se x está no terceiro quadrante e tang(x)=3/4 calcule o valor de cos(x)
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Solução!
![Tang(x)= \dfrac{3}{4}\\\\\\
\dfrac{sen(x)}{Cos(x)}= \dfrac{3}{4} \\\\\\\
4sen(x)=3cos(x) \\\\\\\
sen(x)= \dfrac{3cos(x)}{4}](https://tex.z-dn.net/?f=Tang%28x%29%3D+%5Cdfrac%7B3%7D%7B4%7D%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7Bsen%28x%29%7D%7BCos%28x%29%7D%3D+%5Cdfrac%7B3%7D%7B4%7D+%5C%5C%5C%5C%5C%5C%5C%0A%0A4sen%28x%29%3D3cos%28x%29+%5C%5C%5C%5C%5C%5C%5C%0A+sen%28x%29%3D+%5Cdfrac%7B3cos%28x%29%7D%7B4%7D+ "Tang(x)= \dfrac{3}{4}\\\\\\
\dfrac{sen(x)}{Cos(x)}= \dfrac{3}{4} \\\\\\\
4sen(x)=3cos(x) \\\\\\\
sen(x)= \dfrac{3cos(x)}{4}")
Relação fundamental.
![Sen^{2}(x)+cos^{2}x=1 \\\\\\\
\left (\dfrac{3cos(x)}{4} \right )^{2}+cos^{2}(x)=1\\\\\\
\dfrac{9cos^{2} (x)}{16} +cos^{2}(x)=1\\\\\\
9cosx^{2} (x)+16cos^{2}(x)=16\\\\\\
25cosx^{2}=16\\\\\\
cosx^{2}= \dfrac{16}{25}\\\\\\
cos(x)= \sqrt{\dfrac{16}{25} }\\\\\\\
cos(x)= \dfrac{4}{5}](https://tex.z-dn.net/?f=Sen%5E%7B2%7D%28x%29%2Bcos%5E%7B2%7Dx%3D1+%5C%5C%5C%5C%5C%5C%5C%0A%5Cleft+%28%5Cdfrac%7B3cos%28x%29%7D%7B4%7D+%5Cright+%29%5E%7B2%7D%2Bcos%5E%7B2%7D%28x%29%3D1%5C%5C%5C%5C%5C%5C++%0A%5Cdfrac%7B9cos%5E%7B2%7D+%28x%29%7D%7B16%7D+%2Bcos%5E%7B2%7D%28x%29%3D1%5C%5C%5C%5C%5C%5C%0A+9cosx%5E%7B2%7D+%28x%29%2B16cos%5E%7B2%7D%28x%29%3D16%5C%5C%5C%5C%5C%5C%0A+25cosx%5E%7B2%7D%3D16%5C%5C%5C%5C%5C%5C%0Acosx%5E%7B2%7D%3D+%5Cdfrac%7B16%7D%7B25%7D%5C%5C%5C%5C%5C%5C%0Acos%28x%29%3D+%5Csqrt%7B%5Cdfrac%7B16%7D%7B25%7D+%7D%5C%5C%5C%5C%5C%5C%5C%0Acos%28x%29%3D+%5Cdfrac%7B4%7D%7B5%7D+++ "Sen^{2}(x)+cos^{2}x=1 \\\\\\\
\left (\dfrac{3cos(x)}{4} \right )^{2}+cos^{2}(x)=1\\\\\\
\dfrac{9cos^{2} (x)}{16} +cos^{2}(x)=1\\\\\\
9cosx^{2} (x)+16cos^{2}(x)=16\\\\\\
25cosx^{2}=16\\\\\\
cosx^{2}= \dfrac{16}{25}\\\\\\
cos(x)= \sqrt{\dfrac{16}{25} }\\\\\\\
cos(x)= \dfrac{4}{5}")
Boa noite!
Bons estudos!
Solução!
Relação fundamental.
Boa noite!
Bons estudos!
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