Solve: Cos2θ = -√3/2
cmon brainshot plz help me :)
huh? @@!
in the book it have 3 formulas
cos 2A = cos^2A - sin^2A
2cos^2A - 1 and 1 - 2 sin^2A
Ha wow, I'm trying to do two things at once, sorry. I'm thinking of this as a calc prob. But back on topic, Cos(30 degrees) = √3/2 and we know that roots can be both negative and positive, right? This means that -√3/2 is a possibility, so it is 30/2 which is 15 degrees.
where 30 degrees coming from ?
Unit circle
and it is 30 degrees because you look at the √3/2 ?
15 degrees is not the answer though :|
The answer are : 75 degrees, 105 degrees, 255 degrees and 285 degrees
but how you get that :|
Hi, yes, ahh I don't really remember this stuff. Isn't there as specific formula for this type of equation? I said 15 since I know that Cos30 = Root3 / 2, so I thought it would be 30 /2 as you are multiplying by two here, but that is incorrect because this is a circle out of 360 degrees, not 100, so you can't just divide by 2.
not sure what formulas your trying to say here :|
those are the 3 formulas that the book gave me
this is gonna be one hell of a test tomorrow :|
Now I feel guilty. Don't worry, I'll get someone to help you and while I do that, i'll try to pull up an old formula sheet with the formula for this on it.
Oh no it's fine, I'm sorry for saying that lol :) It just that, this stuff is really confusing, I'm glad I got someone to help me you know :) so thanks for that haha :D
I actually did manage to find the formula sheet, but it seems as though I was thinking of the Cos(a+b) formulas which do not help. But, I typed it into my calculator and I actually did get 15 as being equal to theta here. Are you sure it isn't correct??
Wait is the equation Cos2x =-Root(3) / 2 or -Root (3/2) I've been going off the basis that it has been -Root(3) / 2
yes
Cos2x =-Root(3) / 2
I'm sorry about that !!
-Root (3/2) is the correct one sorry !
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