# Math Help - need help - 3 various problems

1. ## need help - 3 various problems

Hey guys, Im sorry to bothered you but Im kinda desperate and really need your help.

Im preparing for math exam and Ive got a 3 various problem
I know the right answers but Im not able to find out the method. I would really appreciate your help.

1. if cos x = -2/5 than cos 2x = ?? (i know it is -17/25)
or sin x = 3/5 and cos2x= ?? (7/25) but I have no clue how to do it.

2. the number of X (X is from (pi, 2pi) which is valid for sin^2x-sinx=0
(it is 0)

or X is from (0, 2pi) for sin^3x+sinx=0
(it is 1)

3. the last one
the imaginary part of the number
(-1 -i)^8 is (0)
or (1+i)^8 is (16)
(-2 -2i)^8 is (2^12)
(1+i)^32 is (0)

Much appreciate your help and time!

2. Originally Posted by mika619
Hey guys, Im sorry to bothered you but Im kinda desperate and really need your help.

Im preparing for math exam and Ive got a 3 various problem
I know the right answers but Im not able to find out the method. I would really appreciate your help.

1. if cos x = -2/5 than cos 2x = ?? (i know it is -17/25)
or sin x = 3/5 and cos2x= ?? (7/25) but I have no clue how to do it.
Use $\cos(2x) = 2 \cos^2x-1 = 1-2 \sin^2x$

Originally Posted by mika619
2. the number of X (X is from (pi, 2pi) which is valid for sin^2x-sinx=0
(it is 0)

or X is from (0, 2pi) for sin^3x+sinx=0
(it is 1)
Factor out $\sin x$

Originally Posted by mika619
3. the last one
the imaginary part of the number
(-1 -i)^8 is (0)
or (1+i)^8 is (16)
(-2 -2i)^8 is (2^12)
(1+i)^32 is (0)

Much appreciate your help and time!
$-1-i = \sqrt{2} \: e^{i\frac{5\pi}{4}}$

3. ok, thanks for you help, anyway im dumb as hell so it wasn't actually really usefull for me...
update: OK, it was helpful I got the first two... the 3th one is pure mess.

I just found another I guess it is pretty simple but I can't solve it...

$
\frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} \$

4. Originally Posted by mika619
ok, thanks for you help, anyway im dumb as hell so it wasn't actually really usefull for me...
update: OK, it was helpful I got the first two... the 3th one is pure mess.
$-1-i = \sqrt{2} \: e^{i\frac{5\pi}{4}}$

$(-1-i)^8 = \left(\sqrt{2} \: e^{i\frac{5\pi}{4}}\right)^8 = 16 \: e^{i\frac{8\times5\pi}{4}} = 16$

Originally Posted by mika619
I just found another I guess it is pretty simple but I can't solve it...

$
\frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} \$

Since $\sqrt{3} \leq \sqrt{5}$

$\frac{3 \sqrt{5} \ - 5 \sqrt{3}}{|\sqrt{3}-\sqrt{5}|} = \frac{3 \sqrt{5} \ - 5 \sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Multiply numerator and denominator by $\sqrt{5}+\sqrt{3}$

Expand and simplify

5. I had the test today... I got 85% Anyway could you solve this for me
(1+i)^33 I still don't have a clue. the options were I guess
a)2^16
b)2^7
c)-2^16
d)-2^17
e)none of them
thx

(1 + i)^33 = 65 536 + 65 536 i

6. Hello, mika619!

Recall that: . $i^4 \:=\:+1$

( $(1+i)^{33} \:=$

. . $(a)\;2^{16} \qquad (b)\;2^7 \qquad(c)\;2^{16} \qquad (d)\;-2^{17 } \qquad (e)\;\text{none of these}$

Note that: . $(1 + i)^2 \;=\;1 + 2i + i^2 \;=\;1 + 2i - 1$

. . So we have: . $(1+i)^2 \;=\;2i$

Raise both sides to the 16th power:

. . $\bigg[(1+i)^2\bigg]^{16} \;=\;(2i)^{16}$

. . $(1+i)^{32} \;=\;2^{16}\cdot i^{16} \;=\;65,536\,(i^4)^4 \;=\;65,536\,(+1)^4$

And we have: . $(1+i)^{32} \;=\;65,\!536$

Multiply both side by $(1+i):$

. . $(1+i)^{32}\cdot(1+i) \;=\;65,\!536\,(1+i)$

Therefore: . $(1+i)^{33} \;=\;65,\!536 + 65,\!536i$

7. Hello, mika619!

Does #3 go like this?

3. Find the imaginary part of these numbers:

. . $(a)\;(-1 -i)^8 \qquad (b)\;(1+i)^8 \qquad (c)\;

All four of these turn out to be real numbers,
. . so their imaginary compoenents are all zero.

$(a)\;(-1-i)^8$

We find that: . $(-1-i)^2 \:=\:1 + 2i + i^2 \:=\:1 + 2i - 1$

. . So we have: . $(-1 - i)^2 \:=\:2i$

Raise both sides to the 4th power:

. . $\bigg[(1+i)^2\bigg]^4 \;=\;\bigg[2i\bigg]^4$

. . $(-1-i)^8 \;=\;2^4\cdot i^4 \;=\;16\cdot1$

Therefore: . $(-1-i)^8 \;=\;16$

$(c)\;(-2-2i)^8$

We find that: . $(-2-2i)^2 \;=\;4 + 8i + 4i^2 \;=\;4 + 8i - 4$

. . So we have: . $(-2-2i)^2 \;=\;8i$

Raise both sides to the 4th power:

. . $\bigg[(-2-2i)^2\bigg]^4 \;=\;\bigg[8i\bigg]^4$

. . $(-2-2i)^7 \;=\;8^4\cdot i^4 \;=\;4096\cdot1$

Therefore: . $(-2-2i)^8 \;=\;4096$

Get the idea?