How would I write this trig solution?

[opus]

If I want to find the point where the graphs of $y=cos^2(x)$ and $g=sin^2(x)$ intersect, I would set the equations equal to each other and solve for the $x$. This solution is $x=sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{π}{4}+2nπ,\frac{3π}{4}+2nπ$.
However these the graphs do intersect at point's not in this listed solution, say $\frac{-π}{4}$ and this is because we are squaring the values. So where has my logic gone wrong so that I can list all solutions of $y=cos^2(x)$ and $g=sin^2(x)$?

 

[CWatters]

Perhaps because root 2 = 1.414 and -1.414 ?

 

[opus]

Ah so then are you saying:

$sin^2(x)=\frac{1}{2}$
$sin(x)=\frac{+-\sqrt2}{2}$? Noting the plus or minus

 

[member 587159]

Maybe you made a mistake when solving?

$\cos^2 x = \sin^2 x \iff \cos x = \sin x \lor \cos x = - \sin x$

and the last two equations are standard to solve. Maybe you didn't include one of those 2?

 

[DrClaude]

Perhaps because root 2 = 1.414 and -1.414 ?

That's not correct. $\sqrt{2} \approx 1.414$, never with a minus sign.

$sin^2(x)=\frac{1}{2}$
$sin(x)=\frac{+-\sqrt2}{2}$? Noting the plus or minus

 

[Dullard]

Your logic is fine. n = -1 will produce your 'missing' value.

 

[Mark44]

Perhaps because root 2 = 1.414 and -1.414 ?

Although it's true that 2 has two square roots, the symbol $\sqrt{2}$ is by definition the principal, or positive, square root. As already noted, it is not correct to say that $\sqrt{2} = \pm 1.414...$ This is a mistake that crops up regularly here at this site.

 

[Mark44]

If I want to find the point where the graphs of $y=cos^2(x)$ and $g=sin^2(x)$ intersect, I would set the equations equal to each other and solve for the $x$.

Technically, you're not "setting equations equal to each other," since that makes no sense mathematically. For example, if 2x + 3y= 5 and x + 2y = 3, then "setting the equations equal" would look something like this: 2x + 3y = 5 = x + 2y = 3, and this makes no sense.

For your problem, if one equation is $y = cos^2(x)$ and the other is $y = sin^2(x)$, then when the two functions intersect, the point $(x_i, y_i)$ will be on both graphs. In this problem you're setting the y values equal, which means that $\cos^2(x) = \sin^2(x)$ at all points of intersection.

 

[Charles Link]

Since $\sin^2(x)=1-\cos^2(x)$, you could write $\cos^2(x)=1-\cos^2(x)$, and solve that algebraically.

 

[robphy]

Since $\sin^2(x)=1-\cos^2(x)$, you could write $\cos^2(x)=1-\cos^2(x)$, and solve that algebraically.

Similarly, since $\sin^2(x)=\cos^2(x)$, we can write $\frac{\sin^2(x)}{\cos^2(x)}=1$ as long as $\cos^2(x)\neq 0$.
Thus, solve $\tan^2(x)=1$ when $\cos^2(x)\neq 0$.

 

[opus]

Since $\sin^2(x)=1-\cos^2(x)$, you could write $\cos^2(x)=1-\cos^2(x)$, and solve that algebraically.

This is what I did but instead for sin rather than cos. I would run into the same problem though, since I'd be taking the root of a square.

 

[opus]

Similarly, since $\sin^2(x)=\cos^2(x)$, we can write $\frac{\sin^2(x)}{\cos^2(x)}=1$ as long as $\cos^2(x)\neq 0$.
Thus, solve $\tan^2(x)=1$ when $\cos^2(x)\neq 0$.

This would be $\frac{π}{4}$ and $\frac{5π}{4}$ all +nπ, but I still can't get $\frac{3π}{4}$ with this considering the positive nature of the principal square root because $sin^2$ and $cos^2$ intersect at $\frac{3π}{4}$ but tan is -1 there.

 

[member 587159]

This would be $\frac{π}{4}$ and $\frac{5π}{4}$ all +nπ, but I still can't get $\frac{3π}{4}$ with this considering the positive nature of the principal square root because $sin^2$ and $cos^2$ intersect at $\frac{3π}{4}$ but tan is -1 there.

$\tan^2 x = 1 \iff \tan x = 1 \lor \tan x = -1$. Did you miss the $\tan x = -1$?

 

[Charles Link]

This is what I did but instead for sin rather than cos. I would run into the same problem though, since I'd be taking the root of a square.

You take the square root of the square and you get a plus or minus. That is not at all problematic. Problem solved ! $\\$ Note: The square root sign seems to mean by convention that you take the positive square root. That's why in solving something like $x^2=1$, you say $x=\pm \sqrt{x^2} (= \pm \sqrt{1} )$.

 

[robphy]

Note: The square root sign seems to mean by convention that you take the positive square root. That's why in solving something like $x^2=1$, you say $x=\pm \sqrt{x^2} (= \pm \sqrt{1} )$.

When I learned of this in college, it was described this way: $\sqrt{x^2}=\left| x \right|$.
Then, your expression follows:
$x= \pm \left| x \right| = \pm \sqrt{x^2}$

 

[member 587159]

I wouldn't think about $a^2 = b^2 \iff a = b \lor a = -b$ using square roots. It obscures things, and introduces technicalities as squaring isn't injective on the entire real numbers.

Rather, think about it in the following way. No nasty square roots needed.

$a^2 = b^2 \iff a^2 - b^2 = 0 \iff (a-b)(a+b) = 0 \iff a - b = 0 \lor a + b = 0 \iff a = b \lor a = -b$

 

[robphy]

$a^2 = b^2 \iff a^2 - b^2 = 0 \iff (a-b)(a+b) = 0 \iff a - b = 0 \lor a + b = 0 \iff a = b \lor a = -b$

I like linear chains-of-reasoning, as you've done above.
Recently, I've noticed that some of my students can't parse expressions [especially when they don't interpret "relations" but rather "sequences of (calculator-)operations"].
I've found it helpful to use extra white-space or extra-levels of parentheses to help the novice parse the sentences.
$a^2 = b^2\quad \iff\quad a^2 - b^2 = 0 \quad \iff \quad (a-b)(a+b) = 0\quad \iff \quad (a - b = 0) \lor (a + b = 0) \quad \iff\quad (a = b) \lor (a = -b)$

 

[opus]

This has gotten deep and interesting lol. Thanks everyone. And that V means "or", is this correct?

 

[Mark44]

And that V means "or", is this correct?

Yes.
This one, $\wedge$, means "and".

 

[opus]

Ok thanks all!