Question

2sin(a+пи)-2cos(пи\2+а)\sin(а-2пи)

Answer 1

1) 2sin(π +α) - (2cos(π/2 + α)) : sin(α - 2π) = - 2sinα - (- 2sinα) : sinα = - 2sinα + 2sinα : sinα = - 2sinα + 2 = 2 - 2sinα

Answer 2

To simplify the expression, let's break it down step by step:

1. Starting with the expression 2sin(a+пи), we can use the trigonometric identity sin(x + пи) = -sin(x) to simplify it as follows: 2sin(a+пи) = 2(-sin(a)) = -2sin(a).

2. Next, we'll simplify the expression -2cos(пи\2+а). We can use the trigonometric identity cos(x + пи/2) = -sin(x) to rewrite it as: -2cos(пи/2+а) = -2(-sin(а)) = 2sin(а).

3. Moving on to the denominator, we have sin(a-2пи). We can use the periodicity property of sine, which states that sin(x + 2пи) = sin(x), to rewrite it as: sin(a-2пи) = sin(a).

Now, let's substitute the simplified expressions back into the original expression:

(-2sin(a)) / (2sin(a)) The numerator and the denominator cancel each other out, leaving us with:

-2sin(a) / sin(a)

Finally, we can simplify the expression further by canceling out the common factor of sin(a):

-2

Therefore, the simplified form of the expression 2sin(a+пи)-2cos(пи\2+а)\sin(а-2пи) is -2.