1. ## Trig Help

sin(x)=(1/Square root of 10) and pie/2 < X < pie

cos(x)= ?
tan(x)= ?
sec(x)= ?
csc(x)= ?

2. Originally Posted by jrose011
sin(x)=(1/Square root of 10) and pie/2 < X < pie

cos(x)= ?
tan(x)= ?
sec(x)= ?
csc(x)= ?
you can use the formula $\displaystyle \sin^2 x + \cos^2 x = 1$ to find $\displaystyle \cos x$. once you have that, you can find all others. take note of the quadrant the angle is in, it will dictate the signs the trig functions take.

alternatively, construct a right triangle. knowing that sine = opposite/hypotenuse, you can call one of the acute angles in the triangle x, the hypotenuse of the triangle must be $\displaystyle \sqrt{10}$, while the side opposite x must be 1. you can use Pythagoras' theorem to figure out the other side length. once you have all sides filled out, you can use the trig ratios to find your answers: cos(x) = adjacent/hypotenuse, tangent = sine/cosine = opposite/adjacent, etc

again, watch the signs

3. Originally Posted by jrose011
sin(x)=(1/Square root of 10) and pie/2 < X < pie
I think you mean $\displaystyle \pi/2\, <\, x\, <\,\pi"$...? (A "pie" is a pastry; my favorite is Dutch apple. The number "pi" is approximately 3.14159.)