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(i) Show that, if $\tan^2(x) = 2\tan(x) + 1,$ then $\tan (2x) = -1$

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    $\begingroup$ And if I don't show that, what happens? ;) (kidding) Seriously, though--we do like it when you show your work for a problem, without just copy/pasting a question from some homework assignment. $\endgroup$ – apnorton Oct 15 '13 at 0:40
  • $\begingroup$ okay so I started working backwards starting with tan2(x) and trying to find an identity which would prove it was equal to -1 $\endgroup$ – user100893 Oct 15 '13 at 0:50
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It's always the case that $\tan2x=2\tan x/(1-\tan^2x)$. If $\tan^2x =2\tan x+1$, then $2\tan x=\tan^2x-1$ and hence

$$\tan2x={2\tan x\over 1-\tan^2x}={\tan^2x-1\over1-\tan^2x}=-1$$

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The desired result suggests playing with the double-angle formulas: $$\sin(2x) = 2\sin(x)\cos(x)$$ $$\cos(2x) = 2\cos^2(x) - 1$$

Also, since you're given a square tangent, you may as well try using a pythagorean identity $\tan^2 + 1 = \sec^2$.

Now, $$\tan^2(x) = 2\tan(x) + 1 = \sec^2(x) - 1$$ implies $$\sec^2(x) = 2\tan(x) + 2 = 2(\tan(x) + 1)$$ That is, $$1 = 2(\tan(x) + 1)\cos^2(x) = 2\sin(x)\cos(x) + 2\cos^2(x)$$ Pull the one over to the right side. Do you see how to finish it off?

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  • $\begingroup$ Really sorry, I follow this up until the last line then I'm not sure what you've done. Are you using cos(2x) = 2cos^2(x) - cos(2x) then using (tan(x) + 1) as a substitute for the cos(2x)? $\endgroup$ – user100893 Oct 15 '13 at 1:23
  • $\begingroup$ @user100893 No, I'm cancelling the $\cos(x)$ in $\tan(x)$. $\endgroup$ – Neal Oct 15 '13 at 2:37
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Hint:

$$\tan x={\sin x\over \cos x}$$

Hint 2:

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

Putting these plus one more together, we have

$$\tan^2(x)=2\tan(x)+1$$

$$\implies \sin^2x=2\sin x\cos x+\cos^2x\implies \sin^2x-\cos^2x=\sin 2x$$

$$\implies -\cos 2x=\sin 2x\implies -1=\tan 2x$$

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  • $\begingroup$ so assuming that (tanx - 1)^2 = 0 then it works out, how do you work out/know that the quadratic is equal to zero? $\endgroup$ – user100893 Oct 15 '13 at 1:02
  • $\begingroup$ Actually, see my updated hints, the quadratic was a mistaken hint. Note that I have not given you all the identities you need to complete this. $\endgroup$ – abiessu Oct 15 '13 at 1:06
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This all seems too complicated...

Solve the quadratic in $\tan(x)$ and use the $\tan^{-1}$ button; solutions are $x=67.5^{\circ}$ and $x=-22.5^{\circ}$.

So $2x$ is either $135^{\circ}$ or $-45^{\circ}$, and the tangent is $-1$ QED

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