# Thread: Finding the value of sin x

1. ## Finding the value

Ah, my title ! I'm sorry !
I was looking at the wrong question. Heh.

If f (x/x+1) = 1/x,
x cannot be = 0, 1, and
0 <= theta <= pi/2,
then simplify
f (1/cos^2theta)

2. Originally Posted by ahling
Ah, my title ! I'm sorry !
I was looking at the wrong question. Heh.

If f (x/x+1) = 1/x,
x cannot be = 0, 1, and
0 <= theta <= pi/2,
then simplify
f (1/cos^2theta)

For a non exeptional $x$ , let:

$y=\frac{x}{x+1}$

so:

$x=\frac{-y}{y-1}$

so (for non-exceptional $\theta$):

$f(1/\cos^2(\theta))=\frac{1-[1/\cos^2(\theta)]}{[1/\cos^2(\theta)]}$

Now simplify.

RonL

3. Originally Posted by ahling
Ah, my title ! I'm sorry !
I was looking at the wrong question. Heh.

If f (x/x+1) = 1/x,
x cannot be = 0, 1, and
0 <= theta <= pi/2,
then simplify
f (1/cos^2theta)

f (x/x+1) = 1/x
Now in place of x put 1/x
f (1/x+1) = x
f (1/cos^2theta)=f (1/1-sin^2theta)=-sin^2theta
where sintheta cannot be = 0, 1,

4. Originally Posted by nikhil
f (x/x+1) = 1/x
Now in place of x put 1/x
f (1/x+1) = x
f (1/cos^2theta)=f (1/1-sin^2theta)=-sin^2theta
where sintheta cannot be = 0, 1,
Like I should have told the OP use brackets to remove the ambiguity of expressions like 1/x+1 or x/x+1.

RonL