1. ## pre-calc/trig

My teacher always confuses me when she throws T into the mix. i tried replacing with X, didnt help my problem either. i saw in a thread by yovivo, that the help posted could somewhat be applied. so i tried for the first one how the identities might help. kinda went off and got lost again... so im just exhausted with these four problems. thanks for any help..

Directions: Find all values for t in the interval [0,2PI]
1. cot t = csc t
2. sin^2 t - cos^2 t= 1
3.|csct|=2
4. sin^2 t =3/4
Juan

2. 1. $\cot{t} = \csc{t}$

$\Rightarrow \frac{\cos{t}}{\sin{t}} = \frac{1}{\sin{t}}$

$\Rightarrow \cos{t} \sin{t} =\sin{t}$

$\Rightarrow \sin{t}( \cos{t} - 1) = 0$

Find values of t for $\sin{t} = 0$ and $\cos{t} = 1$ in the interval $[0,~2\pi]$ for your solutions.

2. $\sin^2{t} - \cos^2{t} = \cos{(2t)}
$

So solve $\cos{(2t)} = 1$.

3. absolute of $\csc{t} = 2$

$\Rightarrow$ absolute of $\sin{t} = \frac{1}
{2}$

Solve $\sin{t} = \frac{1}{2} \mbox{and} \sin{t} = -\frac{1}{2}$

4. If $sin^2{t} = \frac{3}{4}$

$\Rightarrow \sin{t} = \frac{\sqrt{3}}{2}
$

3. ok all seems good but.....
for the fist one, i cant see how you go from trig identities to costsint=sint. i tried multiplying but on the left i get 0. so from that point on im lost..

then on the second one, is it as simple as divide by 2?

also for these solutions, do i out pi at the end? the first one is confusing me, do i just say that t is 0 and 2pi? but aren't they the same?

then is it truly like 1/2? so 1/2pi for 3?

Juan

4. I've just realised that it should read:

$\cot{t} = \csc{t}$

$\Rightarrow \frac{\cos{t}}{\sin{t}} = \frac{1}{\sin{t}}$

$\Rightarrow \cos{t} =1$

$\Rightarrow t = 0,~2\pi$

No they are not the same, but there are 2 different solutions to the equation.

For #2,

if $\cos{(2t)} = 1$, then $2t = 0 \mbox{ or } 2\pi$

For #3, solving $\sin{t} = \frac{1}{2} \mbox{ we get } t = \frac{\pi}{6},~ \frac{5\pi}{6}$

Solving $\sin{t} = -\frac{1}{2} \mbox{ we get } t = \frac{7\pi}{6},~\frac{11\pi}{6}$

5. so cos(2t) = cos (t) ??
maybe im just over thinking it?
Juan

6. If you solve $\cos{(2t)} = 1$, you get $2t = 0$ or $2t = 2\pi$ or $2t = 4\pi$, which means your solutions are $t = 0$ and $t = \pi$ and $t = 2\pi$

Look at a cosine graph and you will see that at $0,~2\pi,~4\pi,.... \quad \cos{t} = 1$

The graph of $\cos{(2t)}$ is 2 cosine cycles squashed into the interval $[0,~2\pi]$. So $2t = 0,~2\pi,~4\pi,....$ which means $t = 0,~\pi,~2\pi,....$

7. thanks! i see what you are talking about now!
actually my teacher never showed us a sine cycle and all that good stuff....

Juan

8. Slight typo here:
2.
It should say $\sin^2t-\cos^2t = -\cos(2t)
$