Question

[2(cos35°+isin35°)] ^-12

Answer 1

Надеюсь почерк понятный

Answer 2

To simplify the expression $[2(\cos{35^\circ}+i\sin{35^\circ})]^{-12}$, we can use De Moivre's theorem which states that $(\cos{\theta}+i\sin{\theta})^n = \cos{(n\theta)}+i\sin{(n\theta)}$.

In this case, we have $n = -12$ and $\theta = 35^\circ$. So,

$$\begin{align*} [2(\cos{35^\circ}+i\sin{35^\circ})]^{-12} &= 2^{-12} (\cos{(-12 \cdot 35^\circ)}+i\sin{(-12 \cdot 35^\circ)}) \\ &= 2^{-12} (\cos{-420^\circ}+i\sin{-420^\circ}). \end{align*}$$

Since the cosine and sine functions have a period of $360^\circ$, $\cos{-420^\circ} = \cos{(-420^\circ + 6 \cdot 360^\circ)} = \cos{300^\circ}$ and $\sin{-420^\circ} = \sin{(-420^\circ + 6 \cdot 360^\circ)} = \sin{300^\circ}$.

Now, $300^\circ$ is in the fourth quadrant, where both cosine and sine are negative. So,

$$\cos{300^\circ} = -\frac{\sqrt{3}}{2} \quad \text{and} \quad \sin{300^\circ} = -\frac{1}{2}.$$

Substituting these values back into the expression:

$$2^{-12} \left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right).$$

Finally, $2^{-12} = \frac{1}{2^{12}} = \frac{1}{4096}$, so the simplified expression is:

$$\frac{1}{4096} \left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right).$$

If you'd like a decimal approximation, you can calculate the numerical value of this expression.