$\frac{\sin(360-2x)sin(-x)}{sin(90+x)}$ + 2cos $^{2}$ (180+x)

asked by guest
on Oct 26, 2024 at 10:25 am



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

\[\frac{\sin\left( 360 - 2 x \right) \cdot \sin\left( -x \right)}{\sin\left( 90 + x \right)} + 2 \cdot {\cos\left( 180 + x \right)}^{2} = \frac{- \sin{\left(- x \right)} \sin{\left(2 x - 360 \right)} + 2 \sin{\left(x + 90 \right)} \cos^{2}{\left(x + 180 \right)}}{\sin{\left(x + 90 \right)}}\]

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