4

if $a$ is are real number that $a \neq 0$, and $\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$, $x$ is on which trigonometric quadrants?

Things I have done so far: this problem is mostly different from that I previously solved.My Idea was to powering up both sides,take all to one side and then solve it like equation.which I was not successful. any starting hint would be appreciated as I don't looking for full solution.

UPDATE

thanks to ganeshie8 hints,i reached this right now$$\cos^2a=1+\sin a\times \cos a$$

And I stuck here.

user2838619
  • 3,120
  • squaring and trying to solve $x$ is a good start ! you might have figured out already : $\cos x$ is positive only in $I$ and $IV$ quadrants – AgentS Aug 21 '14 at 19:17
  • @ganeshie8,thanks for your hint.i think there is a problem.this question is originally from a test.so answer could be just one quadrant.but both $I$ and $IV$ are possible because both of them result positive ${\frac{\cot x}{\cot x -a^2}}$ so $\cos x$ would be positive .do i missing something? – user2838619 Aug 21 '14 at 19:38
  • squaring gives you an equation which can be factored easily, solving it should eliminate one of the quadrants – AgentS Aug 21 '14 at 19:40
  • you should end up with $a^2\sin x+\cos x = 0 \implies \cot x = -a^2$ which eliminates quadrant $I$ since $\cot x$ is not negative in quadrant $I$ – AgentS Aug 21 '14 at 19:43
  • small correction : $a^2\cos x + \sin x = 0 \implies \tan x = -a^2 $ – AgentS Aug 21 '14 at 19:52
  • @ganeshie8,thanks,i tried what you said but i stuck.could please you look at the post(i updated it)? – user2838619 Aug 21 '14 at 19:56
  • looks there is a minor mistake in your equation, but the moment you see the form $\cos^2x = 1 + \text{something}$, you should try to replace $1-\cos^2x$ by $\sin^2x$ – AgentS Aug 21 '14 at 19:58
  • 1
    $\cos^2x = \dfrac{\cot x}{\cot x - a^2} \implies \cos x (\cot x - a^2) = 1/\sin x \implies \cos x(\cos x - a^2\sin x) = 1 \implies a^2\sin x \cos x = \cos^2 x - 1 \implies a^2 \cos x = -\sin x \cdots $ – AgentS Aug 21 '14 at 20:00
  • 1
    The first quadrant is not possible. No algebraic manipulation is needed. For if $\cot x\gt a^2$, then $\frac{\cot x}{\cot x-a^2}\gt 1$, so the square root is greater than $1$, impossible for a cosine. And if $a^2\gt \cot x$, we end up taking the square root of a negative number. – André Nicolas Aug 21 '14 at 20:10
  • @AndréNicolas you mean it is impossible for forth quadrant?well $\cot x$ is also negative, so the hole fraction will be positive.did i missed any of your points? – user2838619 Aug 21 '14 at 20:23
  • 1
    The fourth quadrant is not ruled out. For in the fourth quadrant, $\cot x\le 0$, and therefore $0\le \frac{\cot x}{\cot x-a^2}\lt 1$. – André Nicolas Aug 21 '14 at 20:29

2 Answers2

1

The equation has the first quadrant solution $x=\frac{\pi}{2}$. Apart from that, the first quadrant is not possible. Note that if $x\ne \frac{\pi}{2}$, and $x$ is in the first quadrant, and $\cot x$ is defined, then $\cot x\gt 0$.

Thus if $\cot x\gt a^2$, then $0\lt \cot x-a^2\lt \cot x$, and therefore $\frac{\cot x}{\cot x-a^2}\gt 1$. So our square root is greater than $1$, which is impossible for a cosine.

And if $a^2\gt \cot x$, we are taking the square root of a negative number.

André Nicolas
  • 507,029
0

We are given: $$\cos x = \sqrt{\frac{\cot x}{\cot x - a^2}}$$

For this to be true,

$$\sqrt{\frac{\cot x}{\cot x - a^2}} \in [-1,1]$$ Now, for any$\sqrt{\frac{a}{b}} \in \mathbb R, \quad a>0 ,\quad b>0$

For the first condition ($a>0$), $$ \implies \cot x > 0 \implies \frac{\cos x}{\sin x} >0\\ \implies (2n-1)\frac{\pi}{2}<x<(2n+1)\frac{\pi}{2} \text{ and } x \ne 0 \quad \forall \space n \in \mathbb Z$$ This means that $x$ is in Quadrants I or IV.

For the second condition ($b>0$), $$\implies \cot x - a^2 > 0 $$

We can now clearly see that $$\cot x > \cot x - a^2 > 0\\ \implies \text{Numerator} > \text{Denominator}\\ \implies \frac{\cot x}{\cot x - a^2} > 1$$

But this is against our initial assumption.

The only solution is $\frac{\pi}{2} \space\implies x \in $ Ist Quadrant

Nick
  • 6,804
  • this problem driven me crazy.$\frac{- \pi}{2}$ could be a solution also.So $\sin(\frac{- \pi}{2})=sin(\frac{3 \pi}{2})$.So forth is acceptable. – user2838619 Aug 21 '14 at 20:56