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Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$

I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS: $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.

yinivem462
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  • Is the confusion about $3\pi/4$ because you are used to having the arguments of the trigonometrics in degrees? If that is the case $3\pi/4$ is $135^\circ$. – plop Sep 06 '22 at 19:39
  • @user85667, thank you for the response, but I know that it is $135^\circ$. The confusion is about where it comes from and how it relates to the LHS. That's what I am supposed to figure out. – yinivem462 Sep 06 '22 at 19:40
  • You can see it appear. For example replace $\cos(\alpha)$ with $\sin(\pi/2-\alpha)$ and use $\sin(A)-\sin(B)=2\cos((A+B)/2)\sin((A-B)/2)$. – plop Sep 06 '22 at 19:45
  • By the way, if you know complex numbers, it is good to take into account that all the torture of trigonometric identities in school is only a relic from the past that math education hasn't had the guts to eliminate. If you replace $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, the identity becomes a polynomial identity, which you can verify mechanically by multiplying everything and collecting terms of the same degree. No tricks are needed. – plop Sep 06 '22 at 22:34

8 Answers8

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Here’s yet another way:

$$LHS=\frac{1-\cos(\frac{\pi}{2}-2\alpha)}{1+\cos(\frac{\pi}{2}-2\alpha)}$$ $$=\frac{2\sin^2(\frac{\pi}{4}-\alpha)}{2\cos^2(\frac{\pi}{4}-\alpha)}$$ $$=\tan^2(\frac{\pi}{4}-\alpha)$$ $$=\tan^2(\alpha-\frac{\pi}{4}+\pi)$$ $$=RHS$$

David Quinn
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$\sin(\alpha)-\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}-\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(-\sin(\alpha)\cos\left(\frac{3\pi}{4}\right)-\cos(\alpha)\sin\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)$

$\sin(\alpha)+\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}+\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(\sin(\alpha)\sin\left(\frac{3\pi}{4}\right)-\cos(\alpha)\cos\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)$

Finally, $$\frac{\left(\sin(\alpha)-\cos(\alpha)\right)^2}{\left(\sin(\alpha)+\cos(\alpha)\right)^2}=\frac{\left(-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)\right)^2}{\left(-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)\right)^2}=\tan^2\left(\alpha+\frac{3\pi}{4}\right)$$

eMathHelp
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$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\dfrac{(\sin(\dfrac{3\pi}{4}+\alpha))^2}{(\cos(\dfrac{3\pi}{4}+\alpha))^2}=\dfrac{(\sin\dfrac{3\pi}{4}\cos\alpha+\cos\dfrac{3\pi}{4}\sin\alpha)^2}{(\cos\dfrac{3\pi}{4}\cos\alpha-\sin\dfrac{3\pi}{4}\sin\alpha)^2}=\dfrac{(\dfrac{1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}{(\dfrac{-1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}=\dfrac{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha-\sin\alpha\cos\alpha}{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha+\sin\alpha\cos\alpha}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$

David Raveh
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$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha} = \left(\frac{\tan \alpha-1}{\tan \alpha+1}\right)^2= \left(\frac{1-\tan \alpha}{1+\tan \alpha}\right)^2$$

$$=\tan^2\left(\dfrac{\pi}{4}-\alpha\right) =\tan^2\left(\alpha-\dfrac{\pi}{4}\right)$$

Inverse tangent function satisfies two angles in range $0, 2 \pi.\;$We are allowed to add $k\cdot\pi$ to any angle for tangent function value to be same. So for $k=1,$

$$ =\tan^2\left(\alpha-\dfrac{\pi}{4}+\pi \right) =\tan^2\left(\alpha+\dfrac{3\pi}{4}\right)= RHS $$

Narasimham
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  • Can you elaborate why is $\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$ equal to $\left(\frac{\tan \alpha-1}{\tan \alpha+1}\right)^2$? Thanks – yinivem462 Sep 06 '22 at 22:12
2

As an alternative by half-angle formula $\tan \frac{\theta}2=\frac{1-\cos \theta}{\sin \theta}$ we have (note that $\cos \alpha \neq \sin \alpha$):

$$\tan\left(\dfrac{3\pi}{4}+\alpha\right) =\frac{1-\cos \left(\frac{3\pi}{2}+2\alpha\right)}{\sin \left(\frac{3\pi}{2}+2\alpha\right) } =\frac{1-\sin(2\alpha)}{-\cos(2\alpha)}=\frac{(\cos \alpha-\sin \alpha)^2}{-(\cos^2 \alpha-\sin^2 \alpha)}=\frac{\cos \alpha-\sin \alpha}{-(\cos \alpha+\sin \alpha)}$$

and then

$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{(\cos \alpha-\sin \alpha)^2}{(\cos \alpha+\sin \alpha)^2}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$

user
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$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{\sin^2\left(\dfrac{3\pi}{4}+\alpha\right)}{\cos^2\left(\dfrac{3\pi}{4}+\alpha\right)}=\frac{\left(\frac{-\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha\right)^2}{\left(\frac{-\sqrt{2}}{2}\cos\alpha-\frac{\sqrt{2}}{2}\sin \alpha\right)^2}=\frac{\frac{1}{2}\cos^2\alpha+\frac{1}{2}\sin^2 \alpha-\frac{1}{2}\sin{\alpha}\cos{\alpha}}{\frac{1}{2}\cos^2 \alpha+\frac{1}{2}\sin^2\alpha+\frac{1}{2}\sin{\alpha}\cos{\alpha}}=\frac{1-\sin{2\alpha}}{1+\sin{2\alpha}}$$ Use the addition formulas for trigonometric functions.

Kamal Saleh
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One more (why not?):

$$ \cos \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ -\frac{\sqrt2}{2} ·(\sin \alpha \ + \ \cos \alpha) $$ $$ \Rightarrow \ \ \sec \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{-\sqrt2}{(\sin \alpha \ + \ \cos \alpha)} \ \ ; $$ $$ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \sec^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ - \ 1 \ \ = \ \ \frac{2}{(\sin \alpha \ + \ \cos \alpha)^2} \ - \ 1 $$

$$ = \ \ \frac{2 \ - \ \sin^2 \alpha \ - \ 2 \sin \alpha \cos \alpha \ - \ \cos^2 \alpha}{\sin^2 \alpha \ + \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha} \ \ = \ \ \frac{1 \ - \ 2 \sin \alpha \cos \alpha }{1 \ + \ 2 \sin \alpha \cos \alpha } \ \ = \ \ \frac{1 \ - \ \sin (2\alpha) }{1 \ + \ \sin (2\alpha) } \ \ . $$

Your approach would be a "time-reversal" of this, but I don't think it's at all obvious that you would want to replace $ \ \sin^2 \alpha + \cos^2 \alpha \ $ with $ \ 2 - \sin^2 \alpha - \cos^2 \alpha \ $ in your numerator, and then work toward a "secant-squared" expression. (eMathHelp probably has the best way to continue from where you left off.)

ADDENDUM --

We may also apply the "angle-addition" formula for tangent:

$$ \tan \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{\tan \alpha \ + \ \tan \frac{3 \pi}{4}}{1 \ - \ \tan \alpha · \tan \frac{3 \pi}{4}} \ \ = \ \ \frac{\tan \alpha \ + \ (-1)}{1 \ - \ \tan \alpha · (-1)} $$ $$ \Rightarrow \ \ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{ \tan^2 \alpha \ - \ 2 · \tan \alpha \ + \ 1 }{\tan^2 \alpha \ + \ 2 · \tan \alpha \ + \ 1 } $$ [multiplying the numerator and denominator by $ \ \cos^2 \alpha \ \ ] $ $$ = \ \ \frac{ \sin^2 \alpha \ - \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha }{\sin^2 \alpha \ + \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha } \ \ , $$ and so to the left-side expression.

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$$ \begin{aligned} \tan^2 \left(\frac{3 \pi}{4}+\alpha\right) &=\tan ^2\left(\pi-\frac{\pi}{4}+\alpha\right) \\ &=\tan ^2\left(\frac{\pi}{4}-\alpha\right) \\ &=\left[\frac{\sin \left(\frac{\pi}{4}-\alpha\right)}{\cos \left(\frac{\pi}{4}-\alpha\right)}\right]^2 \\ &=\left(\frac{\sin \alpha-\cos \alpha}{\cos \alpha+\sin \alpha}\right)^2 \\ &=\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha} \end{aligned} $$

Lai
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