# Proving trigonometric identities in converse

1. Feb 4, 2014

### sooyong94

1. The problem statement, all variables and given/known data
If $\sec x-\csc x=\pm p$, show that
$p^{2} \sin^2 2x +4\sin 2x-4=0$

Show conversely that if $p^{2} \sin^2 2x +4\sin 2x-4=0$, then $\sec x-\csc x$ is equal to +p and -p.

Find, to the nearest minute, the two values of x in the range of 0 to 360 degrees, the equation
$\sec x-\csc x=(1/2) (√5)$

2. Relevant equations
Double angle formula

3. The attempt at a solution
I have proved the above formula by squaring both sides and rewriting them as sines and cosines. Now how do I prove it backwards? I tried dividing both sides by $\sin^2 2x$ and it doesn't really help much.

2. Feb 5, 2014

### haruspex

Have you tried simply reversing the steps of your first proof? What goes wrong? Please post your working as far as you get.

3. Feb 5, 2014

### sooyong94

I did worked until here:
$\frac{4}{\sin 2x} (\frac{1}{\sin 2x}-1)=p^{2}$

and used the double angle formula, but I got this:
$\csc^2 x \sec^2 x - 2\csc x \sec x=p^{2}$

4. Feb 5, 2014

### haruspex

Right so far. If you compare that with the equation you're trying to get to, what equality do you need to make them the same?

5. Feb 5, 2014

### sooyong94

I have managed to proved that conversely. :D
Now I need to solve for x for that equation... When I substituted p=1/2 sqrt(5) into the equation, I have
$5\sin^2 2x+16 \sin 2x-16=0$

Then I finally arrived at $\sin 2x=\frac{4}{5}$. The problem is I have 4 solutions, instead of 2.

Last edited: Feb 5, 2014
6. Feb 5, 2014

### AlephZero

That's right. From the second part, you proved the solutions of the quadratic equation satisfy sec x - csc x = plus or minus p.

So you have to check which solutions are for +p and which are for -p.

7. Feb 5, 2014

### sooyong94

So all I have to do is to substitute them back?

8. Feb 7, 2014

### sooyong94

So all I have to do is to substitute them back?

9. Feb 7, 2014

### haruspex

Yes. Always a good check on the answer anyway.