Question

Seja f(x) = sen x + cos x. Calcule o valor de $\sqrt{6$ . $f (\frac{ \pi }{12} )$

Answer 1

Identidades utilizadas:

$\begin{array}{rclc} \cos \alpha&=&\mathrm{sen}\left(\frac{\pi}{2}-\alpha \right )&\;\;\;\;\text{(i)}\\ \\ \mathrm{sen\,}\alpha+\mathrm{sen\,}\beta&=&2\cdot \mathrm{sen}\left(\frac{\alpha+\beta}{2} \right )\cdot \cos \left(\frac{\alpha-\beta}{2} \right )&\;\;\;\;\text{(ii)}\\ \\ \cos \left(-\alpha \right )&=&\cos \alpha&\;\;\;\;\text{(iii)} \end{array}$


$f\left(x \right )=\mathrm{sen\,}x+\cos x\\ \\ f\left(x \right )=\mathrm{sen\,}x+\mathrm{sen}\left(\frac{\pi}{2}-x \right )\\ \\ f\left(x \right )=2\cdot \mathrm{sen}\left(\dfrac{x+\left(\frac{\pi}{2}-x \right )}{2} \right )\cdot\cos\left(\dfrac{x-\left(\frac{\pi}{2}-x \right )}{2} \right )\\ \\ f\left(x \right )=2\cdot \mathrm{sen}\left(\dfrac{\left(\frac{\pi}{2} \right )}{2} \right )\cdot\cos\left(\dfrac{2x-\frac{\pi}{2}}{2} \right )\\ \\ f\left(x \right )=2\cdot \mathrm{sen\,}\frac{\pi}{4}\cdot\cos\left(x- \frac{\pi}{4} \right )\\ \\ f\left(x \right )=\diagup\!\!\!\!2\cdot \frac{\sqrt{2}}{\diagup\!\!\!\!2}\cdot\cos\left(x-\frac{\pi}{4} \right )\\ \\ \boxed{f\left(x \right )=\sqrt{2}\cdot\cos\left(x-\frac{\pi}{4} \right )}$


$\sqrt{6}\cdot f\left(\frac{\pi}{12} \right )\\ \\ =\sqrt{6}\cdot \sqrt{2}\cdot\cos\left(\frac{\pi}{12}-\frac{\pi}{4} \right )\\ \\ =\sqrt{6 \cdot 2}\cdot \cos\left(\frac{\pi}{12}-\frac{3\pi}{12} \right )\\ \\ =\sqrt{12}\cdot\cos\left(\frac{\pi-3\pi}{12} \right )\\ \\ =\sqrt{2^{2}\cdot 3}\cdot\cos\left(-\frac{2\pi}{12} \right )\\ \\ =2\sqrt{3}\cdot\cos\left(-\frac{\pi}{6} \right )\\ \\ =2\sqrt{3}\cdot\cos\frac{\pi}{6}\\ \\ =\diagup\!\!\!\!2\sqrt{3}\cdot\frac{\sqrt{3}}{\diagup\!\!\!\!2}\\ \\ =\sqrt{3}\cdot \sqrt{3}\\ \\ =\left(\sqrt{3} \right )^{2}\\ \\ =3\\ \\ \\ \boxed{\sqrt{6}\cdot f\left(\frac{\pi}{12} \right )=3}$