i am at wits end here?

Question

i'm on the final 4 of ever doing trig functions again & i cannot figure these out for the lifeof me, help!!


1. 4tan^4x + tan^2x -3 = sec^2(4tan^4-3)

2. 1+sinx/cox + cosx/1+sinx = 2secx

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

4. sinA/sinA+cosA = tanA / 1+tanA

Answer 1

Try factoring the first: (4tan^2x - 3)(tan^2x + 1)
tan^2(x) + 1 = (sin^2x + cos^2x)/cos^2x = sec^2x, which directly gives you equation 1.

2: straightforward, put over common factor (please use parens when you do this, it makes it much easier to understand)

((1 + sinx)(1 + sinx) + cos^2x)/((1 + sinx)cosx)

3: above has done it right, I won't bother

4: sinx/(sinx + cosx) = tanx/(1 + tanx)
This is easy - just multiply the right by cosx/cosx,:

(tanx*cosx)/((1 + tanx)cosx)
= sinx/(cosx + sinx)

If you get stuck on trig problems, always turn everything into sin's and cos's, it makes them much easier to see.
= (1 + 2sinx + sin^2x + cos^2x)/((1 + sinx)cosx)
= (2 + 2sinx)/((1 + sinx)cosx)
= 2/cosx

Answer 2

I am so sorry, I hated trig proofs and those are some incredibly heinous ones. If you have a TI-89, use the SMG (Symbolic Math Guide) and do a step-by-step transformation until you get to the right answer.

Answer 3

#3 isn't too bad.

cos^3x*sin^2x = cosx*cos^2x*sin^2x = cos x (1-sin^2(x))*sin^2(x)
= cos x (sin^2(x) - sin^4(x)) = (sin^2(x) - sin^4(x))cos x

Answer 4

Ive never been there but I hear the view is great.


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Answer 5

Do your own homework.