1. ## Addition Formulas - Question

Ok, I have the following problem in my practice homework:
If sin(a) = (-5/13) and tan(a) > 0, find the exact value of sin(a - (pi/6))

In the back of the book, the following solution is given:
Since sin(a) = (-15/13) < 0 and tan(a) > 0, a is in QIII, and Cos(a) = -sqr( 1 - (-5/13)^2 ) = (-12/13)

My question is, where did this come from, "-sqr( 1 - (-5/13)^2 ) = (-12/13)"? What formula or identity are are they using on this step? It sort of reminds me of the formula of a circle, but I am just not sure...and the book does not say.

2. Hi Cdoddsy

It uses :

$\displaystyle \cos ^2 x=1-\sin^2 x$

3. Ohh, I get it, because if:

cos^2(x) = 1 + sin^2(x), then
cos(x) = sqr(1+sin^2(x))

Thanks!

4. Oops sorry I made mistake.

It should be :

$\displaystyle \cos ^2 x = 1-\sin^2 x$