Question

atividade do livro Iezzi de trigonometria ​

Answer 1

$\boxed{\sin{x}=\dfrac{15}{17}}$

$\sin^2{x}+\cos^2{x}=1\ \to\ \cos{x}=\pm\sqrt{1-\sin^2{x}}$

$\cos{x}=\pm\sqrt{1-\dfrac{225}{289}}=\pm\sqrt{\dfrac{64}{289}}=\pm\dfrac{8}{17}$

$0<x<\dfrac{\pi}{2}\ \to\ \boxed{\cos{x}=\dfrac{8}{17}}$

$\tan{x}=\dfrac{\sin{x}}{\cos{x}}=\dfrac{\frac{15}{17}}{\frac{8}{17}}\ \to\ \boxed{\tan{x}=\dfrac{15}{8}}$

$\boxed{\sin{y}=-\dfrac{3}{5}}$

$\sin^2{y}+\cos^2{y}=1\ \to\ \cos{y}=\pm\sqrt{1-\sin^2{y}}$

$\cos{y}=\pm\sqrt{1-\dfrac{9}{25}}=\pm\sqrt{\dfrac{16}{25}}=\pm\dfrac{4}{5}$

$\pi<y<\dfrac{3\pi}{2}\ \to\ \boxed{\cos{y}=-\dfrac{4}{5}}$

$\tan{y}=\dfrac{\sin{y}}{\cos{y}}=\dfrac{-\frac{3}{5}}{-\frac{4}{5}}\ \to\ \boxed{\tan{y}=\dfrac{3}{4}}$

i.

$\boxed{\sin{\left(x\pm y\right)=\sin{x}\cos{y}\pm\sin{y}\cos{x}}}$

$\sin{(x+y)=\left(\dfrac{15}{17}\right)\left(-\dfrac{4}{5}\right)+\left(-\dfrac{3}{5}\right)\left(\dfrac{8}{17}\right)\ \therefore$

$\boxed{\sin{(x+y)}=-\dfrac{84}{85}}}$

ii.

$\boxed{\cos{(x\pm y)}=\cos{x}\cos{y}\mp\sin{x}\sin{y}}$

$\cos{(x+y)}=\left(\dfrac{8}{17}\right)\left(-\dfrac{4}{5}\right)-\left(\dfrac{15}{17}\right)\left(-\dfrac{3}{5}\right)\ \therefore$

$\boxed{\cos{(x+y)}=\dfrac{13}{85}}$

iii.

$\boxed{\tan{(x\pm y)}=\dfrac{\tan{x}\pm\tan{y}}{1\mp\tan{x}\tan{y}}}$

$\tan{(x+y)}=\dfrac{\frac{15}{8}+\frac{3}{4}}{1-\left(\frac{15}{8}\right)\left(\frac{3}{4}\right)}=\dfrac{\frac{21}{8}}{1-\frac{45}{32}}=-\dfrac{21}{8}\bigg(\dfrac{32}{13}\bigg)\ \therefore$

$\boxed{\tan{(x+y)}=-\dfrac{84}{13}}$