1. ## parameter

im supposed to eliminate A from the pair of equations
x= sin2A
y= sec4A

ive learnt most trig identities i think, up to the double angle identities etc.
ive done several problems like this but here im really stuck

2. Originally Posted by furor celtica
im supposed to eliminate A from the pair of equations
x= sin2A
y= sec4A

ive learnt most trig identities i think, up to the double angle identities etc.
ive done several problems like this but here im really stuck
Well Remember that $\displaystyle \sin(2A) = 2\cos(A)\sin(A)$, remember that $\displaystyle \sec(x) = \frac{1}{\cos(x)}$, and remember that $\displaystyle \cos(2A) = \cos^2(A) - \sin^2(A)$.

3. Originally Posted by Mush
Well Remember that $\displaystyle \sin(2A) = 2\cos(A)\sin(A)$, remember that $\displaystyle \sec(x) = \frac{1}{\cos(x)}$, and remember that $\displaystyle \cos(2A) = \cos^2(A) - \sin^2(A)$.
I think mentioning the double angle formula for sin(2A) is an unintentional red herring. I'd be inclined to merely note that $\displaystyle \cos(4x) = \cos(2[2x]) = 1 - 2 \sin^2(2x)$ (using one of the well-known forms of the cosine double angle formula).

4. i passed through those already, including the sin2A double angle identity which i already figured might be totally irrelevant to this problem. but where do i go from cos4A? i really didnt need someone to repeat these identities to me. im actually stuck.

5. Originally Posted by furor celtica
[snip] i really didnt need someone to repeat these identities to me. im actually stuck.
Actually, I think you did need the identity being repeated to you. It was being brought to your attention because it is relevant to answering the question. The hope was that you would contemplate how it was linked to the question, see

$\displaystyle x = {\color{red}\sin (2A)}$

$\displaystyle y = \sec (4A) = \frac{1}{\cos (4A)} = \frac{1}{1 - 2 {\color{red}\sin^2 (2A)}}$

and then join the dots ....

6. maybe i just overestimated the difficulty of this problem cos it was the last of the section
anyway i sure know to whom i'll be running when i need help next time!(ironic)