Question

(Ufal) Se x e y são tais que 0≤ x ≤ π\2, 0 ≤ y π\2, sen y = 3/5 e cos x = 3/4, então cos (x-y) é igual a:

a)9-9√7/20
b)9+√7/10
c)3(4+√7)/4
d)3(4-√7)/20
e)3(4+√7)/10

Answer 1

$\mathrm{\cos{x}=\dfrac{3}{4}\ \ \| \ \ \sin{y}=\dfrac{3}{5}\ \ \| \ \ cos(x-y)=\ ?}\\\\\\ \mathbf{Calculando\ \sin{x}:}\\\\ \mathrm{\sin^2{x}+cos^2{x}=1\ \to\ \sin{x}=\pm\sqrt{1-\cos^2{x}}=}\\\\ \mathrm{=\pm\sqrt{1-\bigg(\dfrac{3}{4}\bigg)^2}=\pm\sqrt{\dfrac{16}{16}-\dfrac{9}{16}}=\pm\sqrt{\dfrac{7}{16}}=\pm\dfrac{\sqrt{7}}{4}}\\\\ \mathrm{Como\ 0\ \textless \ x\ \textless \ \dfrac{\pi}{2}\ \to\ \sin{x}=\dfrac{\sqrt{7}}{4}}$

$\mathbf{Calculando\ \cos{y}:}\\\\ \mathrm{\sin^2{y}+\cos^2{y}=1\ \to\ \cos{y}=\pm\sqrt{1-\sin^2{y}}=}\\\\ \mathrm{=\pm\sqrt{1-\bigg(\dfrac{3}{5}\bigg)^2}=\pm\sqrt{\dfrac{25}{25}-\dfrac{9}{25}}=\pm\sqrt{\dfrac{16}{25}}=\pm\dfrac{4}{5}}\\\\ \mathrm{Como\ 0\ \textless \ x\ \textless \ \dfrac{\pi}{2}\ \to\ \cos{y}=\dfrac{4}{5}}$

$\mathbf{Calculando\ \cos{(x-y)}}:}\\\\ \mathrm{\cos{(x-y)}=\cos{x}.\cos{y}+\sin{x}.\sin{y}=}\\\\ \mathrm{=\dfrac{3}{4}.\dfrac{4}{5}+\dfrac{\sqrt{7}}{4}.\dfrac{3}{5}=\dfrac{12}{20}+\dfrac{3\sqrt{7}}{20}=\mathbf{\dfrac{3(4+\sqrt{7})}{20}}}$