$\frac{\sin(360-2x)sin(-x)}{sin(90+x)}$ + 2cos $^{2}$ (180+x)
You asked:
Evaluate the expression: \(\frac{\sin\left( 360 - 2 x \right) \cdot \sin\left( -x \right)}{\sin\left( 90 + x \right)} + 2 \cdot {\cos\left( 180 + x \right)}^{2}\)
MathBot Answer:
Evaluated
\(\displaystyle \frac{\sin\left( 360 - 2 x \right) \cdot \sin\left( -x \right)}{\sin\left( 90 + x \right)} + 2 \cdot {\cos\left( 180 + x \right)}^{2} = \frac{\sin{\left(x \right)} \sin{\left(2 x - 360 \right)}}{\sin{\left(x + 90 \right)}} + 2 \cos^{2}{\left(x + 180 \right)} \)
Factored
\[\frac{\sin\left( 360 - 2 x \right) \cdot \sin\left( -x \right)}{\sin\left( 90 + x \right)} + 2 \cdot {\cos\left( 180 + x \right)}^{2} = \frac{- \sin{\left(- x \right)} \sin{\left(2 x - 360 \right)} + 2 \sin{\left(x + 90 \right)} \cos^{2}{\left(x + 180 \right)}}{\sin{\left(x + 90 \right)}}\]