Question

Encontre o valor de $\sin^{-1} \Big[\sqrt{2} \sin \theta \Big]+\sin^{-1} \Big[\sqrt{\cos 2\theta}\Big]$

Answer 1

$\displaystyle \sf sin^{-1}\left[\sqrt{2}sin\theta \right ]+sin^{-1}\left[\sqrt{cos2\theta} \right ]= k \\\\ \underline{Fa{\c c}amos}}: \\\\ m = sin^{-1}\left[\sqrt{2}sin\theta\right] \to sin (m) = \sqrt{2}sin\theta \\\\\ cos(m) = \sqrt{1-(\sqrt{2}sin\theta)^2} \\\\ cos(m) = \sqrt{1-2sin^2\theta}\\\\ cos(m) = \sqrt{cos2\theta} \\\\\\ n = sin^{-1} \left[\sqrt{cos2\theta}\right] \to sin(n) =\sqrt{cos2\theta} \\\\ cos(n)=\sqrt{1-(\sqrt{cos2\theta})^2} \\\\\ cos(n) =\sqrt{1-cos2\theta }$

$\sf cos(n)=\sqrt{1-(1-2sin^2\theta)} \\\\\ cos(n) = \sqrt{1-1+2sin^2\theta} \\\\\ cos(n)=sin\theta.\sqrt{2}$

Daí, temos :

$\sf m + n = k \\\\ Aplicando \ sin\ em\ ambos\ os\ lados : \\\\ sin(m+n)= sin k \\\\ sin(m).cos(n)+sin(n).cos(m) = sin k \\\\ \sqrt{2}.sin\theta.sin\theta\sqrt{2} +\sqrt{cos2\theta}.\sqrt{cos2\theta } = sin k \\\\ 2sin^2\theta +cos2\theta = sin k \\\\ 2sin^2\theta +1-2sin^2\theta = sin k \\\\ sin k = 1 \\\\\\ \huge\boxed{\displaystyle \sf k = \left(\frac{\pi }{2}\right)\ }\checkmark$