Ensino Médio ⇒ Trigonometria Tópico resolvido

[Valdilenex]

Simplificando-se a expressão
[tex3]\frac{cos (pi-X). cos(pi/2 + X)}{sen(pi+ X)}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{cos (pi-X). cos(pi/2 + X)}{sen(pi+ X)}[/tex3]

Onde X é diferente k.[tex3]\pi [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\pi [/tex3]

e k e Z, obtem-se
resposta :-cosx

[petras]

Valdilenex,
[tex3]\mathsf{cos (\pi - x)}=-cos x\\
sen(\pi+x) =-sen x = -cos (\frac{\pi}{2}-x)\\
cos(\frac{\pi}{2}+x) = -cos(\frac{\pi}{2}-x)\\
\frac{cos(\pi-x)\cdot(cos(\frac{\pi}{2}+x)}{sen(\pi+x)}=\frac{-cosx \cdot\cancel{(-cos(\frac{\pi}{2}-x)}} {\cancel{-cos(\frac{\pi}{2}-x)}}=\boxed{\color{red}-cosx}\\
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{cos (\pi - x)}=-cos x\\
sen(\pi+x) =-sen x = -cos (\frac{\pi}{2}-x)\\
cos(\frac{\pi}{2}+x) = -cos(\frac{\pi}{2}-x)\\
\frac{cos(\pi-x)\cdot(cos(\frac{\pi}{2}+x)}{sen(\pi+x)}=\frac{-cosx \cdot\cancel{(-cos(\frac{\pi}{2}-x)}} {\cancel{-cos(\frac{\pi}{2}-x)}}=\boxed{\color{red}-cosx}\\
[/tex3]

[Valdilenex]

Não entendi como cosseno de pi -x virou -cos x nem as outras transformações

Valdilenex,
[tex3]\mathsf{cos (\pi - x)}=-cos x\\
sen(\pi+x) =-sen x = -cos (\frac{\pi}{2}-x)\\
cos(\frac{\pi}{2}+x) = -cos(\frac{\pi}{2}-x)\\
\frac{cos(\pi-x)\cdot(cos(\frac{\pi}{2}+x)}{sen(\pi+x)}=\frac{-cosx \cdot\cancel{(-cos(\frac{\pi}{2}-x)}} {\cancel{-cos(\frac{\pi}{2}-x)}}=\boxed{\color{red}-cosx}\\
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{cos (\pi - x)}=-cos x\\
sen(\pi+x) =-sen x = -cos (\frac{\pi}{2}-x)\\
cos(\frac{\pi}{2}+x) = -cos(\frac{\pi}{2}-x)\\
\frac{cos(\pi-x)\cdot(cos(\frac{\pi}{2}+x)}{sen(\pi+x)}=\frac{-cosx \cdot\cancel{(-cos(\frac{\pi}{2}-x)}} {\cancel{-cos(\frac{\pi}{2}-x)}}=\boxed{\color{red}-cosx}\\
[/tex3]

[petras]

Valdilenex,
[tex3]\mathsf{cos(a-b)=cosa.cosb+sena.senb\\
cos(π-x)=cos(π).cos(x)+sen(π).sen(x)\\
cos(π-x)=-1.cos(x)+0.sen(x)\\
\boxed{\color{red}cos(π-x)=-cos(x)}\\
cos(a+b)=cosa.cosb-sena.senb\\
cos(\frac{\pi}{2}+x)=cos(\frac{\pi}{2}).cos(x)-sen(\frac{\pi}{2}).sen(x)\\
cos(\frac{\pi}{2}+x)=0.cos(x)-1.sen(x)\\
\boxed{\color{red}cos(\frac{\pi}{2}+x)=-sen(x)}\\
sen(a+b) = senacosb+senbcosa\\
sen(x+\pi) = senx cos\pi+sen\pi cosx\\
sen(\pi+x) = -senx.(-1)+0.cosx\\
\boxed{\color{red}sen(x+\pi)=-senx}\rightarrow cox ~e ~sen x ~são~arcos ~complementares\therefore \boxed{\color{red}sen(x+\pi)=-cos(\frac{\pi}{2}-x)}\\
\boxed{\color{red}-cos(\frac{\pi}{2}-x)=cos(\frac{\pi}{2}+x)}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{cos(a-b)=cosa.cosb+sena.senb\\
cos(π-x)=cos(π).cos(x)+sen(π).sen(x)\\
cos(π-x)=-1.cos(x)+0.sen(x)\\
\boxed{\color{red}cos(π-x)=-cos(x)}\\
cos(a+b)=cosa.cosb-sena.senb\\
cos(\frac{\pi}{2}+x)=cos(\frac{\pi}{2}).cos(x)-sen(\frac{\pi}{2}).sen(x)\\
cos(\frac{\pi}{2}+x)=0.cos(x)-1.sen(x)\\
\boxed{\color{red}cos(\frac{\pi}{2}+x)=-sen(x)}\\
sen(a+b) = senacosb+senbcosa\\
sen(x+\pi) = senx cos\pi+sen\pi cosx\\
sen(\pi+x) = -senx.(-1)+0.cosx\\
\boxed{\color{red}sen(x+\pi)=-senx}\rightarrow cox ~e ~sen x ~são~arcos ~complementares\therefore \boxed{\color{red}sen(x+\pi)=-cos(\frac{\pi}{2}-x)}\\
\boxed{\color{red}-cos(\frac{\pi}{2}-x)=cos(\frac{\pi}{2}+x)}}[/tex3]