i) Tomando $arccos(x^4) = \alpha$ e $arcsen(x^2 − \frac{1}{4}) = \theta$. Teremos que $\alpha + \theta = \frac{\pi}{2}$, ou seja, são arcos complementares, portanto $cos(\alpha) = sen(\theta)$. ii) $\alpha = arccos(x^4) \therefore cos(\alpha) = x^4$ $\theta = arcsen(x^2 − \frac{1}{4}) \therefore sen(\theta) =(x^2 − \frac{1}{4})$ iii) $x^4 = x^2 - \frac{1}{4} \Rightarrow x^4 - x^2 + \frac{1}{4} \therefore (x^2- \frac{1}{2})^2 = 0 \Leftrightarrow x = \frac{\sqrt 2}{2} \therefore arccos x = \frac{\pi}{4}\ ou \arccos x = \frac{3\pi}{4} $