Answer: \(\alpha=99.7^\circ,279.7^\circ\); \(a=-2\); \(\phi=-\dfrac{2\pi}{9},-\dfrac{\pi}{9},\dfrac{\pi}{9},\dfrac{2\pi}{9}\).
(a) Let
\(u=\alpha+45^\circ.\)
Then
\(\tan u=-\frac1{\sqrt2}.\)
The reference angle is
\(\tan^{-1}\left(\frac1{\sqrt2}\right)\approx35.3^\circ.\)
Since tangent is negative in quadrants II and IV,
\(u=144.7^\circ\quad\text{or}\quad u=324.7^\circ.\)
Thus
\(\alpha=144.7^\circ-45^\circ \quad\text{or}\quad \alpha=324.7^\circ-45^\circ.\)
Therefore
\(\alpha=99.7^\circ,\quad279.7^\circ.\)
(b)(i) Combine the fractions:
\(\frac1{\sin\theta-1}-\frac1{\sin\theta+1} =\frac{(\sin\theta+1)-(\sin\theta-1)}{(\sin\theta-1)(\sin\theta+1)}.\)
The numerator is \(2\), and the denominator is
\(\sin^2\theta-1=-\cos^2\theta.\)
So
\(\frac1{\sin\theta-1}-\frac1{\sin\theta+1} =-\frac2{\cos^2\theta}.\)
Therefore
\(\frac1{\sin\theta-1}-\frac1{\sin\theta+1} =-2\operatorname{sec}^2\theta,\)
so
\(a=-2.\)
(b)(ii) Using the result from part (i), with \(\theta=3\phi\),
\(-2\operatorname{sec}^2(3\phi)=-8.\)
So
\(\operatorname{sec}^2(3\phi)=4.\)
Hence
\(\cos^2(3\phi)=\frac14,\)
so
\(\cos(3\phi)=\pm\frac12.\)
Since
\(-\frac{\pi}{3}\leq\phi\leq\frac{\pi}{3},\)
we have
\(-\pi\leq3\phi\leq\pi.\)
In this interval, the solutions are
\(3\phi=-\frac{2\pi}{3},\ -\frac{\pi}{3},\ \frac{\pi}{3},\ \frac{2\pi}{3}.\)
Therefore
\(\phi=-\frac{2\pi}{9},\ -\frac{\pi}{9},\ \frac{\pi}{9},\ \frac{2\pi}{9}.\)
