Question

2cos^2 135° + 6sin150° - 4ctg90°cos141°​

Answer 1

Ответ:

$\displaystyle 2 \cos^{2}( {135}^{ \circ} ) + 6 \sin( {150}^{ \circ} ) - 4 \cot( {90}^{ \circ} ) \cos( {141}^{ \circ} ) = 2 \times { \bigg( - \frac{ \sqrt{2} }{2} \bigg) }^{2} + 6 \times \frac{1}{2} - 4 \times 0 \times \cos( {141}^{ \circ} ) = 2 \times \bigg( - \frac{1}{2} \bigg) + 3 - 0 = - 1 + 3 = 2$