Trig identities help! MATHMETICIANS

dave44

dave44

Active member
Okay, I feel like a retard. Did this shit like a year ago, but I need to recap and can't remember how to do this for the life of me (should be very simple).

Find the values of sin(x) and cos(x) when cos(2x) is 1/9

 
Oops, accidently hit enter. Cos2x=cos^2x-sin^2x Also cos2x=2cos^2x-1. And cos^2x=1-2sin^2x Set the second two both equal to 1/9, should get cosx=square root of 5/9 and sinx=2/3. Substitute answers back into first equation to check answers. 1/9= 5/9-4/9
 
SOHCAHTOA

Cos (2x) = 1/9 = adjacent/ hypotenuse

2x= 1/9, x= (1/9)/2 =1/18

x= 1/18

sin (1/18) = .0556

cos (1/18)= .9985

hope that helps, since I just basically did that whole thing for you. I'm a math nerd.

 
I came here to write the same thing, but just backed you up. And that makes two of us kimber :)
 
How is that correct? The cos of cos2x cannot just disappear, you'd have to take the cosine inverse to get rid of it. So to solve for x, it would be

cos2x=1/9

cos^-1 (cos2x)=cos^-1 (1/9)

2x=cos^-1 (1/9)

x=cos^-1 (1/18)

But the problem can easily be solved without a calculator using double angle identities, so why not just do it that way and get "nice" numbers?
 
lol you are right, I completely forgot afterwards that cosx does not equal cos2x/2

I fail, maybe I should go back to high school
 
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