Here is the textbook solution to a simple trigonometry problem I just completed. The Exercise was to rewrite the LHS in terms of $\cot\theta$.

$$\sqrt{\frac{4}{\tan^6(\theta)}}=\frac{2}{\tan^3(\theta)}=2 \cdot \left( \frac{1}{\tan(\theta)}\right)^3 = 2 \cot^3(\theta)$$

My answer was $\pm 2 \cot^3\theta$. My question is, when is it useful/necessary to write the $\pm$? Do you think I would be penalised for this in an exam?

Thank you!

I wasn't sure what to tag this, so I've tagged it algebra. Any other suggestions?

  • 4
    $\begingroup$ With no further information on the range of $\theta$, the answer of your textbook is wrong. This should read $=2/|\tan^3(\theta)|$. So, I would not use $\pm$ here but rather the absolute sign. $\endgroup$ – Did Apr 13 '11 at 8:17

You're right: $\sqrt{\tan^2\theta}=|\tan\theta|=\pm\tan\theta$. The tangent can be positive or negative, so the (positive) square root of its square can be equal or opposite in sign. Unless there is reason to believe that the tangent is positive in this case, the solution you quote is incorrect. However, a better solution than $\pm2\cot^3\theta$, which leaves open how to choose the sign, is $2|\cot\theta|^3$.

  • $\begingroup$ Thank you! I'm not sure I understand the use of the modulus. Is that because $\tan\theta$ must be positive in order to take the square root? $\endgroup$ – Danny King Apr 13 '11 at 8:40
  • 2
    $\begingroup$ No, it need not. It's because if you square the tangent and then take the square root, you lose the sign, so regardless of whether the tangent is positive or negative, the square root of its square is positive. To make this more concrete, $\sqrt{5^2}=\sqrt{(-5)^2}=5=|5|=|-5|$. $\endgroup$ – joriki Apr 13 '11 at 8:45
  • $\begingroup$ Ah that clears it up, thanks! $\endgroup$ – Danny King Apr 13 '11 at 8:57

$\sqrt a$ always means the single non-negative number $u$ such that $u^2=a$. On the other hand, the equation $x^2=a$ has in general two solutions, $x=\pm \sqrt a$. So, $\sqrt 4=2$, but $x^2=4$ has two solutions, $\pm 2$.


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