Matemática, perguntado por jhoiltoonfelipeneves, 6 meses atrás

atividade do livro Iezzi de trigonometria ​

Anexos:

Soluções para a tarefa

Respondido por niltonjunior20oss764
0

\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}}

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