Question

Se tg x=t, então cos 2x + sen 2x é equivalente a?

Answer 1

 Olá Cristiano,

 Da expressão,

$\cos(2x)+\sin(2x)=\\\cos(x+x)+\sin(x+x)=\\\cos\,x\cdot\cos\,x-\sin\,x\cdot\sin\,x+\sin\,x\cdot\cos\,x+\cos\,x\cdot\sin\,x=\\\cos^2x-\sin^2x+2\cdot\sin\,x\cdot\cos\,x=$

 Do enunciado,

$\tan\,x=t\\\\\frac{\sin\,x}{\cos\,x}=t\\\\\frac{\sin^2x}{\cos^2x}=t^2\\\\\boxed{\sin^2x=t^2\cdot\cos^2x}$


 Sabe-se que, $\sin^2x+\cos^2x=1$

 Segue que,

$\sin^2x+\cos^2x=1\\t^2\cdot\cos^2x+\cos^2x=1\\\cos^2x(t^2+1)=1\\\boxed{\cos^2x=\frac{1}{t^2+1}}$


 Continuando da expressão (lá em cima)...

$\cos(2x)+\sin(2x)=\\\cos(x+x)+\sin(x+x)=\\\cos\,x\cdot\cos\,x-\sin\,x\cdot\sin\,x+\sin\,x\cdot\cos\,x+\cos\,x\cdot\sin\,x=\\\cos^2x-\sin^2x+2\cdot\sin\,x\cdot\cos\,x=\\\cos^2x-t^2\cdot\cos^2x+2\cdot(t\cdot\cos\,x)\cdot\cos\,x=\\\cos^2x-t^2\cdot\cos^2x+2t\cdot\cos^2x=\\\cos^2x(1-t^2+2t)=\\\\\frac{1}{t^2+1}\cdot(1-t^2+2t)=\\\\\boxed{\boxed{\frac{-t^2+2t+1}{t^2+1}}}$