Trigonometry

[omicron]

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

[geometer]

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

Exactly what help do you need?

[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?

[omicron]

Exactly what help do you need?
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?

As for the second problem, do you know that cosec x is defined as 1/sin x?
So are u saying that cosec^2x=1 can be also written as \frac{1}{sin^2x} = 1

[selfAdjoint]

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

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

[omicron]

It is in the 1st and 2nd quadrants?

[rhia]

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

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

:bugeye:

[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.

[omicron]

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