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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?

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    $\begingroup$ From your other questions I think you are perfectly capable of figuring out this one by yourself. $\endgroup$ – mez Mar 13 '13 at 20:27
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    $\begingroup$ For interest: wolframalpha.com/input/?i=cos%5E2x%E2%88%92sin%5E2x%3D1 $\endgroup$ – SimonJGreen Mar 13 '13 at 21:16
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    $\begingroup$ Here we go, 10 answers! $\endgroup$ – Julien Mar 14 '13 at 13:33
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    $\begingroup$ This question get over 1k views within 3 days, incredible! $\endgroup$ – A. Chu Mar 17 '13 at 9:30
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Given

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

Known

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

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

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

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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?

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    $\begingroup$ This is the easiest one, not as general as the others, but it solves the problem at hand. $\endgroup$ – Ross Millikan Mar 13 '13 at 16:38
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    $\begingroup$ 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. $\endgroup$ – MJD Mar 13 '13 at 16:40
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Use the identity $\cos 2x =\cos^2x-\sin^2x$.

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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.

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  • $\begingroup$ 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 $\endgroup$ – dato datuashvili Mar 13 '13 at 16:33
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    $\begingroup$ @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. $\endgroup$ – Cameron Buie Mar 13 '13 at 16:35
  • $\begingroup$ if sin(x)=0,then what is equal to cos(x) from this equation?i meant this $\endgroup$ – dato datuashvili Mar 13 '13 at 16:55
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    $\begingroup$ @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. $\endgroup$ – Cameron Buie Mar 13 '13 at 17:00
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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} $.

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

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    $\begingroup$ Just using the double-angle formula for the cosine will get you the same result with a lot less work. $\endgroup$ – Michael Hardy Mar 13 '13 at 23:16
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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:

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

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

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

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

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

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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 !

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