Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $\sec \theta = k$, $|k| \geq 1$, and that $\theta$ is obtuse, express in terms of $k$: $\cos \theta$, and $\csc \theta$

For $\cos \theta$ I get: $\frac{-1}{k}, (as \theta$ is obtuse, however this is wrong too)

For $\csc \theta$ I get the answer to be: $\frac{k}{\sqrt{k^2 - 1}}$, (this is a wrong answer too).

share|improve this question
1  
How do you know it's wrong? –  Gerry Myerson Mar 1 '13 at 12:02
    
Answer in the back of the book –  Assad Mar 1 '13 at 12:34
    
Yes, and what is the answer in the back of the book? –  Gerry Myerson Mar 1 '13 at 12:37
    
cscθ=-k/√(k^2−1) –  Assad Mar 1 '13 at 12:41
    
Ah. $\theta$ is obtuse, so $\sec\theta$ is negative, so $k$ is negative, but $\csc\theta$ is positive, so you need that $-k$. –  Gerry Myerson Mar 1 '13 at 23:32

1 Answer 1

up vote 3 down vote accepted

The first one is not difficult. By definition $$ \cos\theta=\frac{1}{\sec\theta}=\frac{1}{k}. $$

If by obtuse you mean $\pi / 2\leq \theta \leq \pi$ modulo $2\pi$ as usual, this implies $\cos\theta =k\leq 0$ and $\sin\theta \geq 0$.

Now $$ \sin^2\theta+\cos^2\theta=1\quad\Rightarrow\quad|\sin\theta|=\sqrt{1-\cos^2\theta}=\sqrt{1-1/k^2}=\frac{\sqrt{k^2-1}}{|k|}. $$ Since $\sin\theta\geq 0$, you get $$ \sin\theta = \frac{\sqrt{k^2-1}}{-k}\quad\Rightarrow\quad \csc\theta=-\frac{k}{\sqrt{k^2-1}}. $$

share|improve this answer
    
Thank you for that! –  Assad Mar 2 '13 at 19:05
    
@Assad You're welcome! –  1015 Mar 2 '13 at 19:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.