Como você avalia $cos (\frac{π}{8})$ $cos (pi / 8)$ ?

Responda:

$cos (\frac{π}{8}) = \sqrt{\frac{1}{2} + \frac{\sqrt{2}}{4}}$ $cos(pi/8) = sqrt(1/2+sqrt(2)/4)$

Explicação:

$Use the double-angle formula for cos(x) :$ $"Use the double-angle formula for cos(x) : "$
$cos (2 x) = 2 {cos}^{2} (x) - 1$ $cos(2x) = 2 cos^2(x) - 1$
$\Rightarrow cos (x) = \pm \sqrt{\frac{1 + cos (2 x)}{2}}$ $=> cos(x) = pm sqrt((1 + cos(2x))/2)$
$Now fill in x = \frac{π}{8}$ $"Now fill in x = "pi/8$
$\Rightarrow cos (\frac{π}{8}) = \pm \sqrt{\frac{1 + cos (\frac{π}{4})}{2}}$ $=> cos(pi/8) = pm sqrt((1 + cos(pi/4))/2)$
$\Rightarrow cos (\frac{π}{8}) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$ $=> cos(pi/8) = sqrt((1+sqrt(2)/2)/2)$
$\Rightarrow cos (\frac{π}{8}) = \sqrt{\frac{1}{2} + \frac{\sqrt{2}}{4}}$ $=> cos(pi/8) = sqrt(1/2+sqrt(2)/4)$

$Remarks :$ $"Remarks : "$
$1) cos (\frac{π}{4}) = sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} is a known value$ $"1) "cos(pi/4) = sin(pi/4) = sqrt(2)/2" is a known value"$
$because sin (x) = cos (\frac{π}{2} - x), so$ $"because "sin(x) = cos(pi/2-x)," so "$
$sin (\frac{π}{4}) = cos (\frac{π}{4}) and {sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin(pi/4)=cos(pi/4)" and "sin^2(x)+cos^2(x) = 1$
$\Rightarrow 2 {cos}^{2} (\frac{π}{4}) = 1 \Rightarrow cos (\frac{π}{4}) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} .$ $=> 2 cos^2(pi/4) = 1 => cos(pi/4) = 1/sqrt(2) = sqrt(2)/2.$
$2) because \frac{π}{8} lies in the first quadrant, cos (\frac{π}{8}) > 0, so$ $"2) because "pi/8" lies in the first quadrant, "cos(pi/8) > 0", so"$
$we need to take the solution with the + sign.$ $"we need to take the solution with the + sign."$

Como você avalia cos (pi / 8) cos(π8)cos (pi / 8) ?

Responda:

Explicação:

Como você avalia $cos (\frac{π}{8})$ $cos (pi / 8)$ ?