Sabendo que tg x= 2 e que
,determine o valor de sen x e cos x
Soluções para a tarefa
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Pela fórmula da tangente, obtemos:
![\tan x=\dfrac{\sin x}{\cos x} \tan x=\dfrac{\sin x}{\cos x}](https://tex.z-dn.net/?f=%5Ctan+x%3D%5Cdfrac%7B%5Csin+x%7D%7B%5Ccos+x%7D)
![2=\dfrac{\sin x}{\cos x} 2=\dfrac{\sin x}{\cos x}](https://tex.z-dn.net/?f=2%3D%5Cdfrac%7B%5Csin+x%7D%7B%5Ccos+x%7D)
![\sin x=2\cos x \sin x=2\cos x](https://tex.z-dn.net/?f=%5Csin+x%3D2%5Ccos+x)
Utilizando o encontrado acima na fórmula
, temos:
![\sin^{2}x+cos^{2}x=1 \sin^{2}x+cos^{2}x=1](https://tex.z-dn.net/?f=%5Csin%5E%7B2%7Dx%2Bcos%5E%7B2%7Dx%3D1)
![(2\cos x)^{2}+cos^{2}x=1 (2\cos x)^{2}+cos^{2}x=1](https://tex.z-dn.net/?f=%282%5Ccos+x%29%5E%7B2%7D%2Bcos%5E%7B2%7Dx%3D1)
![4\cos^{2}x+cos^{2}x=1 4\cos^{2}x+cos^{2}x=1](https://tex.z-dn.net/?f=4%5Ccos%5E%7B2%7Dx%2Bcos%5E%7B2%7Dx%3D1)
![5\cos^{2}x=1 5\cos^{2}x=1](https://tex.z-dn.net/?f=5%5Ccos%5E%7B2%7Dx%3D1)
![\cos^{2}x=\dfrac{1}{5} \cos^{2}x=\dfrac{1}{5}](https://tex.z-dn.net/?f=%5Ccos%5E%7B2%7Dx%3D%5Cdfrac%7B1%7D%7B5%7D)
![\cos x=\pm\sqrt{\dfrac{1}{5}}=\pm\dfrac{1}{\sqrt{5}} \cos x=\pm\sqrt{\dfrac{1}{5}}=\pm\dfrac{1}{\sqrt{5}}](https://tex.z-dn.net/?f=%5Ccos+x%3D%5Cpm%5Csqrt%7B%5Cdfrac%7B1%7D%7B5%7D%7D%3D%5Cpm%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D)
Agora, calculando
, temos:
![\sin x=2\cos x \sin x=2\cos x](https://tex.z-dn.net/?f=%5Csin+x%3D2%5Ccos+x)
![\sin x=2\times(\pm\dfrac{1}{\sqrt{5}}) \sin x=2\times(\pm\dfrac{1}{\sqrt{5}})](https://tex.z-dn.net/?f=%5Csin+x%3D2%5Ctimes%28%5Cpm%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%29)
![\sin x=\pm\dfrac{2}{\sqrt{5}} \sin x=\pm\dfrac{2}{\sqrt{5}}](https://tex.z-dn.net/?f=%5Csin+x%3D%5Cpm%5Cdfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D)
Obs: O seno e o cosseno terão os mesmos sinais
Utilizando o encontrado acima na fórmula
Agora, calculando
Obs: O seno e o cosseno terão os mesmos sinais
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