Solution. Comparing with \(a\cos (bx+c)+d\): \(a=2\), \(b=2\), \(c=\frac {\pi }{2}\), \(d=-1\). Hence \[\text {amplitude}=\left |2\right |=2,\quad \text {period}=\frac {2\pi }{2}=\pi ,\quad \text {phase shift}=-\frac {c}{b}=-\frac {\pi }{4},\] a shift of \(\frac {\pi }{4}\) to the left, and a vertical shift of \(1\) unit down.

The curve therefore oscillates between \(d-a=-3\) and \(d+a=1\), completing two full cycles on \([0,2\pi ]\), with its centre line at \(y=-1\). Setting the bracket to zero, \(2x+\frac {\pi }{2}=0\) gives \(x=-\frac {\pi }{4}\), so the cosine peak that normally sits at \(x=0\) has moved there; within \([0,2\pi ]\) the peaks fall at \(x=\frac {3\pi }{4}\) and \(x=\frac {7\pi }{4}\).

Solution. The first move in both is to get a single trigonometric function.

(a). Use \(\cos 2\theta =1-2\sin ^{2}\theta \): \[1-2\sin ^{2}\theta +\sin \theta =1 \implies \sin \theta \left (1-2\sin \theta \right )=0 ,\] so \(\sin \theta =0\) or \(\sin \theta =\frac {1}{2}\). In \((-\pi ,\pi ]\), \[\theta =0,\ \pi \quad \text {or}\quad \theta =\frac {\pi }{6},\ \frac {5\pi }{6}.\]

(b). Use \(\tan ^{2}\theta =\sec ^{2}\theta -1\): \[\sec ^{2}\theta -1-\sec \theta -1=0 \implies \sec ^{2}\theta -\sec \theta -2=0 \implies \left (\sec \theta -2\right )\left (\sec \theta +1\right )=0 .\] So \(\sec \theta =2\), giving \(\cos \theta =\frac {1}{2}\) and \(\theta =\pm \frac {\pi }{3}\); or \(\sec \theta =-1\), giving \(\cos \theta =-1\) and \(\theta =\pi \).

But \(\theta =\pi \) must be rejected: there \(\tan \theta =0\) and \(\sec \theta =-1\), so the left side is \(0+1-1=0\) — it does satisfy the equation. Both values stand: \[\theta =\frac {\pi }{3},\ -\frac {\pi }{3},\ \pi .\]

Solution. The proof is the one given earlier in this chapter: expand \(\sin (A-B)\) and \(\cos (A-B)\), then divide above and below by \(\cos A\cos B\).

For the value, let \(A=\sin ^{-1}\frac {24}{25}\) and \(B=\cos ^{-1}\frac {4}{5}\), both in the first quadrant. A right triangle with opposite \(24\) and hypotenuse \(25\) has adjacent \(\sqrt {625-576}=7\), so \(\tan A=\frac {24}{7}\). One with adjacent \(4\) and hypotenuse \(5\) has opposite \(3\), so \(\tan B=\frac {3}{4}\). Then \[\tan (A-B)=\frac {\frac {24}{7}-\frac {3}{4}}{1+\frac {24}{7}\cdot \frac {3}{4}} =\frac {\frac {96-21}{28}}{\frac {28+72}{28}} =\frac {75}{100}=\frac {3}{4}.\]

Solution. (a). From \(\sin ^{2}\theta +\cos ^{2}\theta =1\), \(\cos ^{2}\theta =1-\frac {9}{25}=\frac {16}{25}\), so \(\cos \theta =\pm \frac {4}{5}\). In the second quadrant the cosine is negative, so \(\cos \theta =-\frac {4}{5}\) and \[\cot \theta =\frac {\cos \theta }{\sin \theta } =\frac {-\frac {4}{5}}{\frac {3}{5}}=-\frac {4}{3}.\]

(b). \(\sec \theta =-\frac {5}{3}\) gives \(\cos \theta =-\frac {3}{5}\), so \(\sin ^{2}\theta =1-\frac {9}{25}=\frac {16}{25}\) and \(\sin \theta =\pm \frac {4}{5}\). The condition \(\operatorname {cosec}\theta <0\) means \(\sin \theta <0\), so \(\sin \theta =-\frac {4}{5}\) and \[\tan \theta =\frac {\sin \theta }{\cos \theta } =\frac {-\frac {4}{5}}{-\frac {3}{5}}=\frac {4}{3}.\]

Solution. Turn the difference of cosines into a product using \(\cos A-\cos B=2\sin \frac {A+B}{2}\sin \frac {B-A}{2}\), with \(A=2x\) and \(B=4x\): \[\cos 2x-\cos 4x=2\sin 3x\sin x .\] The equation becomes \[2\sin 3x\sin x=\sin x \implies \sin x\left (2\sin 3x-1\right )=0 ,\] so \(\sin x=0\) or \(\sin 3x=\frac {1}{2}\).

From \(\sin x=0\):  \(x=n\pi \),  \(n\in \mathbb {Z}\).

From \(\sin 3x=\frac {1}{2}\): the sine takes the value \(\frac {1}{2}\) at \(\frac {\pi }{6}\) and \(\frac {5\pi }{6}\), and repeats every \(2\pi \), so \[3x=\frac {\pi }{6}+2n\pi \quad \text {or}\quad 3x=\frac {5\pi }{6}+2n\pi ,\] giving \[x=\frac {\pi }{18}+\frac {2n\pi }{3} \quad \text {or}\quad x=\frac {5\pi }{18}+\frac {2n\pi }{3},\quad n\in \mathbb {Z}.\]