Inverse trigonometric functions are used to work backward from a trig ratio to find an angle. The answers depend on the restricted ranges of inverse sine, inverse cosine, and inverse tangent. These problems include exact inverse trig values, degrees and radians, and compositions like \(\sin(\cos^{-1}(x))\) using right triangle relationships.
Practice Problems
\(\textbf{1)}\) \( \cos^{-1}(-1) \)The answer is \( \theta= 180^\circ \) or \( x=\pi \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}(-1)\text{ asks for the angle whose cosine is }-1.\)
\(\cos(\pi)=-1\)
\(\cos(180^\circ)=-1\)
\(\text{The answer is }\theta=180^\circ\text{ or }x=\pi.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}(-1)\text{ asks for the angle whose cosine is }-1.\)
\(\cos(\pi)=-1\)
\(\cos(180^\circ)=-1\)
\(\text{The answer is }\theta=180^\circ\text{ or }x=\pi.\)
\(\textbf{2)}\) \( \sin^{-1} \left(-\frac{1}{2}\right) \) The answer is \( \theta=-30^\circ \) or \( x=-\displaystyle\frac{\pi}{6} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac12\right)\text{ asks for the angle whose sine is }-\frac12.\)
\(\sin\left(-\frac{\pi}{6}\right)=-\frac12\)
\(\sin(-30^\circ)=-\frac12\)
\(\text{The answer is }\theta=-30^\circ\text{ or }x=-\frac{\pi}{6}.\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac12\right)\text{ asks for the angle whose sine is }-\frac12.\)
\(\sin\left(-\frac{\pi}{6}\right)=-\frac12\)
\(\sin(-30^\circ)=-\frac12\)
\(\text{The answer is }\theta=-30^\circ\text{ or }x=-\frac{\pi}{6}.\)
\(\textbf{3)}\) \( \tan^{-1} \left(-\sqrt{3}\right) \) The answer is \( \theta=-60^\circ \) or \( x=-\displaystyle\frac{\pi}{3} \)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(-\sqrt3)\text{ asks for the angle whose tangent is }-\sqrt3.\)
\(\tan\left(-\frac{\pi}{3}\right)=-\sqrt3\)
\(\tan(-60^\circ)=-\sqrt3\)
\(\text{The answer is }\theta=-60^\circ\text{ or }x=-\frac{\pi}{3}.\)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(-\sqrt3)\text{ asks for the angle whose tangent is }-\sqrt3.\)
\(\tan\left(-\frac{\pi}{3}\right)=-\sqrt3\)
\(\tan(-60^\circ)=-\sqrt3\)
\(\text{The answer is }\theta=-60^\circ\text{ or }x=-\frac{\pi}{3}.\)
\(\textbf{4)}\) \( \cos^{-1} \left(\frac{1}{2}\right) \)The answer is \( \theta=60^\circ \) or \( x=\displaystyle\frac{\pi}{3} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(\frac12\right)\text{ asks for the angle whose cosine is }\frac12.\)
\(\cos\left(\frac{\pi}{3}\right)=\frac12\)
\(\cos(60^\circ)=\frac12\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(\frac12\right)\text{ asks for the angle whose cosine is }\frac12.\)
\(\cos\left(\frac{\pi}{3}\right)=\frac12\)
\(\cos(60^\circ)=\frac12\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
\(\textbf{5)}\) \( \sin^{-1} \left(\frac{\sqrt{3}}{2}\right) \)The answer is \( \theta=60^\circ \) or \( x=\displaystyle\frac{\pi}{3} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(\frac{\sqrt3}{2}\right)\text{ asks for the angle whose sine is }\frac{\sqrt3}{2}.\)
\(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\)
\(\sin(60^\circ)=\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(\frac{\sqrt3}{2}\right)\text{ asks for the angle whose sine is }\frac{\sqrt3}{2}.\)
\(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\)
\(\sin(60^\circ)=\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
\(\textbf{6)}\) \( \tan^{-1} (1) \)The answer is \( \theta=45^\circ \) or \( x=\displaystyle\frac{\pi}{4} \)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(1)\text{ asks for the angle whose tangent is }1.\)
\(\tan\left(\frac{\pi}{4}\right)=1\)
\(\tan(45^\circ)=1\)
\(\text{The answer is }\theta=45^\circ\text{ or }x=\frac{\pi}{4}.\)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(1)\text{ asks for the angle whose tangent is }1.\)
\(\tan\left(\frac{\pi}{4}\right)=1\)
\(\tan(45^\circ)=1\)
\(\text{The answer is }\theta=45^\circ\text{ or }x=\frac{\pi}{4}.\)
\(\textbf{7)}\) \( \cos^{-1} (0) \)The answer is \( \theta=90^\circ \) or \( x=\displaystyle\frac{\pi}{2} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}(0)\text{ asks for the angle whose cosine is }0.\)
\(\cos\left(\frac{\pi}{2}\right)=0\)
\(\cos(90^\circ)=0\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}(0)\text{ asks for the angle whose cosine is }0.\)
\(\cos\left(\frac{\pi}{2}\right)=0\)
\(\cos(90^\circ)=0\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\textbf{8)}\) \( \sin^{-1} \left(-\frac{\sqrt{2}}{2}\right) \)The answer is \( \theta=-45^\circ \) or \( x=-\displaystyle\frac{\pi}{4} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac{\sqrt2}{2}\right)\text{ asks for the angle whose sine is }-\frac{\sqrt2}{2}.\)
\(\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt2}{2}\)
\(\sin(-45^\circ)=-\frac{\sqrt2}{2}\)
\(\text{The answer is }\theta=-45^\circ\text{ or }x=-\frac{\pi}{4}.\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac{\sqrt2}{2}\right)\text{ asks for the angle whose sine is }-\frac{\sqrt2}{2}.\)
\(\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt2}{2}\)
\(\sin(-45^\circ)=-\frac{\sqrt2}{2}\)
\(\text{The answer is }\theta=-45^\circ\text{ or }x=-\frac{\pi}{4}.\)
\(\textbf{9)}\) \( \tan^{-1} (0) \)The answer is \( \theta=0^\circ \) or \( x=0 \)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(0)\text{ asks for the angle whose tangent is }0.\)
\(\tan(0)=0\)
\(\tan(0^\circ)=0\)
\(\text{The answer is }\theta=0^\circ\text{ or }x=0.\)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(0)\text{ asks for the angle whose tangent is }0.\)
\(\tan(0)=0\)
\(\tan(0^\circ)=0\)
\(\text{The answer is }\theta=0^\circ\text{ or }x=0.\)
\(\textbf{10)}\) \( \cos^{-1} \left(-\frac{\sqrt{3}}{2}\right) \)The answer is \( \theta=150^\circ \) or \( x=\displaystyle\frac{5\pi}{6} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(-\frac{\sqrt3}{2}\right)\text{ asks for the angle whose cosine is }-\frac{\sqrt3}{2}.\)
\(\cos\left(\frac{5\pi}{6}\right)=-\frac{\sqrt3}{2}\)
\(\cos(150^\circ)=-\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=150^\circ\text{ or }x=\frac{5\pi}{6}.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(-\frac{\sqrt3}{2}\right)\text{ asks for the angle whose cosine is }-\frac{\sqrt3}{2}.\)
\(\cos\left(\frac{5\pi}{6}\right)=-\frac{\sqrt3}{2}\)
\(\cos(150^\circ)=-\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=150^\circ\text{ or }x=\frac{5\pi}{6}.\)
\(\textbf{11)}\) \( \sin^{-1} (1) \)The answer is \( \theta=90^\circ \) or \( x=\displaystyle\frac{\pi}{2} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}(1)\text{ asks for the angle whose sine is }1.\)
\(\sin\left(\frac{\pi}{2}\right)=1\)
\(\sin(90^\circ)=1\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}(1)\text{ asks for the angle whose sine is }1.\)
\(\sin\left(\frac{\pi}{2}\right)=1\)
\(\sin(90^\circ)=1\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\textbf{12)}\) \( \tan^{-1} \left(-\frac{\sqrt{3}}{3}\right) \)The answer is \( \theta=-30^\circ \) or \( x=-\displaystyle\frac{\pi}{6} \)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}\left(-\frac{\sqrt3}{3}\right)\text{ asks for the angle whose tangent is }-\frac{\sqrt3}{3}.\)
\(\tan\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{3}\)
\(\tan(-30^\circ)=-\frac{\sqrt3}{3}\)
\(\text{The answer is }\theta=-30^\circ\text{ or }x=-\frac{\pi}{6}.\)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}\left(-\frac{\sqrt3}{3}\right)\text{ asks for the angle whose tangent is }-\frac{\sqrt3}{3}.\)
\(\tan\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{3}\)
\(\tan(-30^\circ)=-\frac{\sqrt3}{3}\)
\(\text{The answer is }\theta=-30^\circ\text{ or }x=-\frac{\pi}{6}.\)
\(\textbf{13)}\) \( \cos^{-1} \left(\frac{\sqrt{2}}{2}\right) \)The answer is \( \theta=45^\circ \) or \( x=\displaystyle\frac{\pi}{4} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(\frac{\sqrt2}{2}\right)\text{ asks for the angle whose cosine is }\frac{\sqrt2}{2}.\)
\(\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)
\(\cos(45^\circ)=\frac{\sqrt2}{2}\)
\(\text{The answer is }\theta=45^\circ\text{ or }x=\frac{\pi}{4}.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\cos^{-1}\left(\frac{\sqrt2}{2}\right)\text{ asks for the angle whose cosine is }\frac{\sqrt2}{2}.\)
\(\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)
\(\cos(45^\circ)=\frac{\sqrt2}{2}\)
\(\text{The answer is }\theta=45^\circ\text{ or }x=\frac{\pi}{4}.\)
\(\textbf{14)}\) \( \sin^{-1} \left(-\frac{\sqrt{3}}{2}\right) \)The answer is \( \theta=-60^\circ \) or \( x=-\displaystyle\frac{\pi}{3} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac{\sqrt3}{2}\right)\text{ asks for the angle whose sine is }-\frac{\sqrt3}{2}.\)
\(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt3}{2}\)
\(\sin(-60^\circ)=-\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=-60^\circ\text{ or }x=-\frac{\pi}{3}.\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-90^\circ\leq \theta \leq 90^\circ\text{ or }-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.\)
\(\sin^{-1}\left(-\frac{\sqrt3}{2}\right)\text{ asks for the angle whose sine is }-\frac{\sqrt3}{2}.\)
\(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt3}{2}\)
\(\sin(-60^\circ)=-\frac{\sqrt3}{2}\)
\(\text{The answer is }\theta=-60^\circ\text{ or }x=-\frac{\pi}{3}.\)
\(\textbf{15)}\) \( \tan^{-1} (\sqrt{3}) \)The answer is \( \theta=60^\circ \) or \( x=\displaystyle\frac{\pi}{3} \)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(\sqrt3)\text{ asks for the angle whose tangent is }\sqrt3.\)
\(\tan\left(\frac{\pi}{3}\right)=\sqrt3\)
\(\tan(60^\circ)=\sqrt3\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
\(\text{For }\tan^{-1}(x),\text{ the range is }-90^\circ\lt \theta \lt 90^\circ\text{ or }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}.\)
\(\tan^{-1}(\sqrt3)\text{ asks for the angle whose tangent is }\sqrt3.\)
\(\tan\left(\frac{\pi}{3}\right)=\sqrt3\)
\(\tan(60^\circ)=\sqrt3\)
\(\text{The answer is }\theta=60^\circ\text{ or }x=\frac{\pi}{3}.\)
Challenge Problems
\(\textbf{16)}\) \( \cos^{-1} \left(\sin{\left(-\frac{\pi}{2}\right)}\right) \)The answer is \( \theta=180^\circ \) or \( x=\pi \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\sin\left(-\frac{\pi}{2}\right)=-1\)
\(\cos^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right)=\cos^{-1}(-1)\)
\(\cos(\pi)=-1\)
\(\text{The answer is }\theta=180^\circ\text{ or }x=\pi.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\sin\left(-\frac{\pi}{2}\right)=-1\)
\(\cos^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right)=\cos^{-1}(-1)\)
\(\cos(\pi)=-1\)
\(\text{The answer is }\theta=180^\circ\text{ or }x=\pi.\)
\(\textbf{17)}\) \( \cos^{-1} \left(\sin{\pi}\right) \) The answer is \( \theta=90^\circ \) or \( x=\displaystyle\frac{\pi}{2} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\sin(\pi)=0\)
\(\cos^{-1}\left(\sin{\pi}\right)=\cos^{-1}(0)\)
\(\cos\left(\frac{\pi}{2}\right)=0\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq 180^\circ\text{ or }0\leq x\leq \pi.\)
\(\sin(\pi)=0\)
\(\cos^{-1}\left(\sin{\pi}\right)=\cos^{-1}(0)\)
\(\cos\left(\frac{\pi}{2}\right)=0\)
\(\text{The answer is }\theta=90^\circ\text{ or }x=\frac{\pi}{2}.\)
\(\textbf{18)}\) \( \sin\left(\cos^{-1} \left(\frac{3}{5}\right)\right) \) The answer is \( \displaystyle\frac{4}{5} \)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq \pi.\)
\(\text{Let }\theta=\cos^{-1}\left(\frac35\right).\)
\(\cos\theta=\frac35\)
\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac35\)
\(\text{Use a right triangle with adjacent }3\text{ and hypotenuse }5.\)
\(\text{The opposite side is }4\text{ by the }3-4-5\text{ triangle.}\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac45\)
\(\sin\left(\cos^{-1}\left(\frac35\right)\right)=\frac45\)
\(\text{For }\cos^{-1}(x),\text{ the range is }0\leq \theta \leq \pi.\)
\(\text{Let }\theta=\cos^{-1}\left(\frac35\right).\)
\(\cos\theta=\frac35\)
\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac35\)
\(\text{Use a right triangle with adjacent }3\text{ and hypotenuse }5.\)
\(\text{The opposite side is }4\text{ by the }3-4-5\text{ triangle.}\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac45\)
\(\sin\left(\cos^{-1}\left(\frac35\right)\right)=\frac45\)
\(\textbf{19)}\) \(\tan\left(\sin^{-1}\left(\frac{3}{5}\right)\right)= \)
The answer is \( \displaystyle\frac{3}{4} \)
\(\text{For }\sin^{-1}(x),\text{ the range is }-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}.\)
\(\text{Let }\theta=\sin^{-1}\left(\frac35\right).\)
\(\sin\theta=\frac35\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac35\)
\(\text{Use a right triangle with opposite }3\text{ and hypotenuse }5.\)
\(\text{The adjacent side is }4\text{ by the }3-4-5\text{ triangle.}\)
\(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac34\)
\(\tan\left(\sin^{-1}\left(\frac35\right)\right)=\frac34\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}.\)
\(\text{Let }\theta=\sin^{-1}\left(\frac35\right).\)
\(\sin\theta=\frac35\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac35\)
\(\text{Use a right triangle with opposite }3\text{ and hypotenuse }5.\)
\(\text{The adjacent side is }4\text{ by the }3-4-5\text{ triangle.}\)
\(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac34\)
\(\tan\left(\sin^{-1}\left(\frac35\right)\right)=\frac34\)
\(\textbf{20)}\) \(\cos\left(\sin^{-1}\left(\frac{5}{13}\right)\right)\)The answer is \(\displaystyle\frac{12}{13}\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}.\)
\(\text{Let }\theta=\sin^{-1}\left(\frac{5}{13}\right).\)
\(\sin\theta=\frac{5}{13}\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{5}{13}\)
\(\text{Use a right triangle with opposite }5\text{ and hypotenuse }13.\)
\(\text{The adjacent side is }12\text{ by the }5-12-13\text{ triangle.}\)
\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{13}\)
\(\cos\left(\sin^{-1}\left(\frac{5}{13}\right)\right)=\frac{12}{13}\)
\(\text{For }\sin^{-1}(x),\text{ the range is }-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}.\)
\(\text{Let }\theta=\sin^{-1}\left(\frac{5}{13}\right).\)
\(\sin\theta=\frac{5}{13}\)
\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{5}{13}\)
\(\text{Use a right triangle with opposite }5\text{ and hypotenuse }13.\)
\(\text{The adjacent side is }12\text{ by the }5-12-13\text{ triangle.}\)
\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{13}\)
\(\cos\left(\sin^{-1}\left(\frac{5}{13}\right)\right)=\frac{12}{13}\)
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\(\bullet\text{ Proving Trigonometric Identities}\)
\(\,\,\,\,\,\,\,\,\sec{x}-\cos{x}=\displaystyle\frac{\tan^2{x}}{\sec{x}}…\)
\(\bullet\text{ Graphing Trig Functions- sin and cos}\)
\(\,\,\,\,\,\,\,\,f(x)=A \sin{B(x-c)}+D \) 
\(…\)
\(\bullet\text{ Solving Trigonometric Equations}\)
\(\,\,\,\,\,\,\,\,2\cos{(x)}=\sqrt{3}…\)