Free Math Worksheets

- 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) \(x=1\)

- 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 \(\alpha=\sin^{-1}(0.45) \approx 26.74^\circ\). 2. Because \(P\) is in Quadrant I, its x-coordinate is positive. Using \(x^2+y^2=1\), \(x=\sqrt{1-0.45^2}=\sqrt{0.7975} \approx 0.89\).

1. \(\alpha \approx 26.74^\circ\) 2. \(x \approx 0.89\)

- The input of \(\arccos\) must be between \(-1\) and \(1\). - A y-intercept has \(x=0\). - Apply cosine to undo \(\arccos\). - Use exact unit-circle values.

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. At \(x=0\), \(f(0)=\arccos(1)=0\). Therefore, the y-intercept is \((0,0)\). 3. Solve \(\arccos(x+1)=0\). Applying cosine gives \(x+1=\cos(0)=1\), so \(x=0\). 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) \((0,0)\) 3) \(x=0\) 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. a) The graph fails the horizontal line test. For example, \(\cos(\frac{\pi}{2})=\cos(\frac{3\pi}{2})=0\). 2. Therefore, cosine is not one-to-one on \([0,2\pi]\) and is not invertible on that entire interval. 3. b) 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. c) The inverse of cosine restricted to \([0,\pi]\) is the inverse cosine function, written \(\arccos(x)\) or \(\cos^{-1}(x)\).

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