Solution. The signs in each quadrant are worth having at hand:

Each condition names a set of two quadrants; the answer is the one they share.

\[\text {(a) } \sin >0:\ \text {I or II};\quad \tan <0:\ \text {II or IV} \quad \implies \quad \textbf {quadrant II}.\] \[\text {(b) } \sin <0:\ \text {III or IV};\quad \cos <0:\ \text {II or III} \quad \implies \quad \textbf {quadrant III}.\] \[\text {(c) } \cos >0:\ \text {I or IV};\quad \tan <0:\ \text {II or IV} \quad \implies \quad \textbf {quadrant IV}.\]

(d). \(\sec \theta <0\) means \(\frac {1}{\cos \theta }<0\), that is \(\cos \theta <0\): \[\tan >0:\ \text {I or III};\quad \cos <0:\ \text {II or III} \quad \implies \quad \textbf {quadrant III}.\]

(e). \(\cot \theta <0\) means \(\tan \theta <0\), since they are reciprocals and share a sign: \[\sin <0:\ \text {III or IV};\quad \tan <0:\ \text {II or IV} \quad \implies \quad \textbf {quadrant IV}.\]

(f). \(\csc >0\) means \(\sin >0\) and \(\sec >0\) means \(\cos >0\): \[\text {I or II};\quad \text {I or IV}\quad \implies \quad \textbf {quadrant I}.\]

Solution. Multiply by \(\frac {\pi }{180}\) and cancel.

\[\text {(a) } -250\times \frac {\pi }{180}=-\frac {250\pi }{180}=-\frac {25\pi }{18} \qquad (\div 10)\] \[\text {(b) } 25\times \frac {\pi }{180}=\frac {25\pi }{180}=\frac {5\pi }{36}\qquad (\div 5)\] \[\text {(c) } 75\times \frac {\pi }{180}=\frac {75\pi }{180}=\frac {5\pi }{12}\qquad (\div 15)\] \[\text {(d) } -330\times \frac {\pi }{180}=-\frac {330\pi }{180}=-\frac {11\pi }{6} \qquad (\div 30)\] \[\text {(e) } 570\times \frac {\pi }{180}=\frac {570\pi }{180}=\frac {19\pi }{6} \qquad (\div 30)\] \[\text {(f) } 635\times \frac {\pi }{180}=\frac {635\pi }{180}=\frac {127\pi }{36} \qquad (\div 5)\]

Solution. Multiply by \(\frac {180}{\pi }\), so the \(\pi \) cancels and only the fraction is left.

\[\text {(a) } \frac {7\pi }{3}\times \frac {180}{\pi }=7\times 60=420^\circ \] \[\text {(b) } -\frac {3\pi }{5}\times \frac {180}{\pi }=-3\times 36=-108^\circ \] \[\text {(c) } \frac {\pi }{13}\times \frac {180}{\pi }=\frac {180}{13} \approx 13.85^\circ \] \[\text {(d) } \frac {5\pi }{3}\times \frac {180}{\pi }=5\times 60=300^\circ \] \[\text {(e) } \frac {7\pi }{6}\times \frac {180}{\pi }=7\times 30=210^\circ \] \[\text {(f) } -\frac {11\pi }{4}\times \frac {180}{\pi }=-11\times 45=-495^\circ \]

Solution. Three tools cover all of these: the addition formulae, the exact values at \(30^\circ \), \(45^\circ \) and \(60^\circ \), and the fact that adding or subtracting a full revolution changes nothing.

(a). \(75^\circ =45^\circ +30^\circ \): \[\sin 75^\circ =\sin 45^\circ \cos 30^\circ +\cos 45^\circ \sin 30^\circ =\frac {\sqrt {2}}{2}\cdot \frac {\sqrt {3}}{2}+\frac {\sqrt {2}}{2}\cdot \frac {1}{2} =\frac {\sqrt {6}+\sqrt {2}}{4}.\]

(b). Cosine is an even function, so \(\cos (-15^\circ )=\cos 15^\circ \), and \(15^\circ =45^\circ -30^\circ \): \[\cos 15^\circ =\cos 45^\circ \cos 30^\circ +\sin 45^\circ \sin 30^\circ =\frac {\sqrt {6}}{4}+\frac {\sqrt {2}}{4}=\frac {\sqrt {6}+\sqrt {2}}{4}.\] The same value as (a), because \(75^\circ \) and \(15^\circ \) are complementary.

(c). \(150^\circ \) is in quadrant II, where cosine is negative, with reference angle \(180^\circ -150^\circ =30^\circ \): \[\cos 150^\circ =-\cos 30^\circ =-\frac {\sqrt {3}}{2}.\]

(d). Sine is odd, so \(\sin (-135^\circ )=-\sin 135^\circ \). And \(135^\circ \) is in quadrant II with reference angle \(45^\circ \): \[\sin (-135^\circ )=-\sin 45^\circ =-\frac {\sqrt {2}}{2}.\]

(e). \(315^\circ \) is in quadrant IV, where tangent is negative, with reference angle \(360^\circ -315^\circ =45^\circ \): \[\tan 315^\circ =-\tan 45^\circ =-1.\]

(f). Subtract a full revolution, \(2\pi =\frac {6\pi }{3}\): \[\cos \frac {7\pi }{3}=\cos \left (\frac {7\pi }{3}-\frac {6\pi }{3}\right )=\cos \frac {\pi }{3} =\frac {1}{2}.\]

(g). Add a full revolution to make the angle positive: \[\cot \left (-\frac {5\pi }{3}\right )=\cot \left (-\frac {5\pi }{3}+2\pi \right ) =\cot \frac {\pi }{3}=\frac {1}{\tan \frac {\pi }{3}}=\frac {1}{\sqrt {3}}=\frac {\sqrt {3}}{3}.\]

(h). \(\frac {7\pi }{6}=210^\circ \), in quadrant III with reference angle \(30^\circ \), where cosine is negative: \[\cos \frac {7\pi }{6}=-\frac {\sqrt {3}}{2}\quad \implies \quad \sec \frac {7\pi }{6}=\frac {1}{-\frac {\sqrt {3}}{2}}=-\frac {2}{\sqrt {3}} =-\frac {2\sqrt {3}}{3}.\]

(i). \(\frac {\pi }{12}=15^\circ \), and \(\sin 15^\circ =\frac {\sqrt {6}-\sqrt {2}}{4}\) by the same addition formula as in (b). So \[\csc \frac {\pi }{12}=\frac {4}{\sqrt {6}-\sqrt {2}} =\frac {4\left (\sqrt {6}+\sqrt {2}\right )}{6-2} =\frac {4\left (\sqrt {6}+\sqrt {2}\right )}{4}=\sqrt {6}+\sqrt {2}.\]

Solution. Step 1 — find \(\sin \theta \). From \(\sin ^2\theta +\cos ^2\theta =1\): \[\sin ^2\theta =1-\left (-\tfrac {4}{5}\right )^2=1-\tfrac {16}{25}=\tfrac {9}{25} \quad \implies \quad \sin \theta =\pm \tfrac {3}{5}.\] In the third quadrant sine is negative, so \[\sin \theta =-\tfrac {3}{5}.\]

Step 2 — form the two required ratios. \[\cot \theta =\frac {\cos \theta }{\sin \theta }=\frac {-\frac {4}{5}}{-\frac {3}{5}} =\frac {4}{3},\] \[\csc \theta =\frac {1}{\sin \theta }=\frac {1}{-\frac {3}{5}}=-\frac {5}{3}.\]

Solution. The reliable method is to convert everything to sines and cosines, combine over a common denominator, and use \(\sin ^2+\cos ^2=1\).

(a). \[\sec x-\sin x\tan x=\frac {1}{\cos x}-\sin x\cdot \frac {\sin x}{\cos x} =\frac {1-\sin ^2x}{\cos x}=\frac {\cos ^2x}{\cos x}=\cos x.\]

(b). \[\csc x-\cos x\cot x=\frac {1}{\sin x}-\cos x\cdot \frac {\cos x}{\sin x} =\frac {1-\cos ^2x}{\sin x}=\frac {\sin ^2x}{\sin x}=\sin x.\]

(c). \[\cos x+\tan x\sin x=\cos x+\frac {\sin ^2x}{\cos x}=\frac {\cos ^2x+\sin ^2x}{\cos x} =\frac {1}{\cos x}=\sec x.\]

(d). Use \(\sin ^2x-1=-\cos ^2x\) and \(\tan ^2x+1=\sec ^2x\): \[\left (-\cos ^2x\right )\left (\sec ^2x\right )=-\cos ^2x\cdot \frac {1}{\cos ^2x}=-1.\]

(e). \[\frac {\sec \theta -\cos \theta }{\tan \theta } =\frac {\frac {1}{\cos \theta }-\cos \theta }{\frac {\sin \theta }{\cos \theta }} =\frac {\frac {1-\cos ^2\theta }{\cos \theta }}{\frac {\sin \theta }{\cos \theta }} =\frac {\sin ^2\theta }{\cos \theta }\cdot \frac {\cos \theta }{\sin \theta }=\sin \theta .\]

(f). \[\frac {\sec \theta +1}{\tan \theta +\sin \theta } =\frac {\frac {1}{\cos \theta }+1}{\frac {\sin \theta }{\cos \theta }+\sin \theta } =\frac {\frac {1+\cos \theta }{\cos \theta }} {\frac {\sin \theta +\sin \theta \cos \theta }{\cos \theta }} =\frac {1+\cos \theta }{\sin \theta (1+\cos \theta )}=\frac {1}{\sin \theta }=\csc \theta .\]

Solution.

(a). Starting from the left and using \(\tan ^2x+1=\sec ^2x\): \[\sec ^2x+\tan x\sec x=\frac {1}{\cos ^2x}+\frac {\sin x}{\cos x}\cdot \frac {1}{\cos x} =\frac {1}{\cos ^2x}+\frac {\sin x}{\cos ^2x}=\frac {1+\sin x}{\cos ^2x}.\qquad \blacksquare \]

(b). Split the fraction: \[\frac {\cos x+\sin x}{\cos x}=\frac {\cos x}{\cos x}+\frac {\sin x}{\cos x} =1+\tan x.\qquad \blacksquare \]

(c). Both sides are fractions, so cross-multiply — legitimate here because we are showing the products are equal, not moving terms: \[(\csc x-1)(\csc x+1)=\csc ^2x-1=\cot ^2x=\cot x\cdot \cot x,\] using \(1+\cot ^2x=\csc ^2x\). The two cross-products agree, so the fractions are equal. \(\blacksquare \)

(d). Work on the left: \[\tan x+\cot x=\frac {\sin x}{\cos x}+\frac {\cos x}{\sin x} =\frac {\sin ^2x+\cos ^2x}{\sin x\cos x}=\frac {1}{\sin x\cos x},\] \[\therefore \quad \frac {1}{\tan x+\cot x}=\sin x\cos x.\qquad \blacksquare \]

(e). Replace \(1+\cot ^2x\) by \(\csc ^2x\) and \(1+\tan ^2x\) by \(\sec ^2x\): \[\frac {\tan x\csc ^2x}{\sec ^2x}=\frac {\sin x}{\cos x}\cdot \frac {1}{\sin ^2x} \cdot \cos ^2x=\frac {\cos x}{\sin x}=\cot x.\qquad \blacksquare \]

(f). Start from the right-hand side, which is the more complicated: \[(\tan x-\sec x)^2=\left (\frac {\sin x}{\cos x}-\frac {1}{\cos x}\right )^2 =\left (\frac {\sin x-1}{\cos x}\right )^2=\frac {(1-\sin x)^2}{\cos ^2x},\] since squaring removes the sign. Now use \(\cos ^2x=1-\sin ^2x=(1-\sin x)(1+\sin x)\): \[=\frac {(1-\sin x)^2}{(1-\sin x)(1+\sin x)}=\frac {1-\sin x}{1+\sin x}. \qquad \blacksquare \]

(g). Start from the left: \[\frac {1}{\sec x(1-\sin x)}=\frac {\cos x}{1-\sin x}.\] Multiply top and bottom by \(1+\sin x\): \[=\frac {\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}=\frac {\cos x(1+\sin x)}{1-\sin ^2x} =\frac {\cos x(1+\sin x)}{\cos ^2x}=\frac {1+\sin x}{\cos x}\] \[=\frac {1}{\cos x}+\frac {\sin x}{\cos x}=\sec x+\tan x.\qquad \blacksquare \]

(h). Work on the denominator of the left side: \[\cot x+\cot y=\frac {1}{\tan x}+\frac {1}{\tan y}=\frac {\tan y+\tan x}{\tan x\tan y}.\] Therefore \[\frac {\tan x+\tan y}{\cot x+\cot y} =(\tan x+\tan y)\cdot \frac {\tan x\tan y}{\tan x+\tan y}=\tan x\tan y. \qquad \blacksquare \]

Solution. All six come from the addition formulae \[\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B,\qquad \sin (A\pm B)=\sin A\cos B\pm \cos A\sin B.\]

(a). Take \(A=B=x\) in the cosine formula: \[\cos 2x=\cos x\cos x-\sin x\sin x=\cos ^2x-\sin ^2x.\] Replacing \(\sin ^2x\) by \(1-\cos ^2x\): \[=\cos ^2x-\left (1-\cos ^2x\right )=2\cos ^2x-1.\qquad \blacksquare \]

(b). From the same starting point, replace \(\cos ^2x\) by \(1-\sin ^2x\) instead: \[\cos 2x=\left (1-\sin ^2x\right )-\sin ^2x=1-2\sin ^2x.\qquad \blacksquare \]

(c). With \(\cos \pi =-1\) and \(\sin \pi =0\): \[\cos (\theta -\pi )=\cos \theta \cos \pi +\sin \theta \sin \pi =\cos \theta (-1)+0 =-\cos \theta .\qquad \blacksquare \]

(d). With \(\cos \frac {\pi }{2}=0\) and \(\sin \frac {\pi }{2}=1\): \[\cos \left (\theta +\tfrac {\pi }{2}\right )=\cos \theta \cdot 0-\sin \theta \cdot 1 =-\sin \theta .\qquad \blacksquare \]

(e). \[\sin \left (\theta +\tfrac {\pi }{2}\right )=\sin \theta \cdot 0+\cos \theta \cdot 1 =\cos \theta .\qquad \blacksquare \]

(f). Using \(\tan (A-B)=\frac {\tan A-\tan B}{1+\tan A\tan B}\) with \(\tan \pi =0\): \[\tan (\theta -\pi )=\frac {\tan \theta -0}{1+0}=\tan \theta .\qquad \blacksquare \]

Solution. (a). \(\sin x=-\frac {1}{2}\). The reference angle is \(\frac {\pi }{6}\), and sine is negative in quadrants III and IV: \[x=\pi +\tfrac {\pi }{6}=\tfrac {7\pi }{6},\qquad x=2\pi -\tfrac {\pi }{6}=\tfrac {11\pi }{6}.\]

(b). \(\cos x=\frac {1}{2}\), reference angle \(\frac {\pi }{3}\), cosine positive in quadrants I and IV: \[x=\tfrac {\pi }{3},\qquad x=2\pi -\tfrac {\pi }{3}=\tfrac {5\pi }{3}.\]

(c). Factorise — do not divide by \(\cos x\): \[\cos x\left (2\cos x-\sqrt {3}\right )=0.\] \[\cos x=0\quad \implies \quad x=\tfrac {\pi }{2},\ \tfrac {3\pi }{2};\] \[\cos x=\tfrac {\sqrt {3}}{2}\quad \implies \quad x=\tfrac {\pi }{6},\ \tfrac {11\pi }{6}.\] \[\therefore \quad x=\tfrac {\pi }{6},\ \tfrac {\pi }{2},\ \tfrac {3\pi }{2},\ \tfrac {11\pi }{6}.\]

(d). \(\sin x=1\), which happens once in the interval: \[x=\tfrac {\pi }{2}.\]

(e). Replace \(\sin ^2x\) by \(1-\cos ^2x\): \[2\cos ^2x+\cos x=1-\cos ^2x\quad \implies \quad 3\cos ^2x+\cos x-1=0.\] This does not factorise, and the quadratic formula gives \[\cos x=\frac {-1\pm \sqrt {1+12}}{6}=\frac {-1\pm \sqrt {13}}{6} \approx 0.4343\ \text { or } -0.7676.\] Neither is a standard value, so the angles cannot be written exactly and the question cannot be finished “without a calculator” as instructed. To three decimal places, \[x\approx 1.121,\ 5.162,\ 2.447,\ 3.836.\]

Almost certainly the intended equation was \[2\cos ^2x+\cos x-1=0,\] which factorises as \((2\cos x-1)(\cos x+1)=0\), giving \[\cos x=\tfrac {1}{2}\ \implies \ x=\tfrac {\pi }{3},\ \tfrac {5\pi }{3};\qquad \cos x=-1\ \implies \ x=\pi .\]

(f). Since \(\cot x=\frac {1}{\tan x}\), \[\tan x=\frac {1}{\tan x}\quad \implies \quad \tan ^2x=1\quad \implies \quad \tan x=\pm 1.\] The reference angle is \(\frac {\pi }{4}\), and tangent takes each value twice in \([0,2\pi ]\): \[x=\tfrac {\pi }{4},\ \tfrac {3\pi }{4},\ \tfrac {5\pi }{4},\ \tfrac {7\pi }{4}.\]

Solution. The interval is two full revolutions, so every solution in \([0,2\pi )\) has a partner \(2\pi \) further on. Find the solutions in one revolution and then add \(2\pi \) to each, keeping any that still fall within \(4\pi \).

(a). Factorise: \[\tan x\left (2\sec x-1\right )=0.\] \[\tan x=0\quad \implies \quad x=0,\ \pi ,\ 2\pi ,\ 3\pi ,\ 4\pi .\] \[2\sec x=1\quad \implies \quad \sec x=\tfrac {1}{2}\quad \implies \quad \cos x=2,\] which is impossible, since \(\cos x\) never exceeds \(1\). That branch gives nothing. \[\therefore \quad x=0,\ \pi ,\ 2\pi ,\ 3\pi ,\ 4\pi .\]

(b). A quadratic in \(\sec x\): \[(\sec x-2)(\sec x+1)=0.\] \[\sec x=2\ \implies \ \cos x=\tfrac {1}{2}\ \implies \ x=\tfrac {\pi }{3},\ \tfrac {5\pi }{3},\ \tfrac {7\pi }{3},\ \tfrac {11\pi }{3};\] \[\sec x=-1\ \implies \ \cos x=-1\ \implies \ x=\pi ,\ 3\pi .\] \[\therefore \quad x=\tfrac {\pi }{3},\ \pi ,\ \tfrac {5\pi }{3},\ \tfrac {7\pi }{3},\ 3\pi ,\ \tfrac {11\pi }{3}.\]

(c). The equation mixes \(2x\) and \(x\), so use the double-angle form that leaves only sines, \(\cos 2x=1-2\sin ^2x\): \[1-2\sin ^2x+3\sin x-2=0\quad \implies \quad -2\sin ^2x+3\sin x-1=0 \quad \implies \quad 2\sin ^2x-3\sin x+1=0.\] \[(2\sin x-1)(\sin x-1)=0.\] \[\sin x=\tfrac {1}{2}\ \implies \ x=\tfrac {\pi }{6},\ \tfrac {5\pi }{6},\ \tfrac {13\pi }{6},\ \tfrac {17\pi }{6};\] \[\sin x=1\ \implies \ x=\tfrac {\pi }{2},\ \tfrac {5\pi }{2}.\] \[\therefore \quad x=\tfrac {\pi }{6},\ \tfrac {\pi }{2},\ \tfrac {5\pi }{6},\ \tfrac {13\pi }{6},\ \tfrac {5\pi }{2},\ \tfrac {17\pi }{6}.\]

(d). Combine the two terms into one. Since \(\sin x+\cos x=\sqrt {2}\sin \left (x+\frac {\pi }{4}\right )\), \[\sqrt {2}\sin \left (x+\tfrac {\pi }{4}\right )=1\quad \implies \quad \sin \left (x+\tfrac {\pi }{4}\right )=\frac {1}{\sqrt {2}}.\] Writing \(u=x+\frac {\pi }{4}\), which runs over \(\left [\frac {\pi }{4},\ 4\pi +\frac {\pi }{4}\right ]\): \[u=\tfrac {\pi }{4},\ \tfrac {3\pi }{4},\ \tfrac {9\pi }{4},\ \tfrac {11\pi }{4},\ \tfrac {17\pi }{4}.\] Subtracting \(\frac {\pi }{4}\) from each: \[\therefore \quad x=0,\ \tfrac {\pi }{2},\ 2\pi ,\ \tfrac {5\pi }{2},\ 4\pi .\]

Solution. (a). The sine wave lifted one unit. It runs between \(0\) and \(2\) instead of \(-1\) and \(1\), touching \(0\) at \(x=\frac {3\pi }{2}\) and reaching \(2\) at \(x=\frac {\pi }{2}\). Starts and ends at \(1\).

(b). \(-\cos x\) is the cosine wave turned upside down, starting at \(-1\); adding \(1\) lifts it to start at \(0\). It runs between \(0\) and \(2\), peaking at \(2\) when \(x=\pi \), and returns to \(0\) at \(x=2\pi \).

(c). \(-\cos x\) starts at \(-1\), rises through \(0\) at \(x=\frac {\pi }{2}\), and at \(x=\frac {2\pi }{3}\) has reached \(-\cos \frac {2\pi }{3}=\frac {1}{2}\). Only this rising portion is drawn.

(d). Amplitude \(\frac {1}{2}\) and period \[\frac {2\pi }{\frac {1}{2}}=4\pi ,\] so exactly one complete wave fits the interval: up to \(\frac {1}{2}\) at \(x=\pi \), back through zero at \(x=2\pi \), down to \(-\frac {1}{2}\) at \(x=3\pi \), and back to zero at \(x=4\pi \).

Solution. Write each in the standard form \[f(x)=a\sin \left [b(x-c)\right ]+d\qquad \text {or}\qquad f(x)=a\cos \left [b(x-c)\right ]+d,\] from which \[\text {amplitude}=|a|,\qquad \text {period}=\frac {2\pi }{b} \ \left (\text {or } \frac {360^\circ }{b}\right ),\qquad \text {phase shift}=c.\] A positive \(c\) shifts right and a negative \(c\) shifts left.

Two worth setting out.

(b). The bracket must be factorised before the shift can be read: \[2x-\frac {\pi }{2}=2\left (x-\frac {\pi }{4}\right ),\] so \(b=2\) and \(c=\frac {\pi }{4}\). The shift is \(\frac {\pi }{4}\), not \(\frac {\pi }{2}\).

(e). The amplitude is \(|-4|=4\), a positive number. The minus sign does not reduce the amplitude; it turns the curve upside down, so the wave starts at \(-4\) instead of \(+4\).

(g). This is (b) with amplitude tripled and the whole wave lifted \(2\) units. It therefore oscillates between \(2-3=-1\) and \(2+3=5\), with midline \(y=2\).