Question

1÷2csc10 - 2cos20
trigonometria​ angulos dobles

Answer 1

SOLUCIÓN

♛ HØlα!! ✌

Recuerda que

                                         $\boxed{\cos(2\alpha) =\cos^2(\alpha)-\sin^2(\alpha)}\\\\\\\boxed{\csc(\alpha)=\dfrac{1}{\sin(\alpha)}}$

En el problema

                                         $\dfrac{1}{2\csc(10\°)}-2\cos(20\°)\\\\\\\dfrac{sen(10\°)}{2}-2[\cos^2(10\°)-\sin^2(10\°)]\\\\\\Operamos\\\\\\\boxed{\boxed{\boldsymbol{-1.7925}}}$