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

Answer 1

Elijah Alexander

#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))#

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Answer 2

Caroline Adams

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:

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

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.