How do you find the value of #cos((7pi)/8)# using the double or half angle formula?

Question

Elijah Alexander · Accepted Answer

#cos ((7pi)/8)= -sqrt ((sqrt2 +1)/(2sqrt2))# Write cos#(7pi)/8# = cos #(pi-pi/8)#= -cos #pi/8# Now using half angle formula, cosx= #2 cos^2 (x/2) -1#, we can write cos #pi/4# = 2 #cos^2 (pi/8) -1# This means #1/sqrt2 +1 = 2 cos^2 (pi/8)# #2cos^2 (pi/8) = (sqrt2 +1)/sqrt2# #cos^2 (pi/8) = (sqrt2 +1)/(2sqrt2)# #cos (pi/8)= sqrt ((sqrt2 +1)/(2sqrt2))# Hence #cos ((7pi)/8)= - cos (pi/8) = -sqrt ((sqrt2 +1)/(2sqrt2))#

Caroline Adams · Answer

undefined Using the half-angle formula for cosine, $\cos(\frac{\theta}{2}) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}$, where the sign depends on the quadrant of $\frac{\theta}{2}$, you can find the value of $\cos(\frac{7\pi}{8})$ as follows:

1. Rewrite $\frac{7\pi}{8}$ as $\frac{14\pi}{16}$.
2. Apply the half-angle formula: $\cos(\frac{7\pi}{16}) = \pm \sqrt{\frac{1 + \cos(\frac{14\pi}{16})}{2}}$.
3. Determine the sign: Since $\frac{7\pi}{16}$ lies in the second quadrant where cosine is negative, the negative sign applies.
4. Substitute the known values: $\cos(\frac{7\pi}{16}) = - \sqrt{\frac{1 + \cos(\frac{14\pi}{16})}{2}}$.
5. Calculate $\cos(\frac{14\pi}{16})$. This can be simplified as $\cos(\frac{\pi}{2}) = 0$.
6. Substitute $\cos(\frac{14\pi}{16}) = 0$ into the formula.
7. Simplify to find the value of $\cos(\frac{7\pi}{8})$.

So, $\cos(\frac{7\pi}{8}) = - \sqrt{\frac{1 + 0}{2}} = - \sqrt{\frac{1}{2}} = -\frac{\sqrt{2}}{2}$.