<p>How do you do this problem:

if i^2 = -1 and if ((i^2)^3)^k = 1, the the least positive interger value of k is..</p>

<p>1,2,4,6,8

a,b,c,d,e</p>

<p>Also, what is the best book that has practice tests for MathIC that are realistic?</p>

<p>How do you do this problem:

if i^2 = -1 and if ((i^2)^3)^k = 1, the the least positive interger value of k is..</p>

<p>1,2,4,6,8

a,b,c,d,e</p>

<p>Also, what is the best book that has practice tests for MathIC that are realistic?</p>

<p>another question.

(sin^2(theta)+cos^2(theta)-3)^4 =

256,81,64,32,16

a,b,c,d,e</p>

<p>Thanks</p>

<p>2nd question- 16 I believe, just plug in any value for theta. or use the law that sin squared theta + cos squared theta =1. </p>

<p>1st question- 2 i believe. plug in -1 for i^2, -1^3= -1. so the question becomes (-1)^k=1, so if you plug in the answers 2 is the lowest value that works.</p>

<p>Most realistic book is probably Princeton Review's Cracking the Math I and Math II Subject Tests. As it has been said before, it will suffice for a 700+ if you use it well. If you're aiming for an 800, Barron's is usually highly recommended just because its practice tests are so hard; so is RUSH.</p>

<p>Question 1: i is the square root of -1 (you should know that already, but you can easily see that if i^2 is -1, i is the square root of -1)</p>

<p>Anyway... (i^2)^3 = i^6 (i^6)^k = i^6k

Now, to get i^6k to equal 1, we need to see what exponents get i to equal 1

i^4 = 1 (either you know that or you can simply do i^2 times i^2 is i^4, and i^2 times i^2 is -1 times -1, which is 1, so i^4 = 1)</p>

<p>Therefore, multiples of 4 as the exponent will get us 1. Why? Because i to a multiple of 4 means we just have i^4 a certain number of times multiplied, meaning we're multiplying 1 by itself a certain number of times, giving us 1 as a product. If that's confusing, I'll give you an example. i^8 is i to a power that is a multiple of 4. 8 is a multiple of 4, and because of this, we simply have i^4 (which is 1) being multiplied by itself a certain number of times as you can see: i^8 = i^4 times i^4 = 1 x 1 = 1

6k must be a multiple of 4, and the smallest multiple of 4 that 6k can equal is clearly 12 (4 and 8 are too small for 6 to go into)

6k = 12

k = 2</p>

<p>Really doesn't take as long as my explanation</p>

<p>Question 2:

sin^2(theta) + cos^2(theta) = 1

This is a Pythagorean identity that you must know (easy to prove also)

So now that we have that, just subtract 3, giving us 1 -3 = -2

Raise that to the 4th power

(-2)^4 = 16</p>

<p>As far as the most accurate review book, of course College Board. Other than that I suppose Princeton. Barrons is much harder but very good at preparing you.</p>

<p>why did u subtract 3 after u got

sin2 +cos2 = 1</p>

<p>Well, look at the problem:

(sin^2(theta)+cos^2(theta)-3)^4

The sin^2(theta)+cos^2(theta) part in the parentheses is 1, but then you're subtracting 3 from it (in the parentheses)</p>