1. 3 pretty easy questions

1.)

2.)

3.)

2. Originally Posted by stiitches
[3 images]
This looks a lot like you asking us to do your homework for you. What have you tried? Where are you stuck?

3. For the third one I did

1/(5/4) since cos=1/sec
so its cos(y)=4/5
then did cos(2y)=8/5
but it was wrong

for the other two, not sure how to start them and what the Qs are exactly asking.

4. Originally Posted by stiitches
For the third one I did

1/(5/4) since cos=1/sec
so its cos(y)=4/5
then did cos(2y)=8/5
but it was wrong

for the other two, not sure how to start them and what the Qs are exactly asking.
Thanks.

Q3, it's right that cos(y)=4/5, but the next step is not right because it's not generally true that cos(2*theta) = 2*cos(theta).

Use identity $\cos(2y) = \cos^2(y)-\sin^2(y) = 2\cos^2(y)-1 = 2\cdot(4/5)^2-1 =\ ?$

Q1 You need to know interval notation to see what form the answer is supposed to be in. And as for getting the answer you can do:

$x^3+3x<4x^2 \ \ \ \iff$

$x^3-4x^2+3x < 0\ \ \ \iff$

$x(x^2-4x+3) < 0 \ \ \ \iff$

$\;?$

Factor the quadratic and see where the inequality holds.

Q2 asks you to find some trig function values based on a single given value. You need some trig identities. One such is that $\sin^2(\theta)+\cos^2(\theta)=1$. Can you continue?

5. Originally Posted by stiitches
1.)
$x^3- 4x^2+ 3x= x(x- 3)(x- 4)$
and a product of numbers is positive if and only if an even number of its factors are negative.

2.)
csc is defined as "hypotenuse over opposite side" so you can imagine a right triangle with one vertex at (0, 0) on a coordinate system, right angle at (3, 0) and hypotenus of length 5. Where is the third vertex?

3.)
cos(2y)= cos^2(y)- sin^2(y). You are told what sin(y) is and, of course, cos(y)= 1/sec(y).