SOLUTION: Evaluate cos((7pi)/8) I have this so far cos^2 theta=(1/2)(1+cos(2theta)) cos((7pi/8))= sqroot((1+cos(7pi/4))/2) = sqroot((1+ ((sqroot2)/2))/2) =sqr

Algebra ->  Trigonometry-basics -> SOLUTION: Evaluate cos((7pi)/8) I have this so far cos^2 theta=(1/2)(1+cos(2theta)) cos((7pi/8))= sqroot((1+cos(7pi/4))/2) = sqroot((1+ ((sqroot2)/2))/2) =sqr      Log On


   

Question 205030: Evaluate cos((7pi)/8)
I have this so far
cos^2 theta=(1/2)(1+cos(2theta))
cos((7pi/8))= sqroot((1+cos(7pi/4))/2)
= sqroot((1+ ((sqroot2)/2))/2)
=sqroot((2+sqroot2)/4)
=(1/2)(sqroot(2+sqroot2))
but it is marked wrong, is there supposed to be a negative somewhere?
Thank You
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Evaluate
cos%28%287pi%29%2F8%29

I have this so far
cos^2 theta=(1/2)(1+cos(2theta))
cos((7pi/8))= sqroot((1+cos(7pi/4))/2)
       
That step is wrong. You must take the NEGATIVE square root,
not the POSITIVE square root, because 7pi%2F8 is in
QII (the second quadrant), and the cosine is negative in
the second quadrant. So you should have had a negative
sign in front of the right side:

    = sqroot((1+ ((sqroot2)/2))/2)
           =sqroot((2+sqroot2)/4)
           
Those are OK except for the sign.  You must recognize
what quadrant you are in whenever you take a square
root, because square roots can be positive or negative.

correct answer:  -%281%2F2%29%28sqrt%282%2Bsqrt%282%29%29%29  

Edwin