This units really killing me, I have a hard time remembering every little thing :/

I have a lot of questions so just help me out on whatever you can, if you will.

Verify that the x-values are solutions:

1. 3tan^2 (2x)-1 = 0

2. 2 sin^2 (x) - sin x - 1

Solve:

1. 3 Sec^2 (x) - 4
2. sin^2(x) = 3cos^2 (x)

Find all solutions:

1. cos^3(x)=cos(x) [I have this one but my answers don't match the book]
2. sec^2(x)-sec(x) - 2
3. cos (x/2) = Root 2 / 2
4. 2 sec^2(x) + tan^2(x) - 3

Verify the identity:

1. Sin^1/2(x) Cos(x) - sin^5/2(x) Cos(x) = Cos^3(x) [Square root of sin(x)]
2. Cos(-x)/(1+sin[-x]) = sec (x) + tan (x)

You may want to write this one out :/
3. [Sin(x) Cos(Y) + Cos(x) Sin(y)] / [Cos(x) Cos (Y) - Sin(X) Sin(Y)] = [Tan(x) + Tan(y)] / [1-Tan(x) Tan(y)]
4. [Tan(x) + Cot(Y)] / [Tan(x) Cot(Y)] = Tan (Y) + Cot(x)

Verify the identity algebraically:
1. 2 + cos^2(x) = 3Cos^4(x) = sin^2(x)(2 + 3cos^2 x)
2. Sin X / (1-cos X) = (1 + cos X) / sin X

Use logs and trig identities to verify the identity:
1. -ln(1 + cos x) = ln(1 - cos x) - 2 ln |sin x|

This post was edited by drhong on Mar 31 2014 09:37pm

only one i'll respond to since you have the answer that differs from the book

consider y^3 = y, and what values of y makes that true. off the top of my head, y = -1, 0, 1. so now let y = cos(x), and solve cos(x) = {-1, 0, 1}.

This post was edited by carteblanche on Mar 31 2014 10:09pm

I'll be working on the left hand side only, which is:

[Sin(x) Cos(Y) + Cos(x) Sin(y)] / [Cos(x) Cos (Y) - Sin(X) Sin(Y)]

The numerator is well known identity:
Sin(x) Cos(Y) + Cos(x) Sin(y) = Sin(x+y)

So is the denominator:
Cos(x) Cos(Y) - Sin(X) Sin(Y) = Cos(x+y)

Which gives us:
Sin(x+y)/Cos(x+y)

Which is clearly:
Tan(x+y)

Which is also an identity:
Tan(x+y) = [Tan(x) + Tan(y)]/[1-Tan(x)Tan(y)]

Which makes our left hand side:
[Tan(x) + Tan(y)]/[1-Tan(x)Tan(y)]

Which matches the right hand side exactly (which we never touched).

reminds me of a quote:
-Ramamurti Shankar, Fundamentals of Physics.