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Prove $$ \frac{-2+2\tan A+2\cos B\cdot\sin B+\cot^2 A\cdot({\sec^4A-\operatorname{cosec}^2A-2)}}{2+\tan^2A-2\sin^2A} =(\sin A+\cos A)^2 $$

if and only if B is the double angle of A, or $B=2A+2k\pi$, $k=0,1,2,3...$

Advice is welcome as to improve notation and format.

i understand the RHS can be simplified to 1+sin(2A), but that doesn't seen to go anywhere .

Also, I have been able to simplify the left hand side of the expression down a little, but it's not taking me anywhere at all. I have a feeling that the difficulty of this proof lies in using the 2's in the LHS.

again, many thanks, Yun Fei

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I would suggest leaving out the periods for multiplication. Also you can format fractions as $\frac {abc}{def}$, which gives $$\frac {\mathsf{abc}}{\mathsf{def}}$$ –  Karl Kronenfeld May 13 '13 at 8:51
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What you're trying to prove can't be true, the equation holds for $B=2A$ if and only if it holds for $B = 2a + 2\pi$. –  Hurkyl May 13 '13 at 8:55
    
I have a feeling that this identity holds only for particular values of $A$ and $B$. For example, if $A=\pi/4$, the equation simplifies to $\sin 2B =4$. –  TZakrevskiy May 13 '13 at 13:56
    
Has anyone got an idea of how to split the 2's into something useful? Perhaps splitting it will help factorise the LHS and some solution might emerge from there. I know i'm being vague but i haven't had a breakthrough in the question myself. –  Yun Fei Ou Yang May 14 '13 at 7:53
    
@YunFeiOuYang It's clear that your statement isn't true (see solutions below. Can you check your equation? –  Calvin Lin Jul 6 '13 at 0:16
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2 Answers 2

The equation can be written as

$\sin 2B =$ (complicated rational function of $\sin A$ and $\cos A$).

The left side is bounded and the right side appears to have an uncancelled 6th-order pole at $\sin A = 0$ from the $\cot^2 A \csc^4 A$ term. [correction from comments: a 4th order pole from the $\cot^2 A \sec^4 A$ term]

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The right-hand side of your equation is (I think) $$\cos 2A (1+\sin 2A) + \frac{1}{\sin^4 A}$$ which is kinda-sorta close to $\sin 4A$ (in that it includes the product of $\cos 2A$ and $\sin 2A$, if not the requisite factor of $2$). Perhaps OP's equation has a typo or some other error(s), and the thing is really intended to boil down to $\sin 2B = \sin 4A$. Even then, the solution set is more than just $B = 2A + 2k\pi$. (Of course, no identity is going to hold only for non-negative integer $k$, but perhaps that, too, is an error on OP's part.) –  Blue May 14 '13 at 5:51
    
Thanks, I corrected it. Agree with all of your comments. @Blue –  zyx May 14 '13 at 6:03
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Abbreviating your equation as

$$\frac{P + 2\sin B \cos B}{Q} = R$$

we have

$$\sin 2B = Q R - P$$

where

$$\begin{align} R &= \left(\cos A + \sin A \right)^2 = \cos^2 A + \sin^2 A + 2 \sin A \cos A \\ &= 1 + \sin 2A \\[4pt] Q &= 2 - 2 \sin^2 A + \tan^2 A = 2 \cos^2 A + \frac{\sin^2 A}{\cos^2 A} \\ &= \frac{1}{\cos^2 A}\left(2 \cos^4 A + \sin^2 A\right) \end{align}$$

For $P$, we'll conveniently add and subtract $\csc^4 A$:

$$\begin{align} P &= 2 \tan A - 2 + \cot^2 A \left(- 2 + \sec^4 A - \csc^2 A \right) + \left( \csc^4 A - \csc^4 A \right)\\ &= 2 \tan A - 2 \left(1 +\cot^2 A \right) + \frac{\cos^2 A}{\sin^2 A}\left(\frac{1}{\cos^4 A} - \frac{1}{\sin^2 A} \right) + \frac{1}{\sin^4 A} &- \csc^4 A\\ &= 2 \tan A - 2 \csc^2 A + \frac{1}{\sin^2 A \cos^2 A} + \left(-\frac{\cos^2 A}{\sin^4 A} + \frac{1}{\sin^4 A}\right) &-\csc^4 A\\[4pt] &=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{2}{\sin^2 A} + \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt] &=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt] &=\frac{2\sin A}{\cos A}+ \frac{1-\cos^2 A}{\sin^2 A\cos^2 A} &-\csc^4 A \\[4pt] &= \frac{2\sin A \cos A}{\cos^2 A} + \frac{1}{\cos^2 A} &-\csc^4 A \\[4pt] &= \frac{1+\sin 2A}{\cos^2 A} &-\csc^4 A \end{align}$$

Therefore,

$$\begin{align} QR-P &= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A \right)\left( 1 + \sin 2A \right) - \frac{1}{\cos^2 A}\left( 1 + \sin 2 A \right) + \csc^4 A \\[4pt] &= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt] &= \frac{1}{\cos^2A}\left( 2 \cos^4 A - \cos^2 A\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt] &=\left( 2 \cos^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt] &=\cos 2A \left( 1 + \sin 2A \right) + \csc^4 A \\[4pt] \end{align}$$

So, your equation reduces to

$$\sin 2 B = \cos 2A ( 1 + \sin 2 A ) + \csc^4 A$$

which is decidedly not equivalent to $B = 2 A + 2k\pi$. Perhaps your original equation has a typo or something.

Note that, even had the equation become

$$\sin 2B = \sin 4A$$

(which seems closest to what you might be anticipating), we'd have $2B = 4A + 2k\pi$ OR $2B = \pi - 4A + 2k\pi$, whence $B = 2A+k\pi$ OR $B = \frac{\pi}{2}-2A + k\pi$.

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