Last minute explanations on Trig Identities.

I don’t know how to do 1–3 and 5 or maybe I forgot, but I can help with 4.
THE ZERO WITH THE BELT ACROSS=X
4. (1-cotX)^2 = csc^2 X -2cotX
(1-cotX)(1-cotX) = csc^2 X -2cotX
1-cot -cot +cot^2 X = csc^2 X -2cotX
You rearrange the first part of the equation.
(1+ cot^2 X) -cot -cot = (csc^2 X) -2cotX
Csc^2 X -2 cotX = csc^2 X -2cotX
Also, did you take notes? Can you use your notes or cheat-sheets on your final? You can always wake up early tomorrow to ask your teacher for help or stay after school…

You have the right idea for 1, but cos 60 = .5.

For 2, just divide both sides by cosx to get 2sinx=1. You should be able to handle it from there.

For 3, first find arc tan of 1.2. That gives 2 theta. Then divide by 2 to get theta.

4 was done by @cyndihugs

For 5, start by expanding the divisions to get
sin/cos + 1 – 1 + cos/sin = sin/cos + cos/sin
The common denominator is sincos, giving
(sin^2+cos^2))/sincos = 1/sincos

Let me add:

for 1, remember to divide your 360n by 2 as well.
for 2, cosx could be 0, and in that case you can’t divide it out. And when is cosx=0? That’ll give you another solution.