1
$\begingroup$

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.

$\endgroup$
5
  • 3
    $\begingroup$ It seems like you should rather contact your teacher and ask him why your answer is wrong, instead of questioning here. $\endgroup$
    – Listing
    Commented Jul 7, 2012 at 17:59
  • $\begingroup$ I am curious. What was the official answer? $\endgroup$ Commented Jul 7, 2012 at 18:02
  • $\begingroup$ It's an online course. The teacher hasn't been good about responding to questions. $\endgroup$
    – user18260
    Commented Jul 7, 2012 at 18:35
  • $\begingroup$ There was no official answer from the teacher $\endgroup$
    – user18260
    Commented Jul 7, 2012 at 18:36
  • $\begingroup$ 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 $\pi$. $\endgroup$ Commented Jul 7, 2012 at 20:52

2 Answers 2

6
$\begingroup$

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 $$\dfrac1{\csc^2(x)} + \dfrac1{\sec^2(x)} = 1 \implies \csc(x) = \pm\sqrt{\dfrac{\sec^2(x)}{\sec^2(x)-1}}$$

$\endgroup$
1
  • 1
    $\begingroup$ This looks like something that teacher could have wanted...+1 $\endgroup$
    – DonAntonio
    Commented Jul 7, 2012 at 18:02
5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you! I found you're explanation easiest to follow. $\endgroup$
    – user18260
    Commented Jul 7, 2012 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .