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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.

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    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 $\pi$.2012-07-07

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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.

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    Thank you! I found you're explanation easiest to follow.2012-07-07
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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 $\dfrac1{\csc^2(x)} + \dfrac1{\sec^2(x)} = 1 \implies \csc(x) = \pm\sqrt{\dfrac{\sec^2(x)}{\sec^2(x)-1}}$

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    This looks like something that teacher could have wanted...+12012-07-07