# Solve $\cos^2x-\sin^2x= 1$

I am trying to solve for $x$ in the following equation:
$$\cos^2x-\sin^2x= 1$$

Given that $\cos^2x+\sin^2x= 1$, is this something I could use to solve it?

• From your other questions I think you are perfectly capable of figuring out this one by yourself.
– mez
Mar 13, 2013 at 20:27
• Mar 13, 2013 at 21:16
• Here we go, 10 answers! Mar 14, 2013 at 13:33
• This question get over 1k views within 3 days, incredible!
– JSCB
Mar 17, 2013 at 9:30

Observe that $\cos^2 x$ never gets bigger than 1. So if there is a solution, it must have $\cos^2 x = 1$ and $\sin^2 x = 0$. Does that help?

• This is the easiest one, not as general as the others, but it solves the problem at hand. Mar 13, 2013 at 16:38
• On the down side, it depends on "observe that…", and sometimes these solutions that begin with "observe that …" are not so helpful. What if you didn't notice? On the other hand, learning to notice these kinds of things is important too. I upvoted Cameron's answer, since it's the only one that makes use of OP's suggestion.
– MJD
Mar 13, 2013 at 16:40

Use the identity $\cos 2x =\cos^2x-\sin^2x$.

You could indeed use that, since then $$\cos^2x-\sin^2x=\cos^2x+\sin^2x.$$ Gather all your trig terms on one side, and the rest is almost trivial.

• from which you get that $2*sin^2(x)=0$ or sin(x)=0,put into first equation you get cos(x)=1 or cos(x)=-1 Mar 13, 2013 at 16:33
• @dato: What? You do, indeed, get $\sin(x)=0$ from the zero product property, but why would you then put that back into the first equation? All that remains is to say what form $x$ must have as a zero of the sine function. Mar 13, 2013 at 16:35
• if sin(x)=0,then what is equal to cos(x) from this equation?i meant this Mar 13, 2013 at 16:55
• @dato: You have correctly concluded that $\cos(x)=\pm1$, but that's extraneous information. We wish to solve for $x$, and knowing that $\sin(x)=0$ tells us all we need to know. Mar 13, 2013 at 17:00

Given

$$\cos^2x-\sin^2x= 1\tag1$$

Known

$$\cos^2x+\sin^2x= 1\tag2$$

$(1)\quad+\quad(2)$

$$\Rightarrow 2\cos^2x= 2$$ $$\Rightarrow \cos^2x= 1$$ $$\Rightarrow \cos x= \pm1$$ $$x = n\pi$$

Subtract two equation you get $2(\sin^2x)=0 \implies \sin^2x=0 \implies \sin x=0 \implies x = \pi\cdot n, n\in \mathtt{Z}$.

• $2\sin(2x) = 0$? Don't you mean $2\sin^2(x) = 0$? Mar 13, 2013 at 19:51
• @DilipSarwate: I assume that is what Civa meant.
– robjohn
Mar 13, 2013 at 21:27
• @robjohn I was hoping that Civa would read my comment and make the correction himself/herself. Mar 13, 2013 at 22:56
• You have a typo in the rhs: it should be $\sin x=0$. And maybe you could even use $\Leftrightarrow$ instead of $\Rightarrow$. Mar 14, 2013 at 13:34

I set up the problem like this:

$\cos^{2}(x) - \sin^2(x) = 1 \rightarrow$ I used the half angle rule,

$\frac{1}{2}\cdot(1+\cos(2x)) - \frac{1}{2}(1- \cos(2x)) = 1$

$\frac{1}{2}\cdot((1 + (\cos(2x) - 1 + \cos(2x)) = 1$

$(2\cdot \cos(2x)) = 2$

$\cos(2x) = 1$

$x = 0$ , $\pi$, $2\pi$, and so on . . .  $x =\pi n$.

• Just using the double-angle formula for the cosine will get you the same result with a lot less work. Mar 13, 2013 at 23:16

If you don't know the identity $\cos{2x} =\cos^2x-\sin^2x$, you can often solve the problem by expressing $\sin$ and $\cos$ through the exponential function:

$$\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2 - \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2 = 1$$

$$\frac{e^{2ix}+ 2e^{-2ix}+e^{-2ix}}{4} + \frac{e^{2ix} - 2e^{-2ix}+e^{-2ix}}{4} =1$$

$$\frac{e^{2ix}+e^{-2ix}}{2} = 1$$

$$\cos(2x) = 1$$

$$\cos^2x-\sin^2x= 1$$
$$1-\sin^2x-\sin^2x= 1$$
$$-2\sin^2x=0\iff\sin^2x=0\iff\sin x=0\iff x=k\pi$$

Maximum value of $\cos^2(x)$ = $1$

Minimum value of $\sin^2(x)$ = $0$

So this is the only case where you get $\cos^2(x) - \sin^2(x) = 1$.

Hence $\cos^2(x) = 1$ and $\sin^2(x) = 0$ => $x = nπ$

Or you could have used the formula : $\cos^2(x) - \sin^2(x) = \cos(2x)$

Hope the answer is clear !