# Math Help - Trig help.

1. ## Trig help.

I need help on these questions please.

1. given that sin^2 x + cos^2 x = 1 show that 1+cot^2 x = cosec^2 x

2. Solve for values x in the interval -pi<x<pi for the following in 2 decimal places:

a. 2cosec(x/2) - 5 =0
b. 2cot^2 x - cot x -7 = 0

Thanks

2. For the first question I got:

sin^2 x + cos^2 x = 1

so (sin^2 x)/(sin^2 x) + (cos^2 x)/(cos^2 x) = 1/(sin^2 x)

so 1 + (cos x/ sin x)^2 = (1/ sin x)^2

so 1 + cot^2 x = cosec ^2 x.

Is that right?

But im really stuck on the other two questions

3. 1) Divide by $\sin^{2}(x)$ and see what happens.

2a) sin(x) = 1/csc(x)

2b) Solve this first: 2Y^2 - Y - 7 = 0. I see four solutions in your future.

4. for 2a I got:

2cosec(x/2) - 5 = 0

so 2(1/sin(x/2)) - 5 = 0

so 2(2/sin x) - 5 = 0

so 4/sin x -5 = 0

so 4 = 5sin x

so sin x = 4/5

so x = inverse sin(4/5)

So the only solution I got for x is 53.13 to 2 decimal places.

Is this right?

5. sorry tk hunny but 2b does not factorise.

6. I wasn't suggesting factoring, unless that happens to work.

Solve it, nonetheless.