## anonymous one year ago Is the following equation fully simplified? Equation Below:

1. anonymous

$f(x)= \sqrt{(\sin^2(x)-2)^2}-\sqrt{(\sin^2(x)+1)^2}$

2. anonymous

hmmm what do you think?

3. anonymous

I think it is because i can't just cancel the radical and the square (because of the possiblity that the inside is negative

4. anonymous

I was puzzled because of something like this: $\sqrt{(-2)^2}\neq-2$

5. anonymous

well... hmm kinda, yes but one could say $$\sqrt{(-2)^2}\to \begin{cases} -2\\ +2 \end{cases}$$

6. anonymous

if you use the +/- part but, for an expression, it depends on context in this expression, I'd get rid of the radical and exponent though and yes you're correct, the radical IS a constraint on the function but the question is, can it be simplified further and I think it can, you can have a simplified version, keeping in mind the constraints

7. anonymous

but if i complete that simplification, i get $f(x)=-3$ which fails for all x

8. anonymous

hmm

9. anonymous

let's try one value $$\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2}$$ what would that give?

10. anonymous

-1

11. anonymous

hmm -1?

12. anonymous

ahh, i believe I've solved it. Since the first radicand will always be negative before it is squared, I will simplify take its absolute value, by flipping the terms, to get $\sqrt{(\sin^2(x)-2)^2}=\sqrt{(2-\sin^2(x))^2}=2-\sin^2(x)$ from here, I can simplify the second radicand easily because it is always positive and thus is simply $\sqrt{(\sin^2(x)+1)^2}=\sin^2(x)+1$ putting it all together, i get $f(x)=2-\sin^2(x)-(\sin^2(x)+1)=1-2\sin^2(x)$ which equals, finally, $\cos(2x)$ does that work?

13. anonymous

hmmm the first radicand, actually both, are raised to the 2nd power, and thus, even if you get a negative value from the sine addition, the exponent will make it positive anyway -* - = +

14. anonymous

but with respect to eliminating the radical entirely, doesn't the flipping work?

15. anonymous

ahemm nope 2-x $$\ne$$ x-2

16. anonymous

hmm actually... shoot got a mistake there.....

17. anonymous

wheres the mistake?

18. anonymous

hmm well. my -3 is... mistaken for one

19. anonymous

hmmm

20. anonymous

either way.. .you can't simply flip the terms in the radicand though, unless you multiply them by a -1

21. anonymous

Look at this, attempted with x=pi/2$\sqrt{(\sin^2(\pi/2)-2)^2}=1$ but also,$\sqrt{(2-\sin^2(\pi/2))^2}=1$ So if i rewrite it in the second way, I can remove the radical.

22. anonymous

hmmm somehow... I do see the difference, expression wise and yes, the 2nd power exponent, will make, whatever value inside, negative or not, positive

23. anonymous

$$\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2} \\ \quad \\ f\left( \frac{\pi }{2} \right)=\sqrt{[1^2-2]^2}-\sqrt{[1^2+1]^2}\implies f\left( \frac{\pi }{2} \right)=\sqrt{[-1]^2}-\sqrt{[2]^2} \\ \quad \\ f\left( \frac{\pi }{2} \right)=-1-2\implies f\left( \frac{\pi }{2} \right)=-3 \\ \quad \\ f(0)=\sqrt{[sin^2(0)-2]^2}-\sqrt{[sin^2(0)+1]^2} \\ \quad \\ f(0)=\sqrt{[0-2]^2}-\sqrt{[0+1]^2}\implies f(0)=-2-1\implies -3$$ so.... I"d think the -3 kinda checks out though

24. anonymous

but yes, one could say, it CAN be simplified further by getting rid of the radical and exponent BUT the simplified version will still maintain the constraints of the original one

25. anonymous

and yes, I'm aware that $$\bf \sqrt{(-2)^2}\ne -2\qquad or\qquad \sqrt{(-1)^2}\ne -1$$ but all that, depends on the context at hand

26. anonymous

I mean... the same can be said... say. of an inverse trigonometric function their range is $$\textit{Inverse Trigonometric Identities} \\ \quad \\ \begin{array}{cccl} Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\ \hline\\ {\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2} \\ \quad \\ {\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi \\ \quad \\ {\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2} \end{array}$$ but depending on context, you can take the arcSine or arcCosine of a value and get just a "reference angle", and use it on 2nd, 3rd, or 4th quadrants even though the inverse function, may not extend to it

27. anonymous

Wow, i thought there was just a solid rule explaining $\sqrt{(-2)^2}$ I learned that a radical can never be negative no matter what the conditions inside it, so i don't know, I might just put cos(2x), because it checks out on all my tries both positive and negative

28. anonymous

hehe bear in mind that notations and formulas, are only phenomena representations some results in math are called "math fallacies" or "extraneous" because the procedure is correct, bu the answer is ODD so, they're correct mathematically, but wrong logically, and the logic throws them out in this case, is an extraneous case is correct mathematically, logicaly it makes no much sense, you're correct thus is "extraneous" but depending on the context they're used, extraneous results can be valid or not in this case, we don't have an explicit context or phenomena for this expression thus the extraneous result is ok

29. anonymous

ok, thank you