## Precalculus: Mathematics for Calculus, 7th Edition

$sin(2x) = -24/25$ $cos(2x) = -7/25$ $tan(2x) = 24/7$
$$tan(2x) = \frac{2 \space tan(x)}{1 - tan ^2(x)} = \frac{2 (-4/3)}{1 - (-4/3)^2 }$$ $$tan(2x) = \frac{-8/3}{1 - 16/9} = \frac{-8/3}{-7/9} = 24/7$$ - To find cos(x) and sin(x): $$tan^2(x) + 1 = sec^2(x)$$ $$tan^2(x) + 1 = \frac{1}{ cos^2(x)}$$ $$cos^2(x) = \frac{1}{tan^2(x) + 1 }$$ $$cos(x) = \pm \sqrt{ \frac{1}{tan^2(x) + 1 }}$$ x in Quadrant II: cosine values are negative. $$cos(x) = - \sqrt{ \frac{1}{tan^2(x) + 1 }} =-\sqrt{ \frac{1}{16/9 + 1 }}$$ $$-\sqrt{ \frac{1}{25/9 }} = - \sqrt{\frac{9}{25}} = - 3/5$$ $$sin^2(x) + cos^2(x) = 1$$ $$sin^2(x) + (-3/5)^2 = 1$$ $$sin^2(x) + 9/25 = 1$$ $$sin^2 (x) = 1 - 9/25 = 16/25$$ $$sin(x) = 4/5$$ $$sin(2x) = 2sin(x)cos(x) = 2(4/5)(-3/5) = -24/25$$ $$cos(2x) = cos^2(x) - sin^2(x) = (-3/5)^2 - (4/5)^2$$ $$cos(2x) = 9/25 - 16/25= -7/25$$