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My friends say, it is some what difficult to know, which trigonometric function has to be substituted in the inverse trigonometric equations, to get the correct solution. So, I thought to take up this issue.

Consider the below equation, which has to be reduced to it's simplest form.

$$\begin{equation}\arctan\frac{\sqrt{1+x^2}-1}{x},x\neq0\end{equation}$$
We can see that, $x$ can take any value other than $0$. So, $x$ can be replaced with any trig function, which gives all the values fetched by $x$. We have six trig functions, which one should we substitute? Should we try with all six trig functions one by one? We can make some thought process to answer the above questions.

  • $\cos\theta$ and $\sin\theta$ values are restricted to the interval $[-1,1 ]$, but $x$ can have any value other than $0$, which can be greater than $1$ or it can be lesser than $-1$. So, cos and sine functions can't be substituted for $x$.

  • $cosec\theta$ and $\sec\theta$ vaues are restricted to the interval $R-(-1,1)$, but $x$ can have values in the open interval $(-1,1)$ except $0$. So, cosec and sec functions can't be substituted for $x$.

  • $\cot\theta$ and $\tan\theta$ values cover the entire real numbers. So, cot and tan are the only two functions which can be substituted in the equation for $x$.

We can try for other problems, we get best results . I thought this is the better place to discuss about this, so that any one can leave their ideas/techiniques of how they solved problems using substitution.

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  • $\begingroup$ en.wikipedia.org/wiki/Trigonometric_substitution $\endgroup$ – lab bhattacharjee Nov 6 '13 at 10:55
  • $\begingroup$ Why on earth is $x$ not allowed to be $1$? Do you mean $0$? Even if that were the case, the value of the argument of the arctan would still be zero in the limit as $x \to 0$. I guess without further context, I do not understand what you are asking. $\endgroup$ – Ron Gordon Dec 1 '13 at 13:12
  • $\begingroup$ Thank you for the comment. I had typed wrong in hurry, I have corrected it. $\endgroup$ – Immortal Player Dec 1 '13 at 13:34
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Generally, trig substitution is done when you have a $\sqrt{\pm a^2 \pm x^2}, a > 0$, and you want to remove the square root by exploiting trig identities. If you have a $-$, then $x = a \sin t$ (or $a \cos t$) works best: $$\sqrt{a^2 - x^2} = \sqrt{a^2 - \left(a \sin t \right)^2} = \sqrt{a^2 - a^2 \sin^2 t} = a \sqrt{1 - \sin^2 t} = a |\cos t|$$

If you have a $+$, use tangent (or cotangent): $$\sqrt{a^2 + x^2} = \sqrt{a^2 + \left(a \tan t \right)^2} = \sqrt{a^2 + a^2 \tan^2 t} = a \sqrt{1 + \tan^2 t} = a |\sec t|$$

If you have $\sqrt{x^2 - a^2}$, which seems to be much rarer, use secant (or cosecant): $$\sqrt{x^2 - a^2} = \sqrt{ \left( a \sec t \right)^2 - a^2} = \sqrt{a^2 \sec^2 t - a^2} = a \sqrt{\sec^2 t - 1} = a |\tan t|$$

If you have $\sqrt{- x^2 - a^2}$... that's odd.

If you use them in the 'wrong' situation, you'll get expressions like $a\sqrt{1 + \sin^2 t}$, which aren't particularly helpful in most cases.


Note that the ranges match up correctly. In the first case, $|x| \le a$, and indeed, $|a \sin t | \le a$. In the second, $x$ can take any range, which is consistent with the tangent and cotangent. In the third, $|x| \ge a$, and $|a \sec t| \ge a$, so that checks out too.

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  • $\begingroup$ Thank you for sharing your ideas. I have learnt some substitutions from your answer. $\endgroup$ – Immortal Player Dec 2 '13 at 15:07
  • $\begingroup$ I know it is difficult to fully address this question but although the points mentioned in this answer are golden to remember, the intuition behind the "why?" of this is not trivial to the beginner. $\endgroup$ – Nick Mar 12 '15 at 15:18

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