Question

Daca cos C= 3/4 ... sin C = ?

Answer 1

$\displaystyle cos ~C= \frac{3}{4} ~~~~~~~~~~~sin~C=? \\ \bold{sin^2x+cos^2x=1\Rightarrow sin~x= \sqrt{1-cos^2x} } \\ sin~C= \sqrt{1-cos^2C} \Rightarrow sin~C= \sqrt{1-\left( \frac{3}{4} \right)^2} \Rightarrow sin~C= \sqrt{1- \frac{9}{16} } \Rightarrow \\ \Rightarrow sin~C= \sqrt{ \frac{16-9}{16} } \Rightarrow sin~C= \sqrt{ \frac{7}{16} } \Rightarrow sin~C= \frac{ \sqrt{7} }{ \sqrt{16} } \Rightarrow sin~C= \frac{ \sqrt{7} }{4}$