$$ (\sin x - 1)(\sin x + 1) = 0 $$ \begin{align} \sin x - 1 & = 0 & \text{ or }\phantom{00000} \sin x + 1 & = 0 \\ \sin x & = 1 & \sin x & = - 1 \end{align}

\begin{align} x = 90^\circ \phantom{00} & \text{or} \phantom{00} x = 270^\circ \end{align}

$$ (\cos x - 1)(\cos x + 1) = 0 $$ \begin{align} \cos x - 1 & = 0 & \text{ or }\phantom{00000} \cos x + 1 & = 0 \\ \cos x & = 1 & \cos x & = - 1 \end{align}

\begin{align} x = 0^\circ, 360^\circ \phantom{00} & \text{or} \phantom{00} x = 180^\circ \\ \\ \therefore x = 0^\circ, &180^\circ, 360^\circ \end{align}

\begin{align} \sin x (2 \cos x - 3) & = 0 \\ \\ \sin x = 0 \phantom{00} & \text{or} \phantom{000} 2 \cos x - 3 = 0 \\ & \phantom{or000-3} 2\cos x = 3 \\ & \phantom{or000-32} \cos x = {3 \over 2} \phantom{0} (\text{Reject, as } \cos x \le 1) \\ \\ \\ \sin x & = 0 \end{align}

$$x = 0^\circ, 180^\circ, 360^\circ$$

$$ \tan x (2 \cos x - 1) = 0 $$ \begin{align} \tan x & = 0 & \text{ or }\phantom{00000} 2 \cos x - 1 & = 0 \\ & & 2\cos x & = 1 \\ & & \cos x & = {1 \over 2} \end{align}
\begin{align} \tan x & = 0 \end{align}

\begin{align} x = 0^\circ, & 180^\circ, 360^\circ \\ \\ \\ \cos x & = {1 \over 2} \phantom{000000} [\text{1st or 4th quadrant since } \cos x > 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \end{align}

\begin{align} x & = 60^\circ, 360^\circ - 60^\circ \\ & = 60^\circ, 300^\circ \\ \\ \\ \therefore x & = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ \end{align}

$$ \sin^2 x (\tan x + 4) = 0 $$ \begin{align} \sin^2 x & = 0 & \text{ or }\phantom{00000} \tan x + 4 & = 0 \\ \sin x & = \pm\sqrt{0} & \tan x & = -4 \\ \sin x & = 0 \end{align}
$$ \sin x = 0$$

\begin{align} x & = 0^\circ, 180^\circ, 360^\circ \\ \\ \\ \tan x & = -4 \phantom{000000} [\text{2nd or 4th quadrant since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(4\right) \\ & = 75.96^\circ \end{align}

\begin{align} x & = 180^\circ - 75.96^\circ, 360^\circ - 75.96^\circ \\ & = 104.04^\circ, 284.04^\circ \\ & \approx 104.0^\circ, 284.0^\circ \\ \\ \\ \therefore x & = 0^\circ, 104.0^\circ, 180^\circ, 284.0^\circ, 360^\circ \end{align}

\begin{align} (3 \sin x - 1) (\tan x + 1) & = 0 \\ \\ 3\sin x - 1 = 0 \phantom{-00} & \text{or} \phantom{000} \tan x + 1 = 0 \\ 3 \sin x = 1 \phantom{000.} & \phantom{or000+1} \tan x = - 1 \\ \sin x = {1 \over 3} \phantom{0{1 \over 3}} & \\ \\ \\ \sin x & = {1 \over 3} \phantom{000000} [\text{1st or 2nd quadrant since } \sin x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 3\right) \\ & = 19.47^\circ \end{align}

\begin{align} x & = 19.47^\circ, 180^\circ - 19.47^\circ \\ & = 19.47^\circ, 160.53^\circ \\ & \approx 19.5^\circ, 160.5^\circ \\ \\ \\ \tan x & = -1 \phantom{0000000} [\text{2nd or 4th quadrant since } \tan x < 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1) \\ & = 45^\circ \end{align}

\begin{align} x & = 180^\circ - 45^\circ, 360^\circ - 45^\circ \\ & = 135^\circ, 315^\circ \\ \\ \\ \therefore x & = 19.5^\circ, 135^\circ, 160.5^\circ, 315^\circ \end{align}

\begin{align} y & = 3\cos (2x) + 2 \\ \\ \text{Amplitude} & = 3 \\ \\ \text{Centre line: } & y = 2 \\ \\ \text{Max. value} & = 2 + 3 = 5 \\ \text{Min. value} & = 2 - 3 = -1 \\ \\ \text{Period} & = {360^\circ \over 2} \\ & = 180^\circ \\ \\ \text{No. of cycles} & = {360^\circ \over 180^\circ} \\ & = 2 \end{align}

\begin{align} y & = 5 - \sin 2x \\ & = -\sin 2x + 5 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Centre line: } & y = 5 \\ \\ \text{Max. value} & = 5 + 1 = 6 \\ \text{Min. value} & = 5 - 1 = 4 \\ \\ \text{Period} & = {2\pi \over 2} \\ & = \pi \\ \\ \text{No. of cycles} & = {2\pi \over \pi} \\ & = 2 \end{align}

\begin{align} y & = 4\sin 8x \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Centre line: } & y = 0 \\ \\ \text{Max. value} & = 4 \\ \text{Min. value} & = -4 \\ \\ \text{Period} & = {360^\circ \over 8} \\ & = 45^\circ \\ \\ \text{No. of cycles} & = {90^\circ \over 45^\circ} \\ & = 2 \end{align}

\begin{align} y & = 6 \cos \left(x \over 2\right) - 4 \\ & = 6 \cos \left({1 \over 2}x \right) - 4 \\ \\ \text{Amplitude} & = 6 \\ \\ \text{Centre line: } & y = -4 \\ \\ \text{Max. value} & = -4 + 6 = 2 \\ \text{Min. value} & = -4 - 6 = -10 \\ \\ \text{Period} & = {2\pi \over {1 \over 2}} \\ & = 4\pi \\ \\ \text{No. of cycles} & = {4\pi \over 4\pi} \\ & = 1 \end{align}

\begin{align} y & = 2 \sin (3x) - 1 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Centre line: } & y = -1 \\ \\ \text{Max. value} & = -1 + 2 = 1 \\ \text{Min. value} & = -1 - 2 = -3 \\ \\ \text{Period} & = {360^\circ \over 3} \\ & = 120^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 120^\circ} \\ & = 1{1 \over 2} \end{align}

\begin{align} y & = 2(2 - \cos 5x) \\ & = 4 - 2\cos 5x \\ & = -2\cos (5x) + 4 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Centre line: } & y = 4 \\ \\ \text{Max. value} & = 4 + 2 = 6 \\ \text{Min. value} & = 4 - 2 = -2 \\ \\ \text{Period} & = {360^\circ \over 5} \\ & = 72^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 72^\circ} \\ & = 2{1 \over 2} \end{align}

\begin{align} y & = 1 - 2\sin {x \over 3} \\ & = 1 - 2 \sin \left({1 \over 3}x \right) \\ & = -2\sin \left({1 \over 3}x \right) + 1 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Centre line: } & y = 1 \\ \\ \text{Max. value} & = 1 + 2 = 3 \\ \text{Min. value} & = 1 - 2 = -1 \\ \\ \text{Period} & = {360^\circ \over {1 \over 3}} \\ & = 1080^\circ \\ \\ \text{No. of cycles} & = {810^\circ \over 1080^\circ} \\ & = {3 \over 4} \end{align}

\begin{align} y & = 1.3 + 1.2 \cos 2t \\ & = 1.2 \cos (2t) + 1.3 \\ \\ \text{Amplitude} & = 1.2 \\ \\ \text{Centre line: } & y = 1.3 \\ \\ \text{Max. value} & = 1.3 + 1.2 = 2.5 \\ \text{Min. value} & = 1.3 - 1.2 = 0.1 \\ \\ \text{Period} & = {2\pi \over 2} \\ & = \pi \\ \\ \text{No. of cycles} & = {2\pi \over 2} \\ & = 2 \end{align}

\begin{align} y & = p \cos \left({1 \over 2}x \right) - 1 \\ \\ \text{Since } p < q, &\phantom{0} p = -4 \\ \\ y & = - 4 \cos \left( {1 \over 2}x \right) - 1 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Center line: } & y = -1 \\ \\ \text{Maximum value} & = 3 \\ \text{Minimum value} & = -5 \\ \\ \text{No. of cycles} & = {4\pi \over 4\pi} \\ & = 1 \end{align}

\begin{align} y & = 4 \sin 2x \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Max. value} & = 4 \\ \text{Min. value} & = -4 \\ \\ \text{Period} & = {360^\circ \over 2} \\ & = 180^\circ \\ \\ \text{No. of cycles} & = {360^\circ \over 180^\circ} \\ & = 2 \end{align}

\begin{align} y & = 2 \cos x - 1 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Center line: } & y = -1 \\ \\ \text{Max. value} & = -1 + 2 = 1 \\ \text{Min. value} & = -1 - 2 = -3 \\ \\ \text{Period} & = {360^\circ \over 1} \\ & = 360^\circ \\ \\ \text{No. of cycles} & = {360^\circ \over 360^\circ} \\ & = 1 \end{align}

\begin{align} y & = 5 \tan x \\ \\ \text{Period} & = {180^\circ \over 1} \\ & = 180^\circ \\ \\ \text{No. of cycles} & = {360^\circ \over 180^\circ} \\ & = 2 \\ \\ \text{Vertical asymptotes: } & x = 90^\circ \text{ and } x = 270^\circ \end{align}

\begin{align} y & = -4 \tan 3x \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Period} & = {180^\circ \over 3} \\ & = 60^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 60^\circ} \\ & = 3 \\ \\ \text{Vertical asymptotes: } & x = 30^\circ, x = 90^\circ, \text{ and } x = 120^\circ \end{align}

\begin{align} y & = 3 \tan 2x \\ \\ \text{Period} & = {\pi \over 2} \\ \\ \text{No. of cycles} & = {2\pi \over {\pi \over 2}} \\ & = 4 \\ \\ \text{Vertical asymptotes: } & x = {\pi \over 4}, x = {3\pi \over 4}, x = {5\pi \over 4} \text{ and } x = {7\pi \over 4} \end{align}

\begin{align} y & = -1.5 \tan 4x \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Period} & = {\pi \over 4} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 4}} \\ & = 4 \\ \\ \text{Vertical asymptotes: } & x = {\pi \over 8}, x = {3\pi \over 8}, x = {5\pi \over 8} \text{ and } x = {7\pi \over 8} \end{align}

\begin{align} I & = 20 \sin 4t \\ \\ \text{Maximum value} & = 20 \text{ amperes} \\ \\ \text{When } & I = 20, \\ 20 & = 20 \sin 4t \\ {20 \over 20} & = \sin 4t \\ 1 & = \sin 4t \end{align}

\begin{align} 4t & = {\pi \over 2} \\ t & = {\pi \over 8} \\ t & = 0.39269 \\ t & \approx 0.393 \\ \\ \therefore \text{First inst} & \text{ant is } 0.393 \text{ ms} \end{align}

\begin{align} I & = 20 \sin 4t \\ \\ \text{Maximum} & = 20 \\ \\ \text{Minimum} & = -20 \\ \\ \text{Period} & = {2\pi \over 4} \\ & = {\pi \over 2} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 2}} \\ & = 2 \end{align}

\begin{align} \text{Consider } y & = 6 \sin 3x + 2 \\ \\ \text{Amplitude} & = 6 \\ \\ \text{Center line: } & y = 2 \\ \\ \text{Maximum} & = 2 + 6 = 8 \\ \\ \text{Minimum} & = 2 - 6 = -4 \\ \\ \text{Period} & = {2\pi \over 3} \\ \\ \text{No. of cycles} & = {\pi \over {2\pi \over 3}} \\ & = 1{1 \over 2} \end{align}

Reflect the portion below the x-axis to obtain the modulus graph:

\begin{align} \text{Consider } y & = 4 \cos \left(x \over 2\right) - 1 \\ y & = 4 \cos \left( {1 \over 2}x \right) - 1 \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Center line: } & y = -1 \\ \\ \text{Maximum} & = -1 + 4 = 3 \\ \\ \text{Minimum} & = -1 - 4 = -5 \\ \\ \text{Period} & = {2\pi \over {1 \over 2}} \\ & = 4 \pi \\ \\ \text{No. of cycles} & = {\pi \over 4 \pi} \\ & = {1 \over 4} \end{align}

Reflect the portion below the x-axis to obtain the modulus graph:

\begin{align} \text{Consider } y & = 2 \tan 3x \\ \\ \text{Period} & = {\pi \over 3} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 3}} \\ & = 3 \\ \\ \text{Vertical asymptotes: } & x = {\pi \over 6}, {\pi \over 2}, {5\pi \over 6} \end{align}

Reflect the portion below the x-axis to obtain the modulus graph:

\begin{align} \text{Consider } y & = 5 + 3 \sin x \\ y & = 3 \sin x + 5 \\ \\ \text{Amplitude} & = 3 \\ \\ \text{Center line: } & y = 5 \\ \\ \text{Maximum} & = 5 + 3 = 8 \\ \\ \text{Minimum} & = 5 - 3 = 2 \\ \\ \text{Period} & = {2\pi \over 1} \\ & = 2 \pi \\ \\ \text{No. of cycles} & = {\pi \over 2\pi} \\ & = {1 \over 2} \end{align}

\begin{align} \text{Consider } y & = \sin 2x + 1 \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Center line: } & y = 1 \\ \\ \text{Maximum} & = 1 + 1 = 2 \\ \\ \text{Minimum} & = 1 - 1 = 0 \\ \\ \text{Period} & = {2\pi \over 2} \\ & = \pi \\ \\ \text{No. of cycles} & = {\pi \over \pi} \\ & = 1 \end{align}

\begin{align} \text{Consider } y & = 1.5 \cos 4x - 2 \\ \\ \text{Amplitude} & = 1.5 \\ \\ \text{Center line: } & y = -2 \\ \\ \text{Maximum} & = -2 + 1.5 = -0.5 \\ \\ \text{Minimum} & = -2 - 1.5 = -3.5 \\ \\ \text{Period} & = {2\pi \over 4} \\ & = {\pi \over 2} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 2}} \\ & = 2 \end{align}

Reflect the portion below the x-axis to obtain the modulus graph:

\begin{align} y & = a \sin bx + c \\ y & = 4 \sin 3x - 2 \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Center line: } & y = -2 \\ \\ \text{Maximum} & = 2 \\ \\ \text{Minimum} & = -2 -4 \\ & = -6 \\ \\ \text{Period} & = 120^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 120^\circ} \\ & = 1{1 \over 2} \end{align}

\begin{align} y & = b + c \cos ax \\ y & = c \cos ax + b \\ y & = -2 \cos 4x + 3 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Center line: } & y = 3 \\ \\ \text{Maximum} & = 3 + 2 = 5 \\ \\ \text{Minimum} & = 3 - 2 = 1 \\ \\ \text{Period} & = {360^\circ \over 4} \\ & = 90^\circ \\ \text{No. of cycles} & = {180^\circ \over 90^\circ} \\ & = 2 \end{align}

\begin{align} y & = a \sin \left({1 \over b}x \right) + c \\ y & = 4 \sin {1 \over 2}x + 5 \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Center line: } & y = 5 \\ \\ \text{Maximum} & = 5 + 4 = 9 \\ \\ \text{Minimum} & = 5 - 4 = 1 \\ \\ \text{Period} & = {2\pi \over {1 \over 2} } \\ & = 4 \pi \\ \\ \text{No. of cycles} & = {6\pi \over 4\pi} \\ & = 1{1 \over 2} \end{align}

\begin{align} \text{Since Rajan starts from point } A, & \text{ the graph can be modelled by } y = a \sin bt^\circ + c. \\ \\ \text{Period} & = 360 \text{ seconds} \\ {360^\circ \over b} & = 360^\circ \\ 360^\circ & = 360^\circ (b) \\ {360^\circ \over 360^\circ} & = b \\ 1 & = b \\ \\ \text{Radius of wheel} & = 40 \div 2 \\ & = 20 \text{ m} \\ \\ \text{Amplitude} & = 20 \\ a & = 20 \\ \\ \text{Maximum} & = 20 + 25 \text{ (Highest point from the ground)} \\ & = 45 \text{ m} \\ \\ \text{Minimum} & = 25 - 20 \text{ (Lowest point from the ground)} \\ & = 5 \text{ m} \\ \\ c & = \text{Maximum} - \text{Amplitude} \\ & = 45 - 20 \\ & = 25 \\ \\ \therefore y & = a \sin bt^\circ + c \\ y & = (20) \sin [(1)(t^\circ)] + 25 \\ y & = 20 \sin t^\circ + 25 \end{align}

\begin{align} y & = 20 \sin t^\circ + 25 \\ \\ \text{When } & y = 40, \\ 40 & = 20 \sin t^\circ + 25 \\ 40 - 25 & = 20 \sin t^\circ \\ 15 & = 20 \sin t^\circ \\ {15 \over 20} & = \sin t^\circ \\ {3 \over 4} & = \sin t^\circ \phantom{00000000} \text{[1st or 2nd quadrant]} \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left({3 \over 4}\right) \\ & = 48.59^\circ \end{align}

\begin{align} t^\circ & = 48.59^\circ, 180^\circ - 48.59^\circ \\ & = 48.59^\circ, 131.41^\circ \\ & \approx 48.6^\circ, 131.4^\circ \\ \\ \therefore & \phantom{0} 48.6 \text{ s}, 131.4 \text{ s} \end{align}

\begin{align} y & = 2 \cos 2x \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Max. value} & = 2 \\ \text{Min. value} & = -2 \\ \\ \text{Period} & = {2\pi \over 2} \\ & = \pi \\ \\ \text{No. of cycles} & = {2\pi \over \pi} \\ & = 2 \\ \\ \\ y & = 1 + \sin x \\ y & = \sin x + 1 \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Center line: } & y = 1 \\ \\ \text{Max. value} & = 1 + 1 = 2 \\ \text{Min. value} & = 1 - 1 = 0 \\ \\ \text{Period} & = {2\pi \over 1 }\\ & = 2\pi \\ \\ \text{No. of cycles} & = {2\pi \over 2\pi} \\ & = 1 \end{align}

\begin{align} 2|\cos 2x| & = 1 + \sin x \\ |2| |\cos 2x| & = 1 + \sin x \\ |2 \cos 2x| & = 1 + \sin x \end{align}

Obtain the modulus of the first graph by reflecting the parts below the x-axis:

$$ \therefore \text{7 distinct values of } x $$

\begin{align} y & = \sin x \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Max. value} & = 1 \\ \\ \text{Min. value} & = -1 \\ \\ \text{Period} & = {2\pi \over 1} \\ & = 2\pi \\ \\ \text{No. of cycles} & = {2\pi \over 2\pi} \\ & = 1 \\ \\ \\ y & = 2 \cos x \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Max. value} & = 2 \\ \\ \text{Min. value} & = -2 \\ \\ \text{Period} & = {2\pi \over 1} \\ & = 2\pi \\ \\ \text{No. of cycles} & = {2\pi \over 2\pi} \\ & = 1 \end{align}

From the graph in (i), both functions are decreasing between $x = {\pi \over 2}$ to $x = \pi$.

$$ \therefore {\pi \over 2} < x < \pi $$

\begin{align} y & = \tan 3x \\ \\ \text{Period} & = {\pi \over 3} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 3}} \\ & = 3 \\ \\ \text{Vertical asymptotes: } & x = {\pi \over 6}, x = {\pi \over 2} \text{ and } x = {5\pi \over 6} \\ \\ \\ y & = -{5 \over 2} \tan 2x \\ \\ \text{Period} & = {\pi \over 2} \\ \\ \text{No. of cycles} & = {\pi \over {\pi \over 2}} \\ & = 2 \\ \\ \text{Vertical asymptotes: } & x = {\pi \over 2} \text{ and } x = {3\pi \over 2} \end{align}

\begin{align} 6 \text{ solutions} \phantom{000} \text{ [including } (0, 0) \text{ and } (\pi, 0)] \end{align}

\begin{align} y & = 3 \cos x + 2 \\ \\ \text{Amplitude} & = 3 \\ \\ \text{Center line: } & y = 2 \\ \\ \text{Max. value} & = 2 + 3 = 5 \\ \text{Min. value} & = 2 - 3 = -1 \\ \\ \text{Period} & = {2\pi \over 1} = 2\pi \end{align}

\begin{align} y & = 2 + \cos x \\ y & = \cos x + 2 \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Center line: } & y = 2 \\ \\ \text{Max. value} & = 2 + 1 = 3 \\ \text{Min. value} & = 2 - 1 = 1 \\ \\ \text{Period} & = {2\pi \over 1} = 2\pi \\ \\ \\ y & = 4 \sin {1 \over 2}x \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Max. value} & = 4 \\ \text{Min. value} & = -4 \\ \\ \text{Period} & = {2\pi \over {1 \over 2}} = 4\pi \\ \\ \text{No. of cycles} & = {2\pi \over 4\pi} = {1 \over 2} \end{align}

\begin{align} y & = -3.1 \tan 2x \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Period} & = {180^\circ \over 2} = 90^\circ \\ \\ \text{Vertical } & \text{asymptotes are } x = 45^\circ, x = 135^\circ \\ \\ \\ y & = \sqrt{2} \cos 4x \\ \\ \text{Amplitude} & = \sqrt{2} \\ \\ \text{Max. value} & = \sqrt{2} \\ \text{Min. value} & = - \sqrt{2} \\ \\ \text{Period} & = {360^\circ \over 4} = 90^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 90^\circ} = 2 \end{align}

(After the first point of intersection, the two graphs will then intersect every 90°)

\begin{align} \text{From graph, other roots} & = 90^\circ + \alpha, 180^\circ + \alpha, 270^\circ + \alpha \end{align}

\begin{align} \text{Principal values of } \cos^{-1} x: & \phantom{.} 0^\circ \le \cos^{-1} x \le 180^\circ \\ \\ \text{Let } A & = \cos^{-1} \left(-{1 \over 5}\right) \\ \cos A & = -{1 \over 5} \phantom{000000} [\text{2nd quadrant given principal values}] \\ {Adj \over Hyp} & = {-1 \over 5} \end{align}

\begin{align} \sin A & = {\sqrt{24} \over 5} \\ & = {\sqrt{4} \sqrt{6} \over 5} \\ & = {2 \sqrt{6} \over 5} \end{align}

\begin{align} \text{Principal values of } \sin^{-1} x: & \phantom{.} -90^\circ \le \sin^{-1} x \le 90^\circ \\ \\ \text{Let } B & = \sin^{-1} \left(-{2 \over 3}\right) \\ \sin B & = -{2 \over 3} \phantom{000000} [\text{4th quadrant given principal values}] \\ {Opp \over Hyp} & = {-2 \over 3} \end{align}

\begin{align} \tan B & = {-2 \over \sqrt{5}} \\ & = -{2 \over \sqrt{5}} \end{align}

\begin{align} \text{Graph resembles graph of } & y = \sin x \\ \\ \text{Conjecture: } \cos x \tan x & = \sin x \end{align}

($y = \sin^2 x$ in red, $y = \cos^2 x$ in blue, $y = \sin^2 x + \cos^2 x$ in green)

$$ \sin^2 x + \cos^2 x = 1 $$