions.

Arcsine is increasing on its domain \([-1, 1]\) because sine itself is increasing on the restricted interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), and inverting a monotonic function preserves its direction of monotonicity. The choice \(\text{Decreasing}\) is wrong because that behavior actually describes the arccosine function instead. Recognizing which inverse trig functions are increasing versus decreasing helps predict the sign of derivatives and the shape of their graphs.

Since \(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}\), the expression becomes \(\arcsin\left(\frac{\sqrt{2}}{2}\right)\), which equals \(\frac{\pi}{4}\) because that is the angle in the restricted range with the same sine value. The choice \(\frac{3\pi}{4}\) is wrong because although it produces the correct sine value, it lies outside arcsine's range, so the inverse function cannot return it. When the input angle is outside the arcsine range, find the reference angle within \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) that shares the same sine value.

Since \(\arctan(1) = \frac{\pi}{4}\), evaluating secant gives \(\sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos(\pi/4)} = \sqrt{2}\). The choice \(2\) is wrong because it confuses secant with the value obtained from a \(30\)-\(60\)-\(90\) triangle rather than the \(45\)-\(45\)-\(90\) triangle relevant here. Constructing the correct reference triangle based on the exact angle avoids mixing up trig ratios from different special triangles.

Since arcsine only accepts inputs in \([-1, 1]\), the value \(1.5\) falls outside this domain and is therefore invalid. The choice \(-1\) is wrong as a candidate because it is exactly the left endpoint of the valid domain and produces a defined output of \(-\frac{\pi}{2}\). Always verify that a given input lies between \(-1\) and \(1\) inclusive before attempting to evaluate arcsine or arccosine.

Since \(\tan\left(\frac{3\pi}{4}\right) = -1\), the expression becomes \(\arctan(-1)\), which equals \(-\frac{\pi}{4}\) because that angle lies within the restricted range of arctangent. The choice \(\frac{3\pi}{4}\) is wrong because although it produces tangent equal to \(-1\), it lies outside the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Because tangent has period \(\pi\), arctangent compositions often require finding the equivalent angle inside its narrower restricted range.

Since \(\arcsin\left(\frac{1}{3}\right)\) is the angle whose sine equals \(\frac{1}{3}\), cosecant, being the reciprocal of sine, equals \(3\) directly. The choice \(\frac{1}{3}\) is wrong because it repeats the sine value instead of taking its reciprocal to find cosecant. When the outer function is a reciprocal trig function, applying the reciprocal directly to the given ratio often skips the need to build a full triangle.

Letting \(\theta = \arccos\left(\frac{2}{3}\right)\) gives an adjacent side of \(2\) and hypotenuse \(3\) in a right triangle, so the opposite side is \(\sqrt{9-4} = \sqrt{5}\), making \(\sin\theta = \frac{\sqrt{5}}{3}\). The choice \(\frac{2}{3}\) is wrong because that value is the cosine itself, not the sine, of the constructed angle. Building a reference right triangle from the ratio inside an inverse trig function is the reliable method for evaluating these compositions exactly.

Using the double angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\) with \(\sin\theta = \frac{1}{3}\) gives \(1 - 2\left(\frac{1}{9}\right) = \frac{7}{9}\). The choice \(\frac{8}{9}\) is wrong because it results from an arithmetic slip in computing \(2\sin^2\theta\) rather than correctly subtracting from \(1\). Double angle formulas expressed purely in terms of sine or cosine avoid needing to find the angle explicitly when composing with double angle expressions.

Constructing a right triangle with opposite side \(5\) and hypotenuse \(13\) gives an adjacent side of \(\sqrt{169-25} = 12\), so \(\tan\theta = \frac{5}{12}\). The choice \(\frac{12}{5}\) is wrong because it is the reciprocal of the correct tangent ratio, mistakenly swapping opposite and adjacent sides. Using a Pythagorean triple like the \(5\)-\(12\)-\(13\) triangle simplifies these compositions without needing decimal approximations.

Since \(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\), the expression becomes \(\arcsin\left(-\frac{1}{2}\right)\), which equals \(-\frac{\pi}{6}\) because that angle lies in arcsine's restricted range. The choice \(\frac{7\pi}{6}\) is wrong because it lies far outside the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so arcsine can never output it. For angles outside the standard range, always reduce to the numeric sine value first, then find the equivalent angle inside the restricted interval.

With \(\arctan\left(\frac{3}{4}\right)\) giving \(\sin = \frac{3}{5}\), \(\cos = \frac{4}{5}\), and \(\arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}\) giving \(\sin = \frac{\sqrt{3}}{2}\), \(\cos = \frac{1}{2}\), the sum formula \(\sin(a+b) = \sin a\cos b + \cos a\sin b\) yields \(\frac{3}{10} + \frac{4\sqrt{3}}{10} = \frac{3+4\sqrt{3}}{10}\). The choice \(\frac{7}{10}\) is wrong because it incorrectly adds the sine values directly instead of applying the angle sum identity. When two different inverse trig expressions are added inside a trig function, always convert each to a triangle ratio first, then apply the appropriate sum or difference identity.

Because arcsine and arccosine are complementary angle functions built from co-function identities, their sum is always the constant \(\frac{\pi}{2}\) for any valid input \(x\). The choice \(\pi\) is wrong because that would only occur if the two angles were supplementary rather than complementary. This identity is a useful shortcut for quickly rewriting one inverse trig expression in terms of the other on the exam.

Constructing a right triangle with opposite side \(x\) and adjacent side \(1\) gives hypotenuse \(\sqrt{1+x^2}\), so cosine of the angle equals adjacent over hypotenuse, \(\frac{1}{\sqrt{1+x^2}}\). The choice \(\frac{x}{\sqrt{1+x^2}}\) is wrong because that expression instead represents the sine of the same angle, not cosine. Building a generic reference triangle with a variable side length lets you derive formulas for compositions of inverse trig functions with any real input.

With \(\cos\theta = \frac{3}{5}\) and \(\sin\theta = \frac{4}{5}\), the double angle formula \(\sin(2\theta) = 2\sin\theta\cos\theta\) gives \(2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}\). The choice \(\frac{7}{25}\) is wrong because that value actually equals \(\cos(2\theta)\) using the identity \(1-2\sin^2\theta\), not the sine double angle formula requested here. Keeping the sine and cosine double angle formulas distinct is essential when a problem specifically asks for one or the other.

Setting the two inverse functions equal and using the identity \(\arcsin(x) + \arccos(x) = \frac{\pi}{2}\) forces each to equal \(\frac{\pi}{4}\), and since \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), that must be the value of \(x\). The choice \(\frac{1}{2}\) is wrong because \(\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\) while \(\arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}\), which are not equal. Combining the complementary angle identity with the equal-angle condition is a clean way to solve equations relating arcsine and arccosine directly.

Since \(\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\), which lies in the second quadrant, computing tangent there gives \(\tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}\) because tangent is negative in that quadrant. The choice \(\sqrt{3}\) is wrong because it ignores the sign change that occurs when the angle from arccosine falls in the second quadrant rather than the first. Since arccosine outputs range over both quadrants I and II, always check the sign of tangent based on which quadrant the resulting angle lies in.

Both \(\arcsin\left(\frac{3}{5}\right)\) and \(\arccos\left(\frac{4}{5}\right)\) describe the same first-quadrant angle with sine \(\frac{3}{5}\) and cosine \(\frac{4}{5}\), so their difference is \(0\), and \(\cos(0) = 1\). The choice \(\frac{24}{25}\) is wrong because it results from mistakenly assuming the two angles are different and applying the cosine difference formula unnecessarily. Recognizing when two seemingly different inverse trig expressions actually represent the same angle can dramatically simplify a composition problem.

Since arcsine requires its input to satisfy \(-1 \le 2x-1 \le 1\), solving this compound inequality by adding \(1\) and dividing by \(2\) gives \(0 \le x \le 1\). The choice \([-1, 1]\) is wrong because that describes the domain of arcsine itself, not the transformed domain after solving for \(x\). When a linear expression is substituted into arcsine or arccosine, solve the compound inequality \(-1 \le (\text{expression}) \le 1\) to find the valid domain of the composite function.

With \(\arctan(2)\) giving \(\sin = \frac{2}{\sqrt{5}}\), \(\cos = \frac{1}{\sqrt{5}}\) and \(\arctan(3)\) giving \(\sin = \frac{3}{\sqrt{10}}\), \(\cos = \frac{1}{\sqrt{10}}\), the sum formula gives \(\frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{\sqrt{2}}{2}\). The second choice, though algebraically equal after simplification, is written in an unsimplified misleading form and should not be selected as the intended clean answer of \(\frac{\sqrt{2}}{2}\). Combining two arctangent values with the sine sum identity often produces surprisingly clean results, so simplifying radicals fully is essential before matching an answer choice.