Question

Calcule aproximadamente sin 61° aplicando o diferencial.​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Diferencial}\\\sf f(x\pm\Delta~\!x)=f(x)\pm f'(x)\cdot dx\\\sf sen(61^\circ)=sen (45^\circ+16^\circ)\\\sf x=45^\circ~\Delta~\! x=16^\circ~~dx=16^\circ\\\sf f(x)= sen(45^\circ)\implies f'(x)=cos(45^\circ)\\\sf f(45^\circ+16^\circ)= sen(45^\circ)+ cos(45^\circ)\cdot 16^\circ\\\sf 16^\circ=\dfrac{16}{180}\pi~rad=\dfrac{4\pi}{45}\\\sf f(45^\circ+16^\circ)=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{4\pi}{45}\\\\\sf f(45^\circ+16^\circ)=0,707+0,707\cdot0,27\\\sf f(45^\circ+16^\circ)=0,707(1+0,27)\\\sf f(45^\circ+16^\circ)=0,707\cdot1,27\\\sf f(45^\circ+16^\circ)=0,89\\\sf sen(61^\circ)\approxeq 0,89\end{array}}$