Please reply with what you tried for at least one of your attempts.
When you reply, please clarify if you are indeed supposed to "solve" the "equation" (for some value of "a"), or if you are supposed to "prove" the "identity" (by showing, by logical steps, how to go from one side to the other).
Thank you!
The solution is on the right. Maybe I should have written it slightly differently.
I want to know how to factor the left so it equals what is on the right.
I have tried the following:
1. I used the two angle formula for sin(2a - b) = sin(2a)cos(b)-cos(2s)sin(b). I did this both before multiplying through outer brackets and after.
2. To get rid of the double angles by using the double angle formulas i.e. sin(2a) = 2sin(a)cos(a) and cos(2a)=1-2sin^2(a) and tan(2a)=2tan(a)/(1-tan^2(a)) .
When I do this I end up getting 9 terms. I then try to simplify.
I ended up with:
cos(a)/tan(a) - cos(a)tan(a) - tan^2(b)/(2tan(a)sin(a)) + tan^2(a)tan^2(b)/(2tan(a)sin(a)) - sin^2(a)tan^2(b)/(2tan(a)sin(a)) + sin^2(a)tan^2(a)tan^2(b)/(2tan(a)sin(a)) - 2sin(a)cos(a)/sin(a) + tan(b)/sin(a) - 2 sin(a)tan(b)
I have done some reduction....
The two left terms can become:
cos(a)(1/tan(a)-tan(a))
cos(a)(cos(a)/sin(a)-sin(a)/cos(a))
cos(a)(cos^2(a)-sin^2(a))/sin(a)cos(a)
which seems to be leading back to the double angle formula which I don't want...
starting again for the two left terms and just factoring
cos^2(a)/sin(a) - sin(a)
sub in for cos^2(a) = 1 - sin^2(a):
(1-sin^2(a))/sin(a) - sin(a)
dividing by sin(a) on the left
1/sin(a) - sin(a) - sin(a)
1/sin(a) - 2sin(a)
These seems to be leading me back to double angles again.
Stopped there with the two left terms.
for the third to the 3 through the 6th term I can factor out like:
tan^2(b) / ( 2tan(a)sin(a) ) ( 1 - tan^2(a) +2sin^2(a) - 2sin^2(a)tan^2(a) )
which becomes
tan^2(b) / (2tan(a)sin(a) ) ( (1 - sin^2(a)cos^2(a) + 2sin^2(a) - sin^4(a))/cos^2(a) )
I then factor out cos^2(a) from the denominator
tan^2(b) / (2tan(a)sin(a)cos^2(a)) ( cos^2(a) - sin^2(a) + 2sin^2(a)cos^2(a) - 2sin^4(a) )
foil backwards
tan^2(b) / ( 2tan(a)sin(a)cos^2(a) ) ( 1 + 2sin^2(a) ) ( cos^2(a) - 2sin^2(a) )
This is staring to feel double anglee again. There is a symmetry to the double angle identities with all three factored groups in that it seems like a double angle identify but with the second term doubled... Anyway... spinning my wheels I think. I tried a bunch of other factoring but no dice. It seems I am carrying the legacy of the double angles through and I need some sort of possibly only valid in one quadrant identity.. Or I can factor out some sin^2(a) + cos^2(a) and adding them up into 1s and 2s .... Anyway I will stop there hoping someone else can see something.... for now.
The 7th through 9th terms seems like something I should keep to subtract away from the results of factoring of 3 through 7... Or possibly as the 8th and 9th terms are just:
-2cos(a) + 1/cos(a)
the seem oddly familiar to what the results of what the first two terms factoring has become ...
Anybody see something?
I will put more time in once I get a bit of mental clear space.
M
When in doubt, get everything in terms of sines and cosines, and get rid of double angles. So expand out the sin(2a - b) and tan(2a) and put it all in terms of sines and cosines.
I'm not saying that's the best way here, as I haven't tried to solve it, just some general advice that usually helps in the long run.
-Dan
This is what I have been trying.
