Free math worksheets

- The domain of an inverse function comes from the range of the restricted original function. - The range of an inverse function is the interval chosen to make the original trigonometric function one-to-one. - Compare the standard restricted branches of sine, cosine, and tangent rather than their full periodic graphs.

1. Sine is restricted to \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) to define its principal inverse. On that interval its outputs cover \([-1,1]\). Therefore, \(\arcsin\) has domain \([-1,1]\) and range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). 2. Cosine is restricted to \([0,\pi]\) for its principal inverse, and its outputs there cover \([-1,1]\). Therefore, \(\arccos\) has domain \([-1,1]\) and range \([0,\pi]\). 3. Tangent is restricted to \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), where its outputs cover all real numbers. Therefore, \(\arctan\) has domain \(\mathbb{R}\) and range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

a) Domain: \([-1,1]\); range: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) b) Domain: \([-1,1]\); range: \([0,\pi]\) c) Domain: \(\mathbb{R}\); range: \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

- Recall the principal range of the arcsine function. - Look for a standard angle whose sine is the given value. - Check that the angle you choose lies in the principal arcsine range.

1. The principal range of arcsine is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). 2. The angle in that range whose sine is \(\frac{1}{2}\) is \(\frac{\pi}{6}\).

\(\frac{\pi}{6}\)

- Focus on the allowable inputs of the real-valued inverse cosine function. - Compare the given input directly with the endpoints of that domain.

1. The real-valued function \(\arccos(x)\) has domain \([-1,1]\). 2. Since \(1.2\notin[-1,1]\), \(\arccos(1.2)\) is not defined as a real number.

No. The real-valued domain of \(\arccos(x)\) is \([-1,1]\), and \(1.2\) is outside that interval.

- Apply the corresponding trigonometric function to undo each inverse function. - Check the principal-value range of each inverse trigonometric function. - Isolate the inverse tangent before applying tangent. - Use exact unit-circle values.

1. a) Since \(\frac{3\pi}{4}\in[0,\pi]\), apply cosine: \(x=\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}\). 2. b) First divide by \(2\): \(\arctan(x)=-\frac{\pi}{4}\). This angle lies in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), so \(x=\tan\left(-\frac{\pi}{4}\right)=-1\). 3. c) Since \(\frac{\pi}{2}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), apply sine: \(2x-1=\sin\left(\frac{\pi}{2}\right)=1\). Thus, \(x=1\), and the inverse-sine input is \(2(1)-1=1\), which is valid.

a) \(x=-\frac{\sqrt{2}}{2}\) b) \(x=-1\) c) \(x=1\)

- Distinguish between applying sine after arcsine and applying arcsine after sine. - Check the principal output range of arcsine before assuming two functions cancel. - For part b), first identify the sine value of the given standard angle. - Ask which angle in the principal arcsine range has that same sine.

1. The expression \(\arcsin\left(-\frac{3}{5}\right)\) is an angle whose sine is \(-\frac{3}{5}\), so applying sine gives \(-\frac{3}{5}\). 2. Since \(\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}\), the second expression becomes \(\arcsin\left(\frac{1}{2}\right)=\frac{\pi}{6}\). 3. Arcsine returns values only in its principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The angle \(\frac{5\pi}{6}\) lies outside that range, while \(\frac{\pi}{6}\) lies inside it and has the same sine.

a) \(-\frac{3}{5}\) b) \(\frac{\pi}{6}\) c) Arcsine must return an angle in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so it returns the principal angle with the same sine rather than \(\frac{5\pi}{6}\).

- Ask what question \(\arcsin\) answers about a sine value. - Recall the reciprocal relationship that defines cosecant. - Compare the type of input and the type of output in the two expressions.

1. \(\arcsin\left(\frac{1}{2}\right)\) asks for the principal angle whose sine is \(\frac{1}{2}\). Therefore, \(\arcsin\left(\frac{1}{2}\right)=\frac{\pi}{6}\). 2. Cosecant is the reciprocal of sine. Since \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\), \(\csc\left(\frac{\pi}{6}\right)=2\). 3. Arcsine is an inverse function: it takes a sine value in \([-1,1]\) and returns a principal angle. Cosecant is a reciprocal trigonometric function: it takes an angle and returns the reciprocal of that angle's sine value, when the sine is nonzero.

a) \(\frac{\pi}{6}\) b) \(2\) c) Arcsine maps a sine value to a principal angle; cosecant maps an angle to the reciprocal of its sine value. They are not the same operation.

- Which trigonometric function is represented by the y-coordinate on the unit circle? - How can an inverse trigonometric function recover an angle from a sine value? - What equation relates the coordinates of every point on the unit circle?

1. On the unit circle, the y-coordinate equals \(\sin(\alpha)\). Thus, \(\sin(\alpha)=0.45\), so \(\alpha=\arcsin(0.45)\approx26.74^\circ\). 2. Because \(P\) is in Quadrant I, its x-coordinate is positive. Using \(x^2+y^2=1\), \(x=\sqrt{1-0.45^2}=\sqrt{0.7975}\approx0.89\).

1. \(\alpha\approx26.74^\circ\) 2. \(x\approx0.89\)

- Start with the allowable inputs of the real-valued arccosine function. - For the range, determine what values the inside expression actually covers over the domain. - A y-intercept has \(x=0\). - For the final part, connect the inverse-cosine value to the corresponding exact cosine value.

1. The input of \(\arccos\) must lie in \([-1,1]\). Thus, \(-1\le x+1\le1\), which gives \(-2\le x\le0\). The domain is \([-2,0]\). 2. As \(x\) runs through \([-2,0]\), the input \(x+1\) runs through all of \([-1,1]\). Therefore, \(f\) takes the full principal arccosine range \([0,\pi]\). 3. At \(x=0\), \(f(0)=\arccos(1)=0\). Therefore, the y-intercept is \((0,0)\). 4. Solve \(\arccos(x+1)=\frac{\pi}{3}\). Then \(x+1=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), so \(x=-\frac{1}{2}\).

1) \([-2,0]\) 2) \([0,\pi]\) 3) \((0,0)\) 4) \(x=-\frac{1}{2}\)

- Recall the shape of one full cosine cycle. - Find the intervals where the graph moves only downward or only upward. - The standard inverse cosine uses the decreasing half-cycle from \(0\) to \(\pi\).

1. The graph fails the horizontal line test. For example, \(\cos\left(\frac{\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0\). 2. Therefore, cosine is not one-to-one on \([0,2\pi]\) and is not invertible on that entire interval. 3. Cosine is strictly decreasing on \([0,\pi]\) and strictly increasing on \([\pi,2\pi]\). 4. These are the two largest subintervals on which the restrictions are invertible. 5. The inverse of cosine restricted to \([0,\pi]\) is the inverse cosine function, written \(\arccos(x)\).

a) The function is not invertible because it fails the horizontal line test; for example, \(f\left(\frac{\pi}{2}\right)=f\left(\frac{3\pi}{2}\right)=0\). b) \([0,\pi]\) and \([\pi,2\pi]\) c) The inverse cosine function, \(\arccos(x)\)

- For an inverse function, exchange the roles of the original graph's inputs and outputs. - Notice whether each restricted graph actually contains its boundary x-values or only approaches them. - Use brackets for attained boundary values and parentheses for boundaries that are never attained.

1. On the restricted sine branch, the input interval is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), and the graph reaches every output from \(-1\) through \(1\). Reversing inputs and outputs gives \(\arcsin\) domain \([-1,1]\) and range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). 2. On the restricted tangent branch, the input interval is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), and its outputs cover all real numbers. Therefore, \(\arctan\) has domain \(\mathbb{R}\) and range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). 3. The restricted sine graph actually contains the endpoint points \(\left(-\frac{\pi}{2},-1\right)\) and \(\left(\frac{\pi}{2},1\right)\). The tangent graph instead approaches vertical asymptotes at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\) without attaining those x-values. After inversion, that distinction becomes included versus excluded range endpoints.

a) Domain: \([-1,1]\); range: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) b) Domain: \(\mathbb{R}\); range: \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) c) Sine attains the endpoints of its restricted input interval, while tangent only approaches its bounding asymptotes and is never defined there.

- Start with the domain and principal range of the parent arctangent function. - Decide whether the horizontal shift changes which real inputs are allowed. - Apply the outside vertical stretch and vertical shift to the endpoints of the parent range. - Preserve whether the parent range endpoints are included or excluded.

1. Arctangent accepts every real input, and replacing \(x\) by \(x-1\) does not restrict the input. Therefore, the domain is all real numbers. 2. The range of \(\arctan(x)\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). 3. Multiplying the output by \(2\) gives \((-\pi,\pi)\). Adding \(\frac{\pi}{4}\) shifts that interval to \(\left(-\frac{3\pi}{4},\frac{5\pi}{4}\right)\).

Domain: \((-\infty, \infty)\) Range: \(\left(-\frac{3\pi}{4},\frac{5\pi}{4}\right)\)

- Do not assume an inverse trigonometric function always returns the angle originally placed inside the direct function. - Write the principal output range for arccosine, arctangent, and arcsine before evaluating. - Find the direct trigonometric value first, then choose an equivalent angle inside the required principal range. - Check each returned angle against both its trigonometric value and its inverse-function range.

1. Since \(\cos\left(\frac{5\pi}{3}\right)=\frac{1}{2}\), arccosine must return the angle in \([0,\pi]\) with cosine \(\frac{1}{2}\), which is \(\frac{\pi}{3}\). 2. Since \(\tan\left(\frac{3\pi}{4}\right)=-1\), arctangent must return the angle in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) with tangent \(-1\), which is \(-\frac{\pi}{4}\). 3. Since \(\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2}\), arcsine must return the angle in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) with sine \(-\frac{1}{2}\), which is \(-\frac{\pi}{6}\). 4. In each case, the inverse function returns its principal representative rather than automatically returning the original input angle.

a) \(\frac{\pi}{3}\); principal arccosine range: \([0,\pi]\) b) \(-\frac{\pi}{4}\); principal arctangent range: \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) c) \(-\frac{\pi}{6}\); principal arcsine range: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

- Compare the horizontal and vertical extents of the two curves; one is restricted in x to the sine interval, while the other is restricted in x to inverse-sine inputs. - Look at corresponding parts of the two curves on opposite sides of the dashed line. - After identifying the inverse curve, read its allowable x-values and y-values from the displayed extents. - Use the geometric relationship from part b) to determine what happens to the coordinates of \(P\).

1. Curve u extends horizontally to \(x=\pm\frac{\pi}{2}\) but has y-values only from \(-1\) to \(1\), so u is the restricted sine graph. Curve v has x-values only from \(-1\) to \(1\) but extends vertically to \(y=\pm\frac{\pi}{2}\), so v is \(y=\arcsin(x)\). 2. The two curves are mirror images across \(y=x\). This is the geometric relationship between a one-to-one function and its inverse. 3. From curve v, the domain of \(\arcsin\) is \([-1,1]\), and its range is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). 4. Reflection across \(y=x\) exchanges the coordinates of \(P\), so the corresponding point on v is \(\left(\frac{1}{2},\frac{\pi}{6}\right)\).

a) u: restricted \(y=\sin(x)\); v: \(y=\arcsin(x)\) b) The curves are reflections across \(y=x\). c) Domain: \([-1,1]\); range: \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) d) \(\left(\frac{1}{2},\frac{\pi}{6}\right)\); the x- and y-coordinates are exchanged.