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.

I was working on trig homework and came across a question that I didn't understand how to even begin to approach. It asked us to use trigonometric identities to write $\csc(x)$ in terms of $\sec(x)$.

I'm not sure what I can do. I had though the cofunction identity $\sec \left(\frac{\pi}{2} - x \right) = \csc x$ was the answer, but I was told I was wrong, with no explanation. This identity was on a handout provided by the teacher.

How would I go about solving this problem and problems like it? Also, why was my initial answer wrong? Any help would be greatly appreciated.

share|improve this question
3  
It seems like you should rather contact your teacher and ask him why your answer is wrong, instead of questioning here. –  Listing Jul 7 '12 at 17:59
    
I am curious. What was the official answer? –  André Nicolas Jul 7 '12 at 18:02
    
It's an online course. The teacher hasn't been good about responding to questions. –  user18260 Jul 7 '12 at 18:35
    
There was no official answer from the teacher –  user18260 Jul 7 '12 at 18:36
    
It's just that the answer should be a little different depending on the location of $x$, for example $\frac{\sec x}{\sqrt{\sec^2 x-1}}$ between $0$ and $\pi/2$, $-\frac{\sec x}{\sqrt{\sec^2 x-1}}$ between $\pi/2$ and $\pi$. –  André Nicolas Jul 7 '12 at 20:52
add comment

2 Answers

up vote 4 down vote accepted

It’s wrong because you didn’t write $\csc x$ in terms of $\sec x$: $\frac{\pi}2-x$ isn’t $x$. I suspect that you were intended to do something like this:

$$\csc x=\frac1{\sin x}\;,$$ and $$\sec x=\frac1{\cos x}\;,$$ so

$$\begin{align*}&\frac1{\csc^2 x}+\frac1{\sec^2x}=1\;,\\ &\frac1{\csc^2x}=1-\frac1{\sec^2x}=\frac{\sec^2x-1}{\sec^2x}\;,\\ &\csc^2x=\frac{\sec^2x}{\sec^2x-1}\;,\\ &\csc x=\pm\frac{\sec x}{\sqrt{\sec^2x-1}}\;. \end{align*}$$

For a complete answer you still have to sort out where $\csc x$ is positive and where it’s negative, which is just a bit messy.

share|improve this answer
    
Thank you! I found you're explanation easiest to follow. –  user18260 Jul 7 '12 at 20:00
add comment

Probably your teacher wanted a relation between $\csc(x)$ and $\sec(x)$ and not $\sec \left(\frac{\pi}2 - x \right) = \csc(x)$. If you want a relation between $\csc(x)$ and $\sec(x)$, then recall that $$\sin^2(x) + \cos^2(x) = 1$$ This gives us $$\dfrac1{\csc^2(x)} + \dfrac1{\sec^2(x)} = 1 \implies \csc(x) = \pm\sqrt{\dfrac{\sec^2(x)}{\sec^2(x)-1}}$$

share|improve this answer
1  
This looks like something that teacher could have wanted...+1 –  DonAntonio Jul 7 '12 at 18:02
add comment

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.