Question

someone help with an arithmetic for trig.

Answer 1

yes.

Answer 2

\[\sqrt{2}^{7}[(-\sqrt{2}/2)+(-\sqrt{2}/2)i]. How do you get -(\sqrt{2})^{6}-(\sqrt{2})^{6}i\]

Answer 3

Having a brain fart, so I don't know how you get that answer. :X

Answer 4

\(\Large\sqrt{2^7}[\frac{\sqrt{2}}{2}+\frac{-\sqrt{2}}{2}i]\)

like that?

Answer 5

distribute!

Answer 6

Yes ^ to studygurl

Answer 7

forgot to put the -

Answer 8

Okay, so first step it to distribute the \(\large\sqrt{2^7}\) as @freckles said

Answer 9

\[(\sqrt{2})^7 \cdot (\sqrt{2})^1=(\sqrt{2})^8=(\sqrt{2})^6 \cdot (\sqrt{2})^2\]
guess what (sqrt(2))^2=?

Answer 10

\[its \sqrt{2}^{7}. \sqrt{2} \to the 7th power\]

Answer 11

square root of 2 to the 7th power****

Answer 12

@freckles is asking what \(\large(\sqrt{2})^2\) equals

Answer 13

equals to 2?

Answer 14

Correct.

Answer 15

What's next?

Answer 16

you have \(\large\sqrt{2^8}\)
\(\large\sqrt{2^8}\rightarrow\sqrt{2^22^22^22^2}\rightarrow\sqrt{2^2}\sqrt{2^2}\sqrt{2^2}\sqrt{2^2}\rightarrow?\)

Answer 17

equals to 16*

Answer 18

correct

Answer 19

Now you have
\(\Large\frac{16}{2}+\frac{-16}{2}i\)

Can you do the rest?

Answer 20

Yea, I can see the picture now. Thanks!

Answer 21

:)