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How would I prove the following double angle identity?

$\frac{1-\sin(2A)}{\cos(2A)}=\frac{1-\tan A}{1+\tan A}$

My work thus far is

$\frac{1-2\sin A\cos A}{\cos^2A-\sin^2A}$

$\frac{1-2\sin A\cos A}{(\cos A+\sin A)(\cos A-\sin A)}$

Sadly I am stuck.

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    I know this is hard for people just starting out, but remember to work from both sides. Don't try and go from left --> right. You should go from Left --> Simple and from Right --> Simple. And then see if Left-Simple is the same as the Right-Simple. If it is, QED.2012-07-26

4 Answers 4

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It's a good idea in this case to rewrite $\tan A=\frac{\sin A}{\cos A}$, then simplify the expression on the right, multiplying top and bottom by $\cos A$ to clear the "little fractions" from the "big fraction".

That will give you $\frac{\cos A-\sin A}{\cos A+\sin A}$ on the right hand side. Then, rewriting the left hand side denominator as $(\cos A+\sin A)(\cos A-\sin A)$--as you already have--it remains only to somehow rewrite $1-2\sin A\cos A$ as $(\cos A-\sin A)^2$, which can be done by using the Pythagorean identity.

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$ \begin{align} \frac{1-\sin(2A)}{\cos(2A)} &=\frac{1-2\sin(A)\cos(A)}{\cos^2(A)-\sin^2(A)}\tag{1}\\ &=\frac{\sec^2(A)-2\tan(A)}{1-\tan^2(A)}\tag{2}\\ &=\frac{1+\tan^2(A)-2\tan(A)}{1-\tan^2(A)}\tag{3}\\ &=\frac{(1-\tan(A))^2}{1-\tan^2(A)}\tag{4}\\ &=\frac{1-\tan(A)}{1+\tan(A)}\tag{5} \end{align} $

  1. double angle formulas

  2. multiply numerator and denominator by $\sec^2(A)$

  3. $\sec^2(A)=1+\tan^2(A)$

  4. collect square of a difference

  5. cancel $1-\tan(A)$ from numerator and denominator

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You could first write the right hand side of your identity as ${1-{\sin A\over \cos A}\over 1+{\sin A\over\cos A}}$. Multiply top and bottom by $\cos A$:

$ {1-{\sin A\over \cos A}\over 1+{\sin A\over\cos A}} ={1-{\sin A\over \cos A}\over 1+{\sin A\over\cos A}}\cdot{\cos A\over\cos A} ={\cos A-{\sin A\over \cos A}\cdot\cos A\over \cos A+{\sin A\over\cos A}\cdot \cos A} = {\cos A-\sin A\over \cos A+\sin A}. $ Multiply top and bottom by $\cos A-\sin A$ : $ {\cos A-\sin A\over \cos A+\sin A}= {\cos A-\sin A\over \cos A+\sin A}\cdot{\cos A-\sin A\over \cos A-\sin A} ={\cos^2 A+\sin^2 A-2\cos A\sin A\over \cos^2 A-\sin^2 A}. $ Finally apply the Pythagorean Identity and the Double Angle formulas to get what you want.

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    I see now there is alternate way of proving the identity.2012-07-26
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$\frac{1-2\sin A\cos A}{\cos^2A-\sin^2A}$

$=\frac{\sin^2A+\cos^2A-2\sin A\cos A}{(\cos A+\sin A)(\cos A-\sin A)}$

$=\frac{(\cos A-\sin A)^2}{(\cos A+\sin A)(\cos A-\sin A)}$

$=\frac{\cos A-\sin A}{\cos A+\sin A}$ assuming ${\cos A-\sin A} ≠ 0$

$=\frac{1-\tan A}{1 +\tan A}$ assuming ${\cos A} ≠ 0$

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    Thanks for your excellent response.2012-07-26