# I need some help with these two calculator trig problems

1. Oct 10, 2004

### omicron

I need some help with these two questions:
$$tan2A=cot40$$(40degrees)
and
$$cosec^2x=1$$

2. Oct 10, 2004

### geometer

Exactly what help do you need?

3. Oct 10, 2004

### HallsofIvy

Can you use your calculator (which looks to me like the only way to do the first one)?
If you can, use to find cot (40) and then to find 2A.

As for the second problem, do you know that cosec x is defined as 1/sin x?

If cosec^2 x= 1, what is sin^2 x?

4. Oct 10, 2004

### omicron

Doing the question. I just don't know.

How do u find cot(40)? Don't u have to change it to $$\frac{1}{tan\theta}$$ or something like that?

So are u saying that $$cosec^2x=1$$ can be also written as $$\frac{1}{sin^2x} = 1$$

5. Oct 10, 2004

Staff Emeritus
Yup. And then what can you conclude from that, concerning x?

6. Oct 11, 2004

### omicron

It is in the 1st and 2nd quadrants?

7. Oct 11, 2004

### rhia

tan 2A = cot40 =tan 50
=> 2A = [50 +n(180)]degrees
=>A=25+90n deg

cosec^2 x = 1 =>sinx=+-1=>x=180n+(-1)^n *(+-90) deg

8. Oct 11, 2004

### misogynisticfeminist

For the 2nd question.....

$$cosec^2x=1$$ can be expressed as,

$$1/sin^2x=1$$ while multiplying $$sin^2x$$ both sides, we have,

$$1=sin^2x$$ and by square rooting both sides, we now have,

$$sinx= \pm1$$

and so, since sin x is both positive and negative, it must lie in all quadrants with alpha 90 degrees.

Last edited: Oct 11, 2004
9. Oct 12, 2004

### omicron

Oh so now i know. I didn't know u could $$\sqrt{sin^2x}$$. Thanks to everyone that helped.