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My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.

$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$$

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  • $\begingroup$ Try yourself using a few trigonometric identities, e.g. the duplication formulas for sin and cos, and also write down explicitly cot in terms of sin and cos $\endgroup$
    – GFR
    May 21, 2015 at 17:17
  • $\begingroup$ Hint: Start with the right hand side, write $cot\theta =\frac{cos\theta }{sin\theta }$, simplify and then multiply top and bottom of your result by the conjugate of the denominator. $\endgroup$
    – Matematleta
    May 21, 2015 at 17:18
  • $\begingroup$ Tried with trig identities, didn't get far. Regarding the hint: I multiplied it by the conjugate of the denominator, now what? $\endgroup$
    – user90572
    May 21, 2015 at 17:26
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    $\begingroup$ For what it’s worth, manipulating what you want to prove until it says $1=1$ is not a proof unless the manipulations are reversible. Any “identity” could be “proved” that way by a) multiplying both sides by zero, then b) adding $1$ to both sides. Showing that $A=B$ implies $1=1$ does not provide that $A=B$ is true. You need to conclude that $A$ equals $B$, not that $1$ equals $1$. $\endgroup$
    – Steve Kass
    May 21, 2015 at 17:45

5 Answers 5

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Let us look at the RHS only: (It is fine to manipulate both sides but it is usually preferred if you manage to make one into the other.)

$$ \frac{\cot{\theta}-1}{\cot\theta +1} = \frac{\frac{\cos\theta}{\sin\theta}-1}{\frac{\cos\theta}{\sin\theta} +1} $$ Now we multiply the numerator and denominator by $\sin\theta$ to get, \begin{equation} \frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}. \end{equation} If we consider the double angle formulas for $\sin$ and $\cos$ i.e. $$ \cos{2\theta} = \cos^2\theta-\sin^2\theta, \\ \sin 2\theta= 2\sin\theta\cos\theta, $$ we can see that we can get a $\cos2\theta$ in the numerator of $\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}$ by multiplying by $\cos\theta+\sin\theta$ as the $\cos2\theta$ formula is a difference of two squares. Proceeding with this idea we get; $$ \left(\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta}\right) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta+2\sin\theta\cos\theta +\sin^2\theta}. $$ Now the numerator is equal to $\cos2\theta$ so that is done and we only need to fix the denominator, by a quick inspection we see that we have a term $2\sin\theta\cos\theta$ which gives our $\sin2\theta$ term we desired and using the fact $\cos^2\theta+\sin^2\theta=1$ we find we get $$ \frac{\cos2\theta}{1+\sin2\theta} $$ which is what we wanted!

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\begin{align}\dfrac{\cot\theta-1}{\cot\theta+1}&=\dfrac{\dfrac{\cos\theta}{\sin\theta}-1}{\dfrac{\cos\theta}{\sin\theta}+1}=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\\\\&=\dfrac{\cos^2\theta-\sin^2\theta}{(\cos\theta+\sin\theta)^2}=\dfrac{1-2\sin^2\theta}{1+2\cos\theta\sin\theta}\\\\ &=\dfrac{\cos{2\theta}}{1+\sin{2\theta}} \end{align}

Using only the definition of the trigonometric functions in terms of $\exp(\theta)$: $$\cot\theta=-\dfrac{i(e^{-i\theta}+e^{i\theta})}{e^{-i\theta}-e^{i\theta}}$$ Thus: \begin{align}\dfrac{\cot\theta-1}{\cot\theta+1}&=\dfrac{-i(e^{-i\theta}+e^{i\theta})-(e^{-i\theta}-e^{i\theta})}{-i(e^{-i\theta}+e^{i\theta})+(e^{-i\theta}-e^{i\theta})} \\\\&=\dfrac{\left[-i(e^{-i\theta}+e^{i\theta})-(e^{-i\theta}-e^{i\theta})\right]\left[-i(e^{-i\theta}+e^{i\theta})+(e^{-i\theta}-e^{i\theta})\right]}{\left[-i(e^{-i\theta}+e^{i\theta})+(e^{-i\theta}-e^{i\theta})\right]^2} \\\\&=\dfrac{-(e^{-2i\theta}+e^{2i\theta}+2)-(e^{-2i\theta}+e^{2i\theta}-2)}{-(e^{-2i\theta}+e^{2i\theta}+2)+(e^{-2i\theta}+e^{2i\theta}-2)-2i(e^{-2i\theta}-e^{2i\theta})} \\\\&=\dfrac{-2(e^{2i\theta}+e^{-2i\theta})}{-4-2i(e^{-2i\theta}-e^{2i\theta})} \\\\&=\dfrac{e^{2i\theta}+e^{-2i\theta}}{1+\frac{i}{2}(e^{-2i\theta}-e^{2i\theta})} \\\\&=\dfrac{\cos{2\theta}}{1+\sin{2\theta}} \end{align}

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  • $\begingroup$ How did you get from step 2 to 3 (first line, last equality). $\endgroup$
    – user90572
    May 21, 2015 at 17:35
  • $\begingroup$ @MagnusQ. $$\dfrac{\cos\theta}{\sin\theta}\pm 1=\dfrac{\cos\theta\pm\sin\theta}{\sin\theta}$$ And you can cancel the denominators. $\endgroup$
    – Demosthene
    May 21, 2015 at 17:37
  • $\begingroup$ @MagnusQ. See the edit. $\endgroup$
    – Demosthene
    May 21, 2015 at 18:11
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Hint: In RHS, use

  1. $\cot\theta=\frac{\sin\theta}{\cos\theta}$,
  2. simplify and multiply by $\sin\theta+\cos\theta\over{\sin\theta+\cos\theta}$

OR

Use

  1. $\cos2\theta=\cos^2\theta-\sin^2\theta$,
  2. $a^2-b^2=(a+b)(a-b)$
  3. $\sin2\theta=2\sin\theta\cos\theta$ and
  4. $\sin^2\theta+\cos^2\theta=1$

in LHS and simplify.

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Use

$\cos(2x)=\cos^2(x) - \sin^2(x)$ and

$1+\sin(2x) = 1+2\sin(x)\cos(x) = \sin^2(x)+\cos^2(x)+2\sin(x)\cos(x)=(\sin(x)+\cos(x))^2$

So we have $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x) - \sin^2(x)}{(\sin(x)+\cos(x))^2}=\frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)}=\frac{\cot(x)-1}{\cot(x)+1}$$

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Use $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$ to get $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x)-\sin^2(x)}{1+2\sin(x)\cos(x)}$$

Multiply top and bottom by $\frac{1}{\sin^2(x)}$ to get $$\frac{\cot^2(x)-1}{\frac{1}{sin^2(x)}+2\cot(x)}$$

Using $\sin^2(x)+\cos^2(x)=1$ we can get $$\frac{\cot^2(x)-1}{\frac{\sin^2(x)+\cos^2(x)}{sin^2(x)}+2\cot(x)}\ \ \ \ \ \ \ or\ \ \ \ \ \ \ \frac{\cot^2(x)-1}{1+\cot^2(x)+2\cot(x)}$$

By using $a^2-b^2=(a-b)(a+b)$ for the numerator and factoring the denominator, we get $$\frac{(\cot(x)-1)(\cot(x)+1)}{(\cot(x)+1)^2}$$ which equals to $$\frac{\cot(x)-1}{\cot(x)+1}$$

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