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$\text { Find the general value of } \theta, \text { when } 9 \sec ^{4} \theta=16$

My work-

Given $\sec ^{4} \theta=\frac{16}{9}$ or, $\sec ^{4} \theta = \frac{(4)^{2}}{(3)^{2}}$

$\implies \sec ^{2} \theta=\frac{4}{3} \Longrightarrow \theta=\sec ^{-1}\left(\sqrt{\frac{4}{3}}\right)$

hence, $\theta=2 n \pi \pm \frac{\pi}{6}$.

I am following Hobson's Trigonometry and it doesn't have the answer or the solution provided, neither any examples regarding inverse trigonometric functions. I got the feedback that my approach is incorrect but I cannot see how to fix it so any hints on where I am going mistaken ?

P.S. Hobson mentions that $x = \sec^{-1}(y)$ is a multi-valued function.

noobman
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2 Answers2

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$$ \begin{aligned} 9 \sec ^{4} \theta &=16 \\ \sec ^{4} \theta &=\frac{16}{9} \\ \sec \theta &=\pm \frac{2}{\sqrt{3}} \\ \cos \theta &=\pm \frac{\sqrt{3}}{2} \\ \theta &=n \pi \pm \frac{\pi}{6}=\frac{(6 n \pm 1) \pi}{6}, \end{aligned} $$ where $n \in \mathbb{Z}.$

Lai
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  • But Hobson in the book mentions that the values of $\cos^{-1}(y)$ are included in $2n\pi \pm x,$ where $x = \cos y.$ So should it be $n \pi$ as you mention or $2 n \pi$ ? – noobman Nov 28 '21 at 07:38
  • Your mistake is: $\implies \sec \theta=\frac{2}{\sqrt 3}$\ Instead, it should be $\implies \sec \theta=\pm \frac{2}{\sqrt 3}$ – Lai Nov 28 '21 at 07:44
  • But I did take a $\pm$ right ? I mean I don't evaluate $\sqrt{\frac{4}{3}},$ and directly use $\pm \frac{\pi}{6}.$

    Still I didn't get why $n \pi$ instead of $2n \pi$ ? We are only considering positives or similar reasoning ?

     – noobman Nov 28 '21 at 07:48
  • $\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=2 n \pi \pm \frac{\pi}{6}$ $\cos \theta=-\frac{\sqrt{3}}{2} \Rightarrow \theta=2 n \pi \pm \frac{5 \pi}{6}$ Combining gives $ \theta =n\pi\pm \frac{\pi}{6}.$ – Lai Nov 28 '21 at 07:54
  • $\implies \sec ^{2} \theta=\frac{4}{3} \Longrightarrow \theta=\sec ^{-1}\left(\pm \sqrt{\frac{4}{3}}\right)$ – Lai Nov 28 '21 at 07:59
  • $\sec^{-1}(-\sqrt{\frac{4}{3}})=\frac{5\pi}{6}$ ! – Lai Nov 28 '21 at 08:12
  • Ok I wrapped things around my head, and I see where I went wrong. Thank you for explaining all the steps!

    I also figured that when "combining" the two results, a new variable makes it much easier to grasp, as in $k$ instead of $n$ (infact, in another book I saw the two $n$ written for each value of $\cos \theta$ is written as $2m\pi \pm \alpha$ and $2 n \pi \pm \alpha$ which finally combine as $k \pi \pm \alpha.$)

    Thank you again~

     – noobman Nov 28 '21 at 08:20
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    You are welcome. Be aware that $\sec^{-1}(-\sqrt{\frac{4}{3}})=\frac{5\pi}{6}\neq -\frac{\pi}{6} $ ! – Lai Nov 28 '21 at 09:02
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$$\sec^2\theta=\dfrac43$$

$$\cos^2\theta=?$$

$$\sin^2\theta=?=\sin^2A\text{(say)}$$

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ and

if $\sin y=0, y=m\pi$

lab bhattacharjee
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