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I have this problem

$$\arccos\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$$

The answer comes out be $\arcsin(x)-\frac{\pi}{4}$

I've realized that this problem can be solved by using something called substitution, but i really dont get the idea of how you can just substitute $x$ with $\cos(x),~\sin(x)$. Or anything else for that matter.

Also how do you know what to substitute? Is there a method for that?

This has been confusing me a lot and i would appreciate if the answer is not just the solution but also an explanation to how substitution works in brief.

Thanks in advance.

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    $\begingroup$ I don't see a problem to be solved. $\endgroup$ – Camilo Andres Escobar Sarmient Jun 13 at 4:00
  • $\begingroup$ You don't? pretty sure its visible $\endgroup$ – MrKhonsu Jun 13 at 4:08
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    $\begingroup$ Look, that's just a function, you are not trying to solve an equation, you are not trying to find something special about this function, you are not asking for anything specific. Maybe try to state your question differently. $\endgroup$ – Camilo Andres Escobar Sarmient Jun 13 at 4:15
  • $\begingroup$ I've listed the answer. How do you even get there using this thing called substitution? $\endgroup$ – MrKhonsu Jun 13 at 4:26
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I assume you want to simplify the expression.


$\sqrt{1-x^2}$ is defined only when $|x|\leq 1$. Hence, if we let $x=\sin \alpha$, note that for every possible value of $x$ we can select a value of $\alpha$.

Let $t=\arccos\left( \dfrac{x+\sqrt{1-x^2}}{\sqrt 2}\right)$.

Now, $$t=\arccos \left(\dfrac{\sin \alpha + \cos \alpha}{\sqrt 2}\right)$$ Or $$t=\arccos \left(\cos (\alpha-\pi/4)\right)=\alpha -\pi/4$$ This gives us $t=\arcsin x -\pi/4$.

Note: the above is valid only for certain values of $\alpha$. I have left this for you as an exercise(the values for which it is valid).

Edit:

$ t = \begin{cases} \arcsin x-\pi/4, & 1\geq x\geq 1/\sqrt 2 \\ \pi/4 -\arcsin x, & 1/\sqrt 2\geq x\geq -1 \end{cases}$

Corresponding Desmos plot: enter image description here

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  • $\begingroup$ what if $$x=-1$$ $\endgroup$ – lab bhattacharjee Jun 13 at 9:46
  • $\begingroup$ @lab bhattacharjee As I said, only some values of $\alpha$: there isn't a value of $\alpha$ for which the above is satisfied for $x=-1$. $\endgroup$ – AryanSonwatikar Jun 13 at 11:54
  • $\begingroup$ $$\sin\left(-\dfrac\pi2\right)=?$$ $\endgroup$ – lab bhattacharjee Jun 13 at 12:06
  • $\begingroup$ Actually, for $1\geq x\geq 1/\sqrt 2$, the expression is equal to $\arcsin x -\pi/4$ and for $1/\sqrt 2\geq x\geq -1$ it is $\pi/4 -\arcsin x$. I'll add that to my answer along with a Desmos plot screenshot. $\endgroup$ – AryanSonwatikar Jun 13 at 13:08
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The solution has been discussed so I'll just try to address your other concerns.

The expression to be simplified contains certain suggestions to the substitution. The $x$ and $\sqrt{1-x^2}$ have a sum of squares of 1. This is reminiscent of either the $\sin x$ function or $ \cos x$ functions. So we try to substitute say, $x=\sin\alpha$.

With inverse trigonometric expressions however this gets a little tricky.

  1. You have to ensure that your substitution satisfies the domain. For example, here plugging $x=\sin\alpha$ would be invalid if $x$ could take all real values because the range of $\sin x$ is $[-1,1]$. We can do that here because $\sqrt{1-x^2}$ requires $\mid x\mid <1$ .

  2. $\mathrm{sin}\alpha$ is a many-one function. In fact for any $\alpha$ you take $n\pi +(-1)^n\alpha$ gives the same value. So if your simplified expression contains $\alpha$ you would get an infinite number of values for the resultant function . But the $\mathrm{arccos}$ function is single valued. So we have to restrict the domain of $\alpha$ to ensure that $\sin\alpha$ takes all values of $x$ and that each $x$ corresponds to a single $\alpha$.(Establish a bijection between $x$ and $\alpha$ so to speak)

  3. The easiest way to do this usually is to assume $\alpha =\sin^{-1}(x)$ this forces $\alpha\in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Another note other people missed on is that $\sqrt{1-sin^2\alpha}$ is $\mid\cos\alpha\mid$. It is only when you consider our restriction on $\alpha$ that you can justify that $ \cos\alpha$ is positive in $\in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Finally, for inverse trig functions, while: $$\mathrm{trig}(\mathrm{trig}^{-1}x)=x$$ Is true but: $$\mathrm{trig}^{-1}(\mathrm{trig}(x))=x$$ This is true only when $x$ lies in the Principal value branch of $\mathrm{trig}$(Denotes any of the six functions.

So, once you have simplified the function to: $\cos^{-1}(\cos(\alpha-\tfrac{\pi}{4}))$ You have to look at what your value of $\alpha$ is before you cancel cos inverse and cos.

Look up the graph for $\cos^{-1}(\cos x)$ and you'll notice that: $$\cos^{-1}(\cos x)= x ;0\leq x\leq \pi$$ $$\cos^{-1}(\cos x)= -x ;-\pi\leq x\leq 0$$

So,

$$\cos^{-1}(\cos (\alpha-\tfrac{\pi}{4}))= \alpha-\tfrac{\pi}{4};\tfrac{\pi}{4}\leq \alpha\leq \tfrac{5\pi}{4}$$ $$\cos^{-1}(\cos (\alpha-\tfrac{\pi}{4}))= \tfrac{\pi}{4}-\alpha ; -\tfrac{3\pi}{4}\pi\leq \alpha\leq \tfrac{\pi}{4}$$

Checking the appropriate ranges on $x$ corresponding to $\alpha$ gives $x\geq \tfrac{1}{\sqrt{2}}$ and $x\leq \tfrac{1}{\sqrt{2}}$ for the first and second case respectively.

Finally a note on choosing substitutions. You should try to find the substitution that has an identity most closely resembling the given expressions.

Examples:(Try to simplify the expressions and identify the corresponding identity.)

  1. $x, \sqrt{x^2-a^2}$ , use $x=a\sec\alpha$ or $\csc\alpha$.
  2. $x, \sqrt{a^2+x^2}$, use $x=a\tan\alpha$
  3. $\sqrt{1-x},\sqrt{1+x}$, use $x=a\cos2\alpha$
  4. $\sqrt{a-x}{x-b}, \sqrt{\dfrac{a-x}{x-b}}$ use $x=a\sin^2\theta +b\cos^2 \theta$. (Slightly esoteric, rare use)
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  • $\begingroup$ +1. In some places MathJax hasn't been correctly formatted. Please see to that :) $\endgroup$ – AryanSonwatikar Jun 14 at 1:57
  • $\begingroup$ Thank you for explaining the method as a whole $\endgroup$ – MrKhonsu Jun 14 at 2:36
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We need to be very careful about the ranges while dealing with Inverse Trigonometric Functions(Statement)

Let $\arccos x=u\implies0\le u\le\pi,$

$x=\cos u,\sin u=+\sqrt{1-x^2}$

$$f(x)=\arccos\left(\dfrac{\cos u+\sin u}{\sqrt2}\right)=\arccos\left(\cos\left(u-\dfrac\pi4\right)\right)$$

Now $-\dfrac\pi4\le u-\dfrac\pi4\le\pi-\dfrac\pi4$

So if $u-\dfrac\pi4\ge0\iff x=\cos u\le\cos\dfrac\pi4=?,$

$$f(x)=u-\dfrac\pi4$$

If $u-\dfrac\pi4<0\iff x=\cos u>\cos\dfrac\pi4=?,$

$$f(x)=-\left(u-\dfrac\pi4\right)$$

Now use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$

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