# B How would I write this trig solution?

#### opus

Gold Member
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

Homework Helper
Gold Member
Perhaps because root 2 = 1.414 and -1.414 ?

#### opus

Gold Member
Ah so then are you saying:

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

#### Math_QED

Homework Helper
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

Mentor
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

Mentor
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

Mentor
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.

Homework Helper
Gold Member
2018 Award
Since $\sin^2(x)=1-\cos^2(x)$, you could write $\cos^2(x)=1-\cos^2(x)$, and solve that algebraically.

#### robphy

Homework Helper
Gold Member
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

Gold Member
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

Gold Member
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 cant 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.

#### Math_QED

Homework Helper
This would be $\frac{π}{4}$ and $\frac{5π}{4}$ all +nπ, but I still cant 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 $\tan x = -1$?

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
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: $\sqrt{x^2}=\left| x \right|$.
$x= \pm \left| x \right| = \pm \sqrt{x^2}$

#### Math_QED

Homework Helper
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

Homework Helper
Gold Member
$a^2 = b^2 \iff a^2 - b^2 = 0 \iff (a-b)(a+b) = 0 \iff a - b = 0 \lor a + b = 0 \iff a = b \lor a = -b$
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

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

#### Mark44

Mentor
And that V means "or", is this correct?
Yes.
This one, $\wedge$, means "and".

#### opus

Gold Member
Ok thanks all!

"How would I write this trig solution?"

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