Can someone please walk me trough this one:
cos2x = 2 cosx sinx :uhh:
You are to find those x for which the equation holds (right?)
What have you thought of thus far?
you might use the fact that 2cos(x)sin(x) = sin(2x) to transform the equation into something easier.
I would have gotten to that eventually..
Seems like I scared the poster off instead by my questioning..
Maria, look closely at what Mathwonk said: your equation has cos(2x)= 2cos(x)sin(x)and you are to find the values of x for which it is true. His equation has sin(2x)= 2cos(x)sin(x) and is an IDENTITY: it is true for all values of x.
I need to find 4 angles..
I have found that cos2x = sin2x
But im not sure why?
Oh sorry.. Mathwonk..I understand.. stupid me..
I think i`ve got it now
cos2x = 2cosx sonx
cos2x = sin2x
sinx = 1
tanx = 1
and i get 4 angles because og tan = 1 ?
You do have cos2x=sin2x
Dividing with cos2x, you get:
of course.. forgot..
then i get
this is one angle..
Certainly; how would you find the others (when restricting 0<x<360)
180+22,5 = 202,5
but how about the last two?
Now, remember that tan(y+180)=tan(y)
Hence, for any integer n, we have:
Set y=45 (i.e, so that tan(y)=1), we may find solutions 0<x360
by looking at various choices n in the equation:
So I can for instance set n=1,but thenI get x = 113
am I right?
No, you get for n=1: x=22.5+90=112.5
and for n=-1 I get 292,5
How du you get from tan(y+180n)=tan(y) to the equation:
Well, you're seeking x-solutions satisfying
tan(2x)=1, or hence:
tan(2x)=tan(45+180n) for some n
By setting 2x=45+180n, you're guaranteed the last equation is fulfilled.
so I don`t need tan since I have it on both sides?
I hope you have accepted that the x-solutions you're looking for must satisfy:
tan(2x)=tan(45+180n), where n is some integer (We call this equation (e)).
Now, to guarantee that (e) holds , requiring 2x=45+180n is evidently enough, since the lefthand side term of (e) (that is tan(2x)) becomes necessarily equal to the righthandside term in (e) (that is tan(45+180n)).
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