Question

ME AJUDEM, URGENTE, POR FAVOR!

A expressão $sen(17 \pi - 2 \alpha ).sen( \frac{19\pi}{2} - \alpha ) + cos( \frac{21 \pi }{2} + \alpha ).sen( \frac{23 \pi }{2}+2 \alpha )$

a) sen α
b) -sen α
c) cos α
d) sen²α
e) -cos²α

Answer 1

Propriedades utilizadas:

$\begin{array}{rclc} \mathrm{sen}\left(-x \right )&=&-\mathrm{sen}\left(x \right )&\;\;\;\;\text{(i)}\\ \\ \mathrm{sen}\left(x+k\cdot 2\pi \right )&=&\mathrm{sen}\left(x \right )&\;\;\;\;\text{(ii)}\\ \\ \cos\left(x+k\cdot 2\pi \right )&=&\cos\left(x \right )&\;\;\;\;\text{(iii)}\\ \\ \mathrm{sen}\left(x \pm \dfrac{\pi}{2}\right )&=&\pm \cos \left(x\right)&\;\;\;\;\text{(iv)}\\ \\ \cos\left(x \pm \dfrac{\pi}{2}\right )&=&\mp \mathrm{\,sen} \left(x \right)&\;\;\;\;\text{(v)}\\ \\ \mathrm{sen}\left(x\pm y \right )&=&\mathrm{sen}\left(x\right)\cdot \cos\left(y \right )\pm \cos\left(x \right )\cdot \mathrm{sen}\left(y \right )&\;\;\;\;\text{(vii)} \end{array}$



$\mathrm{sen}\left(17\pi-2\alpha \right )\cdot\mathrm{sen}\left(\dfrac{19\pi}{2}-\alpha \right)+\cos\left(\dfrac{21\pi}{2}+\alpha \right)\cdot\mathrm{sen}\left(\dfrac{23\pi}{2}+2\alpha \right)\\ \\ \\ =-\mathrm{sen}\left(2\alpha-17\pi \right )\cdot-\mathrm{sen}\left(\alpha-\dfrac{19\pi}{2} \right)+\cos\left(\alpha+\dfrac{21\pi}{2} \right)\cdot\mathrm{sen}\left(2\alpha+\dfrac{23\pi}{2} \right)\\ \\ \\ =\mathrm{sen}\left(2\alpha-\pi-8\cdot2\pi \right )\cdot\mathrm{sen}\left(\alpha-\dfrac{-\pi+5\cdot4\pi}{2} \right)+\cos\left(\alpha+\dfrac{\pi+5 \cdot 4\pi}{2} \right)\cdot\mathrm{sen}\left(2\alpha+\dfrac{-\pi+6 \cdot 4\pi}{2} \right)\\ \\ \\ =\mathrm{sen}\left(2\alpha-\pi \right )\cdot\mathrm{sen}\left(\alpha-\dfrac{-\pi}{2} \right)+\cos\left(\alpha+\dfrac{\pi}{2} \right)\cdot\mathrm{sen}\left(2\alpha-\dfrac{\pi}{2} \right)$

$=-\mathrm{sen}\left(2\alpha \right )\cdot\mathrm{sen}\left(\alpha+\dfrac{\pi}{2} \right)+\cos\left(\alpha+\dfrac{\pi}{2} \right)\cdot\mathrm{sen}\left(2\alpha-\dfrac{\pi}{2} \right)\\ \\ \\ =-\mathrm{sen}\left(2\alpha \right )\cdot\cos\left(\alpha \right)-\mathrm{sen}\left(\alpha \right)\cdot-\cos\left(2\alpha \right)\\ \\ \\ =\mathrm{sen}\left(\alpha \right )\cdot\cos\left(2\alpha \right )-\cos\left(\alpha \right )\cdot \mathrm{sen}\left(2\alpha \right )\\ \\ \\ =\mathrm{sen}\left(\alpha-2\alpha \right )\\ \\ \\ =\mathrm{sen}\left(-\alpha \right )\\ \\ \\ =\boxed{-\mathrm{sen\,}\alpha}$


Resposta: alternativa a).