"Deriving the Identities", need help.  TOPIC_SOLVED

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

"Deriving the Identities", need help.

Postby elphiefae on Mon Sep 14, 2009 8:35 pm

I'm about to finish up the trigonometry chapter in my algebra 2 book, but there is something on the test that I've never seen in the book before.

There is a right triangle, XYZ (X at the top, Y at the bottom right, Z at the bottom left with Z as the right angle) with {side x as the adjacent, side y as the opposite and side z as the hypotenuse}. The equations are: (one over Sec X equals Cos X) and (Sin squared X plus Cos squared X equals one). The only directions are to use Triangle XYZ to derive the identities. If you know what this means, please explain it simply and maybe with an example or two. Thanks!
Posts: 1
Joined: Mon Sep 14, 2009 8:20 pm




Postby stapel_eliz on Mon Sep 14, 2009 10:04 pm

At a guess, you're expected to use the trig ratios in terms of the triangle given...? :confused:

For instance, sec(X) is "hypotenuse over adjacent", or z/y; cos(X) is "adjacent over hypotenuse", or y/z. Then 1/sec(X) = 1/(z/y) = y/z = cos(X).

But that's just a guess. :wink:
User avatar
Posts: 1781
Joined: Mon Dec 08, 2008 4:22 pm

Return to Trigonometry