Please help me solve this problem.
if a=cos2alpha + isin2alpha
b=cos2beta + isin2beta
prove that,
a-b / a+b = itan (alpha-beta)
Yes, it's also equal to by de Moivre's. I don't know if that helps though. Expanding doesn't yield anything useful.
It could also help to play around with the RHS, . Note that this is a purely imaginary number. You can try to reduce the LHS as or something like that. You'll most probably need your trigonometric identities.
Yeah, it's an interesting problem. I haven't solved it yet...there seem to be several possible ways to the solution though. Just try a bunch of methods, see if one of them works. Maybe another user has a nice, elegant solution that doesn't require a lot of brute force (I don't want to brute force algebra/trig identities right now...). There might even be a geometric solution involving two points on a circle and finding the tangent of the angle in between.
let's write, for the time being, cos(2α) = u, sin(2α) = v, cos(2β) = y, sin(2β) = z.
then:
now we can use some trig to simplify.
note that u^{2}+v^{2} = y^{2}+z^{2} = 1. so our messy fraction becomes:
now
and .
so what we have is:
Man ! Thank so very much Deveno.
You made it so simple to understand mate. You can't imagine the time i've spent on this problem.
Thank You again for your help.
Have a nice day !
edit: Thank you to Richard1234 too. I have solved it by your way too. Now I have two solutions.
Hello, ok59!
Out of curiosity, I brute-forced it . . . whew!
We need the Sum-to-Product Identities:
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De Moivre's Theorem is a relatively simple formula for calculating powers of complex numbers.De Moivre's formula states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).(r cisθ)^n = r^n cis(nθ) here n is an integer.
free algebra word problems
Here's another (equivalent) way: I'll begin with a general calculation that I'll then apply to this specific problem.
Suppose , and (i.e. on the unit circle except the leftmost point).
So let , where is not an odd multiple of .
Then
.
With that in hand, the result follows quickly. Let , . Let . (Note that , so .)
Let . Then , and obviously .
Notice that from , it follows that , so that , and hence so can apply the above derivation with this .
Now consider . Have:
as desired.
Thus the claim is established.
Notice that all the "bad" cases amounted to , because then , so , so .
That would've made , and also would've maded , which is where is undefined.