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I am teaching myself math, and I have a question involving writing trigonometric expressions as algebraic expressions:

Write $\cos(\tan^{-1}(u))$ as an algebraic expression. The "correct" answer, according to the student solutions guide that I am using, is: $$\frac{1}{\sqrt{u^2 + 1}}.$$

The answer I came up with, although more messy is: $$\frac{1/u}{\sqrt{(1/u)^2 + 1}}.$$

To make sure my answer was aligned with the book's answer, I used 2 as a test case for $u$ which makes the answer $\sqrt{5}/5$ or $0.447213596$.

Does it matter what answer I choose, is there a grammar guideline I should follow when solving problems such as this and future mathematical identities?

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  • $\begingroup$ I am new to using this forum and I do not know how to properly display numbers in square roots so I used the notaction "sqrt()" to symbolize square rooted numbers or expressions. $\endgroup$ – CarrotPatch Feb 1 '17 at 20:37
  • $\begingroup$ Welcome to Stackexchange. Here is a quick explanation of how to typeset mathematical expressions on math stackexchange using MathJax. meta.math.stackexchange.com/questions/5020/… $\endgroup$ – John Wayland Bales Feb 1 '17 at 20:40
  • $\begingroup$ I have edited your question, please check that I have not butchered any of the expressions. By the way, you can click on the edit button to see how I formatted math. $\endgroup$ – C. Falcon Feb 1 '17 at 20:43
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    $\begingroup$ What happens when you use $u=-2$? $\endgroup$ – John Wayland Bales Feb 1 '17 at 21:18
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The following diagram using the unit circle $x^2+y^2=1$ explains why the preferred expression is

$$\cos(\arctan(u))=\frac{1}{\sqrt{1+u^2}}$$

and the reason the result is always non-negative even when $u$ is negative.

Recall that for all $u$

$$ -\frac{\pi}{2}<\arctan u <\frac{\pi}{2}$$

and that therefore

$$ 0<\cos(\arctan u)<1 $$ cos(arctan u)

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  • $\begingroup$ Thank you for the explanation. I have a difficult time interpreting math and your answer though initially intimidating, provided a clear understanding of my mistake. Thank you for taking the time to demonstrate your solution using the unit circle. $\endgroup$ – CarrotPatch Feb 1 '17 at 22:23
  • $\begingroup$ You are welcome. Sometimes a picture really is worth a thousand words. $\endgroup$ – John Wayland Bales Feb 1 '17 at 22:24
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Usually the simpler expression is preferred.

Your expression, $\dfrac{1/u}{\sqrt{(1/u)^2 + 1}} $ is the same as your book's, as can be seen by multiplying both the numerator and denominator by $u$:

$(1/u)u = 1$ and $\sqrt{(1/u)^2 + 1}u =\sqrt{((1/u)^2 + 1)u^2} =\sqrt{1+u^2} $.

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  • $\begingroup$ I would like to thank you for the immediate answer. You are spot on with your edit. Your answer leads me to another question. Can you explain why I can multiply by 'u'? I understand that you did so to make it match the book's answer, but what if I did not have a predetermined answer to compare. Are there mathematical rules that allow such? $\endgroup$ – CarrotPatch Feb 1 '17 at 20:53
  • $\begingroup$ @CarrotPatch as a general technique, multiply your fractions by whatever necessary to eliminate fractions inside fractions. $\endgroup$ – Simply Beautiful Art Feb 1 '17 at 21:00
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    $\begingroup$ Actually, the two expressions are equivalent only for $u\ge0$ since $\cos(arctan u)$\ge0$. $\endgroup$ – John Wayland Bales Feb 1 '17 at 21:03
  • $\begingroup$ You can always multiply by one, and$\frac{u}{u} = 1$ if $u \ne 0$. $\endgroup$ – marty cohen Feb 1 '17 at 21:05
  • $\begingroup$ @martycohen The point is that OP's result gives a negative value when $u$ is negative and $\cos(\arctan u)$ is never negative. $\endgroup$ – John Wayland Bales Feb 1 '17 at 21:10

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