1. ## Analytic Trig.

1. (sin2x + cos2x) ^2=1
solutions are in the interval [0, 2pi)

using double angle formulas find sin2u , cos2u, and tan2u in:

2. cot u = -6, 3pi/2< u < 2pi

rewrite the expression in terms of the first power of cos:
3. sin^4x*cos^2x

i get stuck in between theses problems, its usually the factoring...

2. using double angle formulas find sin2u , cos2u, and tan2u in:

2. cot u = -6, 3pi/2< u < 2pi
You see that $\displaystyle sin(2u)=sin(u+u)$. Then you just have to apply the well known formula $\displaystyle sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$.
For $\displaystyle cos(a+b)$, we have that it's equal to $\displaystyle cos(a)cos(b)-sin(a)sin(b)$.

3. Hello, >_<SHY_GUY>_<!

$\displaystyle 1.\;\;(\sin2x + \cos2x)^2\:=\:1,\quad x \in [0, 2\pi)$

Square it out: .$\displaystyle \underbrace{\sin^2\!2x + \cos^2\!2x}_{\text{This is 1}} + 2\sin2x\cos2x \:=\:1$

We have: .$\displaystyle 2\sin2x\cos2x\:=\:0 \quad\Rightarrow\quad \sin4x \:=\:0$

Hence: .$\displaystyle 4x\:=\:0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:5\pi,\:6\pi, \:7\pi$

Therefore: .$\displaystyle x \;=\;0,\:\frac{\pi}{4},\:\frac{\pi}{2},\:\frac{3\p i}{4},\:\pi,\:\frac{5\pi}{4},\:\frac{3\pi}{2},\:\f rac{7\pi}{4}$

4. Hello again, >_<SHY_GUY>_<!

$\displaystyle 2.\;\;\cot u \,=\,-6,\quad \frac{3\pi}{2} < u < 2\pi$

Using double angle formulas, find: .$\displaystyle \sin2u,\;\;\cos2u,\;\;\tan2u$

We are given: .$\displaystyle \cot u \:=\:-\frac{6}{1} \:=\:\frac{adj}{opp}\quad\text{ and }u\text{ is in Quadrant 4.}$

Then: .$\displaystyle opp = \text{-}1,\;\;adj = 6\;\;\text{ and }\:hyp = \sqrt{37}$

Hence: .$\displaystyle \sin u \:=\:-\frac{1}{\sqrt{37}},\quad \cos u \:=\:\frac{6}{\sqrt{37}}$

$\displaystyle \sin2u \:=\:2\sin u\cos u \:=\:2\left(-\frac{1}{\sqrt{37}}\right)\left(\frac{6}{\sqrt{37} }\right) \:=\:-\frac{12}{37}$

$\displaystyle \cos2u \:=\:\cos^2\!u - \sin^2\!u \:=\:\left(\frac{6}{\sqrt{37}}\right)^2 - \left(-\frac{1}{\sqrt{37}}\right)^2 \:=\:\frac{35}{37}$

$\displaystyle \tan2u \;=\;\frac{\sin2u}{\cos2u} \;=\;\frac{\text{-}\frac{12}{37}}{\frac{35}{37}} \;=\;-\frac{12}{35}$