sendo x um angulo agudo, e seno de x = 1/3 calcule cosseno de x e tangente de x
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Usaremos a relação fundamental da trigonometria;
![sen^2x+cos^2x=1 \\
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cosx=\sqrt{1-sen^2x} \\
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cosx=\sqrt{1-(\frac{1}{3})^2} \\
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cosx=\sqrt{1-\frac{1}{9}} \\
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cosx=\sqrt{\frac{8}{9}} \\
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cosx=\frac{2\sqrt2}{3} sen^2x+cos^2x=1 \\
\\
cosx=\sqrt{1-sen^2x} \\
\\
cosx=\sqrt{1-(\frac{1}{3})^2} \\
\\
cosx=\sqrt{1-\frac{1}{9}} \\
\\
cosx=\sqrt{\frac{8}{9}} \\
\\
cosx=\frac{2\sqrt2}{3}](https://tex.z-dn.net/?f=sen%5E2x%2Bcos%5E2x%3D1++%5C%5C%0A%5C%5C%0Acosx%3D%5Csqrt%7B1-sen%5E2x%7D++%5C%5C%0A%5C%5C%0Acosx%3D%5Csqrt%7B1-%28%5Cfrac%7B1%7D%7B3%7D%29%5E2%7D+%5C%5C%0A%5C%5C%0Acosx%3D%5Csqrt%7B1-%5Cfrac%7B1%7D%7B9%7D%7D+%5C%5C%0A%5C%5C%0Acosx%3D%5Csqrt%7B%5Cfrac%7B8%7D%7B9%7D%7D+%5C%5C%0A%5C%5C%0Acosx%3D%5Cfrac%7B2%5Csqrt2%7D%7B3%7D)
![tgx=\frac{senx}{cosx}=\frac{\frac{1}{3}}{\frac{2\sqrt2}{3}}=\frac{1}{3}.\frac{3}{2\sqrt2}=\frac{1}{2\sqrt2} tgx=\frac{senx}{cosx}=\frac{\frac{1}{3}}{\frac{2\sqrt2}{3}}=\frac{1}{3}.\frac{3}{2\sqrt2}=\frac{1}{2\sqrt2}](https://tex.z-dn.net/?f=tgx%3D%5Cfrac%7Bsenx%7D%7Bcosx%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%7D%7B%5Cfrac%7B2%5Csqrt2%7D%7B3%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D.%5Cfrac%7B3%7D%7B2%5Csqrt2%7D%3D%5Cfrac%7B1%7D%7B2%5Csqrt2%7D)
zaiden18:
cosseno de x vai dar 1/3
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