URGENT: Trig Identities Help

[lolcano]

merci

 

[LivingMath]

That is a lot of questions for one post. Working out what your original question was and what your steps are is quite difficult because you don't group or use brackets well. It's normal to write cos A as cos(A) or something similar to make thing more clear. For fractions, it is best to write (1 - cos^2(A))/sin(A) so that it is clear what is part of the fraction and what is not.

csc A - cos A cot A = sin A

Yes

SOLVE: Given 0 <= A < 2pi
12 cos^2 A - cos A -6

Substituting y in gives 12y^2 - y - 6.

You get (2y+1)(y-1) which equals 2y^2 - y - 1, which isn't quite right.

SOLVE given 0<= A < 2pi
cos 2 A (2 cos A + 1)

Okay, so via double angle identities I know that the cos 2 a can equal 2 cos^2 A - 1

2 cos^2 A - 1(2 cos x + 1)

From here, where do I go? Or do I need to factor this part to simplify the cos^2?

Yes, factor.

Verify:
sin A/csc A + cos A/sec A = sin A csc A

Subtract sin(a)/csc(a) from both sides first to get the answer more easily.

You are correct that this is not a valid identity.

This one I think I get:
Verify cos (360-x) = cos x

I think you get this one too.

Simplify: cos (x-pi/2) =
...
sin x

Your math looks good.

This one can also be solved just by thinking about the graph of sin and cos.

 

[BigGlenntheHeavy]

\(\displaystyle If \ you \ want \ help \ here, \ then \ have \ the \ decency \ to \ express \ whatever \ is \ causing \ your\)

\(\displaystyle angst \ in \ an \ appropriate \ manner.\)

\(\displaystyle For \ example \ csc(A)-cos(A)cot(A) \ is \ a \ statement, \ and \ tells \ me \ nothing.\)

\(\displaystyle You \ could \ just \ as \ well \ say \ solve \ Johnny \ has \ five \ marbles.\)