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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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    $\begingroup$ In the top line, do you mean $1-\sin(2A)$, instead of $1-2\sin A$? If not, you've made an error in your first step. $\endgroup$ – Cameron Buie Jul 26 '12 at 17:27
  • $\begingroup$ Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles Jul 26 '12 at 17:27
  • $\begingroup$ Yes I meant 1-sin(2A). $\endgroup$ – Fernando Martinez Jul 26 '12 at 17:30
  • $\begingroup$ Fixed.${}{}{}{}$ $\endgroup$ – Cameron Buie Jul 26 '12 at 17:33
  • $\begingroup$ When you have sines and cosines and you want tangents, try dividing by cosine. $\endgroup$ – robjohn Jul 26 '12 at 17:45
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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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  • $\begingroup$ Thanks for your excellent response. $\endgroup$ – Fernando Martinez Jul 26 '12 at 17:37
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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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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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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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  • $\begingroup$ I see now there is alternate way of proving the identity. $\endgroup$ – Fernando Martinez Jul 26 '12 at 17:38

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