Assuming that is in interval evaluate
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U is on interval (-pi/2, 0), tan(u)=-1/3 Evaluate sin(2u)
Originally Posted by bseymore U is on interval (-pi/2, 0), tan(u)=-1/3 Evaluate sin(2u) $\displaystyle sin\ 2u=\frac{2tan\ u}{1+tan^2\ u}$
sin(2u) = 2sin(u)cos(u) sketch a reference triangle in quad IV ... opposite side = -1 , adjacent side = 3 calculate the hypotenuse and determine the values of sin(u) and cos(u)
Originally Posted by alexmahone $\displaystyle sin\ 2u=\frac{2tan\ u}{1+tan^2\ u}$ Do you think you could show how you reached this answer? Thanks
Originally Posted by bseymore Do you think you could show how you reached this answer? Thanks Work backwards. $\displaystyle \frac{2tan\ u}{1+tan^2\ u}=\frac{2tan\ u}{sec^2\ u}$ $\displaystyle =2tan\ u\ cos^2\ u$ $\displaystyle =2sin\ u\ cos\ u$ $\displaystyle =sin\ 2u$
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