1.)http://imgur.com/fDL4B.jpg

2.) http://imgur.com/9RRb1.jpg

3.) http://imgur.com/m6ltM.jpg

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- Sep 6th 2010, 06:49 PMstiitches3 pretty easy questions
- Sep 6th 2010, 07:02 PMundefined
- Sep 6th 2010, 07:08 PMstiitches
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. - Sep 6th 2010, 07:20 PMundefined
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 $\displaystyle \cos(2y) = \cos^2(y)-\sin^2(y) = 2\cos^2(y)-1 = 2\cdot(4/5)^2-1 =\ ?$

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:

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

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

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

$\displaystyle \;?$

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 $\displaystyle \sin^2(\theta)+\cos^2(\theta)=1$. Can you continue? - Sep 7th 2010, 01:03 PMHallsofIvy
$\displaystyle x^3- 4x^2+ 3x= x(x- 3)(x- 4)$

and a product of numbers is positive if and only if an**even**number of its factors are negative.

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